Question 1

In a simultaneous throw of two fair dice, what is the probability of getting a total (sum of the two faces) equal to 7?

Accepted Answer

Introduction / Context:This is a classic probability question about rolling two fair dice. We are interested in the sum of the numbers on the two dice, and specifically we want the probability that this sum equals 7. Given Data / Assumptions: Two fair six sided dice are thrown together. Each die shows an integer from 1 to 6. All 36 ordered pairs of outcomes are equally likely. We require those pairs whose sum is exactly 7. Concept / Approach:The total number of ordered outcomes is 6 * 6 = 36. To find the probability that the sum is 7, we list all pairs (first die, second die) that add up to 7, count them, and divide by 36. This is straightforward counting of favourable outcomes versus total outcomes. Step-by-Step Solution:Total outcomes when two dice are rolled = 6 * 6 = 36.We need pairs (a, b) such that a + b = 7, with a and b each between 1 and 6.List such pairs: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).There are 6 favourable ordered pairs.Therefore, probability that the sum is 7 = number of favourable outcomes / total outcomes = 6 / 36.Simplify 6 / 36 by dividing numerator and denominator by 6 to get 1 / 6. Verification / Alternative check:We can check that the pairs listed are all possible solutions to a + b = 7 with 1 ≤ a, b ≤ 6. There is no other pair that satisfies the equation within the allowed range. Also, note that the number of favourable outcomes is symmetric around (3.5, 3.5), which is consistent for a typical die sum distribution where 7 is the most likely sum. Why Other Options Are Wrong:The probability 1/4 would require 9 favourable outcomes among 36, which is too many. The value 1/20 would require fewer than 2 favourable outcomes, which is impossible because we have 6. The value 3/4 is far too large and would suggest that most outcomes give a sum of 7. Only 1/6 aligns with the correct count of 6 favourable outcomes out of 36 total outcomes. Common Pitfalls:A common error is to forget that the dice are distinguishable, so (1, 6) and (6, 1) are different ordered outcomes even though both give the sum 7. Another pitfall is miscounting the number of valid pairs by skipping one or adding an invalid combination. Carefully writing out all pairs that add to 7 helps avoid mistakes and reinforces understanding of how sums distribute in two dice rolls. Final Answer:The probability that the total is 7 is 1/6.

Question 2

A single card is drawn at random from a well shuffled standard deck of 52 cards. What is the probability of getting either the queen of clubs or the king of hearts?

Accepted Answer

Introduction / Context:This question is a simple probability problem involving a standard deck of cards. We are asked to find the probability that a card drawn at random is either the queen of clubs or the king of hearts. Given Data / Assumptions: A standard deck has 52 cards. There is exactly one queen of clubs in the deck. There is exactly one king of hearts in the deck. The deck is well shuffled, so every card is equally likely to be drawn. The two desired cards are distinct. Concept / Approach:We are dealing with a union of two mutually exclusive events: drawing the queen of clubs or drawing the king of hearts. Since a single card cannot be both at the same time, the probability of their union is the sum of their individual probabilities. Each specific card has probability 1/52 of being drawn. Step-by-Step Solution:Probability of drawing the queen of clubs = 1 / 52.Probability of drawing the king of hearts = 1 / 52.Because these two cards are different, the events are mutually exclusive when drawing a single card.Therefore, probability of drawing either the queen of clubs or the king of hearts = 1/52 + 1/52 = 2/52.Simplify 2/52 by dividing numerator and denominator by 2 to get 1/26. Verification / Alternative check:We can think of the sample space as 52 equally likely cards and the favourable outcomes as a set of 2 specific cards. The probability is simply the size of the favourable set divided by the size of the sample space, which is 2 / 52 = 1 / 26. This aligns with the previous calculation using the addition rule for mutually exclusive events. Why Other Options Are Wrong:The value 1/13 would correspond to 4 favourable cards out of 52, which would be the case if we were looking for all queens of a suit or all kings of a suit, but we are concerned with exactly two specific cards. The fraction 2/13 arises from miscounting favourable cards or misapplying simplification. The probability 1/52 would correspond to only one specific card being favourable. Only 1/26 correctly reflects two favourable cards out of 52. Common Pitfalls:Students sometimes incorrectly think there are four queens of clubs or four kings of hearts, confusing suits and ranks. Another error is to forget to simplify 2/52 to 1/26, although the unsimplified fraction is still numerically correct. Keeping track of the deck structure (4 suits and 13 ranks) is helpful for avoiding such mistakes. Final Answer:The probability of drawing the queen of clubs or the king of hearts is 1/26.

Question 3

Out of the first 20 natural numbers {1, 2, 3, ..., 20}, one number is selected at random. What is the probability that it is either an even number or a prime number?

Accepted Answer

Introduction / Context:This question combines basic number properties with probability. We choose one number at random from the first 20 natural numbers and want the probability that it is even or prime. Because some numbers are both even and prime, we must handle the overlap correctly using set operations. Given Data / Assumptions: Sample space: numbers from 1 to 20 inclusive. Total numbers available = 20. Even numbers are multiples of 2. Prime numbers are positive integers greater than 1 with no positive divisors other than 1 and themselves. Each number from 1 to 20 is equally likely to be selected. Concept / Approach:We use the formula for the size of a union of two sets: |E ∪ P| = |E| + |P| - |E ∩ P|, where E is the set of even numbers and P is the set of primes. The required probability is |E ∪ P| divided by 20. The intersection must be considered because 2 is both even and prime. Step-by-Step Solution:Even numbers from 1 to 20: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (10 numbers).Prime numbers from 1 to 20: 2, 3, 5, 7, 11, 13, 17, 19 (8 numbers).Intersection of even and prime sets: only the number 2.So |E| = 10, |P| = 8 and |E ∩ P| = 1.Size of E ∪ P = |E| + |P| - |E ∩ P| = 10 + 8 - 1 = 17.Total possible outcomes = 20.Hence probability that the selected number is even or prime = 17 / 20. Verification / Alternative check:We can list the union explicitly: {2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20}. There are 17 numbers in this set. Since there are 20 numbers in total from 1 to 20, the probability is 17/20, which matches our earlier calculation using the inclusion exclusion principle. Why Other Options Are Wrong:The value 1 would mean that every number from 1 to 20 is either even or prime, which is false because 9 and 15, for example, are neither even nor prime. The fraction 16/19 and the value 3/2 are impossible probabilities, and 3/2 is greater than 1. Only 17/20 is consistent with the correct counts of the union and sample space size. Common Pitfalls:Students sometimes double count 2 when they add the counts of even and prime numbers, forgetting to subtract the intersection. Others mistakenly include 1 as a prime number, which would make the prime set larger and the answer incorrect. Applying the inclusion exclusion formula carefully helps avoid these issues. Final Answer:The probability that the selected number is either even or prime is 17/20.

Question 4

A brother and a sister appear for an interview for two vacant posts in an office. The probability that the brother is selected is 1/5 and the probability that the sister is selected is 1/3. Assuming independence, what is the probability that exactly…

Accepted Answer

Introduction / Context:This problem illustrates probability with independent events and focuses on the situation where exactly one of two candidates is selected. We know the individual probabilities of selection for the brother and the sister and want the probability that either the brother or the sister, but not both, get selected. Given Data / Assumptions: Probability that the brother is selected, P(B) = 1/5. Probability that the sister is selected, P(S) = 1/3. The selection events for brother and sister are assumed to be independent. We seek P(exactly one is selected) = P(B and not S) + P(S and not B). Concept / Approach:We compute the probability that the brother is selected while the sister is not, plus the probability that the sister is selected while the brother is not. Because of independence, these can be expressed as products of the individual probabilities and their complements. This is a standard pattern for exactly one of two independent events occurring. Step-by-Step Solution:Complement probabilities: P(not B) = 1 - 1/5 = 4/5 and P(not S) = 1 - 1/3 = 2/3.Probability that the brother is selected but the sister is not: P(B and not S) = P(B) * P(not S) = (1/5) * (2/3) = 2 / 15.Probability that the sister is selected but the brother is not: P(S and not B) = P(S) * P(not B) = (1/3) * (4/5) = 4 / 15.Probability that exactly one of them is selected = 2/15 + 4/15 = 6/15.Simplify 6/15 by dividing numerator and denominator by 3 to get 2/5. Verification / Alternative check:We can also compute the probability that neither is selected and that both are selected, and ensure that all four basic possibilities sum to 1. P(neither) = P(not B) * P(not S) = (4/5) * (2/3) = 8/15. P(both) = P(B) * P(S) = (1/5) * (1/3) = 1/15. So P(exactly one) = 1 - P(neither) - P(both) = 1 - 8/15 - 1/15 = 6/15 = 2/5, confirming the result. Why Other Options Are Wrong:The value 1/5 is just the probability of the brother being selected alone and ignores the sister. The value 3/5 would imply 9/15 for exactly one selection, which does not fit with the other probabilities. The probability 3/4 is too high given that both individual selection probabilities are significantly below 1. Only 2/5 correctly sums the two disjoint cases where exactly one of them is selected. Common Pitfalls:Some students add P(B) and P(S) directly to get 1/5 + 1/3 without subtracting the probability that both are selected, which overcounts the overlapping case. Others forget to use the complements P(not B) and P(not S) correctly. Drawing a small table of outcomes (B selected or not, S selected or not) and filling in probabilities can help visualise and avoid such mistakes. Final Answer:The probability that exactly one of the brother and sister is selected is 2/5.

Question 5

A bag contains 7 green balls and 5 black balls. Three balls are drawn one after another without replacement. What is the probability that all the three balls drawn from the bag are green?

Accepted Answer

Introduction / Context: This question tests your understanding of basic probability for selections made without replacement. The key idea is to calculate the probability that all three balls taken from a finite collection are green when each draw changes the composition of the bag. Problems like this are very common in quantitative aptitude exams and help you practise sequential probability and careful fraction handling. Given Data / Assumptions: Total green balls in the bag = 7 Total black balls in the bag = 5 Total balls in the bag initially = 12 Three balls are drawn one after another Draws are without replacement We want the probability that all three drawn balls are green Concept / Approach: The general approach is to treat each draw as a conditional probability step. For draws without replacement, the denominator (total balls) and the numerator (favorable balls) change after every draw. We multiply the probabilities for each draw because all three events must occur together: green on the first draw, green on the second draw, and green on the third draw. This is an example of using the multiplication rule for dependent events. Step-by-Step Solution: Step 1: Initially there are 12 balls in total and 7 of them are green. Step 2: Probability that the first ball is green = 7/12. Step 3: After drawing one green ball, remaining green balls = 6 and total balls = 11. Step 4: Probability that the second ball is green, given that the first was green = 6/11. Step 5: After drawing two green balls, remaining green balls = 5 and total balls = 10. Step 6: Probability that the third ball is green, given that first two were green = 5/10. Step 7: Multiply all three probabilities to get the overall probability. Overall probability = (7/12) * (6/11) * (5/10). Step 8: Simplify: (7 * 6 * 5) / (12 * 11 * 10) = 210 / 1320. Step 9: Divide numerator and denominator by 30 to simplify: 210/1320 = 7/44. Verification / Alternative check: Another way is to count combinations. The number of ways to choose any 3 balls from 12 is 12C3. The number of ways to choose 3 green balls from the 7 green balls is 7C3. The probability is 7C3 / 12C3. We have 7C3 = 35 and 12C3 = 220, so probability = 35/220 = 7/44, which matches the stepwise multiplication result. This confirms that our calculation is consistent and correct. Why Other Options Are Wrong: Option 21/88 is obtained if the denominators or simplification are handled incorrectly, so it is too small. Option 7/24 is larger than the true probability and effectively assumes more favorable outcomes than actually exist. Option 5/22 does not arise from any correct systematic calculation with either the stepwise or combination method. Option 3/11 is also an overestimate and does not correspond to the true reduced fraction of 7/44. Common Pitfalls: Many learners mistakenly treat draws as independent and keep the denominator fixed at 12 for all three draws, which is wrong for selection without replacement. Another frequent error is to use combinations for the numerator and simple powers like 12^3 for the denominator, which would correspond to ordered draws with replacement. Students also sometimes forget to simplify fractions correctly or cancel common factors in the wrong way. Always track how many favorable balls and how many total balls remain after each draw, and be disciplined in your simplification steps. Final Answer: The probability that all three balls drawn are green is 7/44.

Question 6

One card is drawn at random from a standard pack of 52 playing cards, with each card equally likely to be chosen. What is the probability that the card drawn is red?

Accepted Answer

Introduction / Context: This problem checks your familiarity with the structure of a standard deck of playing cards and simple probability. You need to recall how many cards are red and then compare that count to the total size of the deck. It is a classic single event probability question that appears frequently in aptitude and competitive exams. Given Data / Assumptions: A standard deck has 52 cards. There are 4 suits: hearts, diamonds, clubs, and spades. Hearts and diamonds are red suits. Clubs and spades are black suits. One card is drawn at random and each card is equally likely to be drawn. We want the probability that the card is red. Concept / Approach: For a simple probability event, the probability of an outcome is given by: Probability = (Number of favorable outcomes) / (Total number of possible outcomes). Here, favorable outcomes are all red cards. The total possible outcomes are all cards in the deck. Because the selection is random and each card is equally likely, we can safely use this basic ratio formula. Step-by-Step Solution: Step 1: In a standard deck, hearts and diamonds are red suits. Step 2: Each suit contains 13 cards. Step 3: Total number of red cards = 13 hearts + 13 diamonds = 26 cards. Step 4: Total number of cards in the deck = 52. Step 5: Apply probability formula: Probability of drawing a red card = 26 / 52. Step 6: Simplify the fraction 26 / 52 by dividing numerator and denominator by 26 to get 1 / 2. Verification / Alternative check: You can also think of the deck as evenly split between red and black. There are 2 red suits and 2 black suits, each with 13 cards. Hence exactly half of the deck is red and half is black. Therefore, before any card is drawn, the probability of getting a red card is exactly 1 out of 2, which again confirms the answer 1/2. Why Other Options Are Wrong: Option 1/4 would be correct only if there were 13 red cards in a 52 card deck, which is not true. Option 1/3 does not correspond to any natural split of the deck. Option 3/4 would mean that most cards in the deck are red, which contradicts the equal division between red and black suits. Option 2/13 is much smaller than the real probability and does not match the actual card distribution. Common Pitfalls: A common mistake is to forget that there are two red suits and to count only one of them. Another error is to confuse the number of suits with the number of cards in a suit. Always verify how many cards are in each suit and how many suits share the property you care about. Remember that probability must lie between 0 and 1, and for a balanced deck, many events like red or black will have probability 1/2. Final Answer: The probability that the card drawn is red is 1/2.

Question 7

A bag contains 8 red balls and 4 green balls. Three balls are drawn at random from the bag without replacement. What is the probability that exactly two of the drawn balls are red and the remaining one ball is green?

Accepted Answer

Introduction / Context: This question focuses on probability with combinations and selections without replacement. Instead of the order of drawing, we care only about the composition of the drawn set: exactly two red balls and one green ball. Such problems test your ability to count combinations and to form a correct probability ratio based on favorable and total outcomes. Given Data / Assumptions: Number of red balls in the bag = 8 Number of green balls in the bag = 4 Total balls in the bag = 12 Three balls are drawn at random without replacement. We want exactly 2 red balls and 1 green ball among the three drawn. Concept / Approach: When order does not matter, combinations are used. The probability is computed as: Probability = (Number of favorable combinations) / (Total number of possible combinations). Favorable combinations correspond to choosing 2 red balls from 8 and 1 green ball from 4. The total combinations correspond to choosing any 3 balls from 12, irrespective of color. This is a direct application of the combination formula nCr. Step-by-Step Solution: Step 1: Compute total ways to choose any 3 balls from 12 using 12C3. 12C3 = 12 * 11 * 10 / (3 * 2 * 1) = 220. Step 2: Compute ways to choose 2 red balls from 8 using 8C2. 8C2 = 8 * 7 / (2 * 1) = 28. Step 3: Compute ways to choose 1 green ball from 4 using 4C1. 4C1 = 4. Step 4: Multiply to get total favorable combinations. Favorable combinations = 8C2 * 4C1 = 28 * 4 = 112. Step 5: Form the probability ratio using combinations. Probability = Favorable / Total = 112 / 220. Step 6: Simplify the fraction 112 / 220 by dividing numerator and denominator by 4 to get 28 / 55. Verification / Alternative check: You can verify using stepwise conditional probability. For example, choose an order such as red, red, green and compute (8/12) * (7/11) * (4/10). Then multiply by the number of distinct permutations of the pattern R R G, which is 3 because the green ball can appear in any of the three positions. That gives: (8/12 * 7/11 * 4/10) * 3 = (8 * 7 * 4 * 3) / (12 * 11 * 10) = 672 / 1320 which simplifies to 28 / 55, confirming the previous result. Why Other Options Are Wrong: Option 21/55 is smaller and comes from incorrect counting or missing some favorable arrangements. Option 32/55 overestimates the probability and could result from including outcomes with three red balls or other incorrect combinations. Option 14/55 is exactly half of the correct probability, which can happen if the factor from permutations is missed. Option 7/33 does not match any correct simplification from the standard methods used here. Common Pitfalls: A common mistake is to treat the events as ordered and forget to multiply by the number of permutations, or the reverse: to use combinations for the numerator and a power such as 12^3 for the denominator, which corresponds to ordered draws with replacement. Another frequent error is to misread the phrase exactly two red balls and one green ball and accidentally include the case of three red balls. Always check whether the problem cares about exact counts or at least counts, and whether selection is with or without replacement. Final Answer: The probability that exactly two drawn balls are red and one is green is 28/55.

Question 8

If the probability of an event A is P(A) = 4/9, what are the odds against the occurrence of event A?

Accepted Answer

Introduction / Context: This question checks whether you can convert a probability into odds against an event. Many aptitude tests use both probability notation and odds notation, so understanding the relationship between them is essential. Odds against an event compare the chance that the event does not happen with the chance that it does happen. Given Data / Assumptions: Probability of event A occurring, P(A) = 4/9. Therefore, probability that event A does not occur, P(A complement) = 1 - 4/9. We need odds against event A, written in the form x:y. Concept / Approach: For any event A, the relationship between probability and odds is: P(A) = favorable / total. Odds in favor of A are: Odds in favor of A = P(A) : P(A complement). Odds against A are: Odds against A = P(A complement) : P(A). So to find odds against, first compute the probability of A not happening and then express the ratio between this and P(A) in simplest form. Step-by-Step Solution: Step 1: Given P(A) = 4/9. Step 2: Compute P(A complement) = 1 - 4/9. Step 3: 1 can be written as 9/9, so P(A complement) = 9/9 - 4/9 = 5/9. Step 4: Odds against A are given by P(A complement) : P(A). Step 5: So the raw ratio is (5/9) : (4/9). Step 6: Multiply both sides of the ratio by 9 to remove denominators, giving 5 : 4. Step 7: Therefore, the odds against event A are 5:4. Verification / Alternative check: Imagine there are 9 equally likely outcomes. For 4 of them event A occurs, and for 5 of them it does not occur. Odds against A compare the count of failures to the count of successes, which would be 5 unfavorable outcomes and 4 favorable outcomes. This directly produces odds against A equal to 5:4, matching the calculation using formal probability complements. Why Other Options Are Wrong: Option 4:5 represents odds in favor of A, not odds against A. Option 4:9 incorrectly uses the original probability with total outcomes; it is not a comparison of success and failure. Option 9:4 would correspond to a situation where the event is more likely to fail than succeed by a larger margin, which does not match P(A) = 4/9. Option 1:2 is unrelated to the given probability and does not come from any correct calculation based on P(A) = 4/9. Common Pitfalls: Students often confuse odds with probabilities. Odds are always a ratio of two counts or probabilities, not a single fraction between 0 and 1. Another mistake is reversing odds in favor and odds against. Always remember: odds against A compare failure to success (not success to failure). Also, do not forget to simplify ratios into the smallest whole number form so that they are easy to interpret in exam conditions. Final Answer: The odds against the occurrence of event A are 5:4.

Question 9

From a standard pack of 52 playing cards, one card is drawn at random. What is the probability that the card drawn is either a spade or an ace?

Accepted Answer

Introduction / Context: This question tests your understanding of probability with overlapping events. You are asked to find the probability of drawing a card that is either a spade or an ace from a full deck. Since one card, the ace of spades, belongs to both categories, you must correctly apply the addition rule for probabilities with overlap. Given Data / Assumptions: Total number of cards in the deck = 52. Number of spade cards = 13. Number of aces in the deck = 4. The ace of spades is counted in both the spade set and the ace set. One card is drawn at random with all cards equally likely. Concept / Approach: For two events A and B, the probability that at least one occurs is: P(A or B) = P(A) + P(B) - P(A and B). Here: A = card is a spade. B = card is an ace. A and B = card is the ace of spades. We will first count how many cards are in each category and then convert those counts into probabilities. Step-by-Step Solution: Step 1: Number of spades in the deck = 13. Step 2: Number of aces in the deck = 4. Step 3: The ace of spades is common to both sets, so the intersection has size 1. Step 4: Convert counts to probabilities. P(spade) = 13 / 52. P(ace) = 4 / 52. P(ace of spades) = 1 / 52. Step 5: Apply the addition rule for overlapping events. P(spade or ace) = 13/52 + 4/52 - 1/52. Step 6: Simplify the numerator: (13 + 4 - 1) / 52 = 16 / 52. Step 7: Simplify 16 / 52 by dividing numerator and denominator by 4 to get 4 / 13. Verification / Alternative check: You can verify by direct counting. Count all spades (13 cards) and all non spade aces. The non spade aces are hearts, diamonds, and clubs, which are 3 additional cards. So the total favorable cards = 13 + 3 = 16. Total cards = 52. Probability = 16 / 52 = 4 / 13, which matches the previous calculation using the formula with intersection. Why Other Options Are Wrong: Option 5/13 is larger than the correct value and could arise from adding 13/52 and 4/52 without subtracting the overlap, which double counts the ace of spades. Option 1/4 would be correct if there were 13 favorable cards instead of 16. Option 3/13 underestimates the probability and does not correspond to any correct tally of favorable cards. Option 7/26 also does not simplify to the correct fraction and does not align with the actual count of 16 favorable cards. Common Pitfalls: A very common mistake is to ignore intersections when using the addition rule. Whenever two conditions can occur together in at least one outcome, you must subtract the intersection once. Another error is forgetting the exact composition of a standard deck or miscounting the number of aces. Always double check how many cards satisfy each condition and whether there is any overlap between those sets before you convert the counts into probabilities. Final Answer: The probability that the card drawn is either a spade or an ace is 4/13.

Question 10

A woman has two children and it is known that at least one of the children is a boy. Under this information, what is the probability that both of her children are boys?

Accepted Answer

Introduction / Context: This is a classic conditional probability problem related to the gender of children in a family. The key subtlety is the phrase at least one child is a boy, which changes the sample space compared to choosing two children without any extra information. Many students answer 1/2 by intuition, but that is not correct under the usual assumptions stated in such questions. Given Data / Assumptions: The family has exactly two children. Each child is equally likely to be a boy or a girl. The gender of one child does not affect the gender of the other (independence). It is known that at least one of the children is a boy. We want the probability that both children are boys, given this information. Concept / Approach: We will list all equally likely possibilities for the two children and then condition on the information that at least one child is a boy. If we denote a boy by B and a girl by G, then ordered pairs (older, younger) cover all scenarios. We then remove any outcomes that contradict the given information and compute conditional probability: P(both boys | at least one boy) = (number of outcomes with both boys) / (number of outcomes with at least one boy). Step-by-Step Solution: Step 1: List all possible ordered pairs of genders for the two children: BB, BG, GB, GG. Step 2: Under the usual assumption, each of these four outcomes is equally likely. Step 3: The information at least one child is a boy rules out the outcome GG, where both are girls. Step 4: The reduced sample space is now {BB, BG, GB}. Step 5: Among these three remaining outcomes, only BB has both children boys. Step 6: Count favorable outcomes for both boys: there is 1 such outcome (BB). Step 7: Count all outcomes that satisfy at least one boy: there are 3 such outcomes (BB, BG, GB). Step 8: Conditional probability = favorable / total = 1 / 3. Verification / Alternative check: You can think of this in terms of long run frequencies. Imagine many families with two children, and every boy and girl arrangement is equally likely. Among 4 families on average, you expect 1 with GG, 1 with BB, and 2 with mixed genders BG and GB. If you now choose only those families that have at least one boy, you are choosing from 3 types: BB, BG, GB. Out of those 3 types, only 1 type, BB, has both boys. Therefore the probability is 1/3 rather than 1/2, confirming the conditional probability calculation. Why Other Options Are Wrong: Option 1/2 would be correct only if we started by knowing that exactly one child is a boy or if we looked at a completely different conditioning scenario. Option 2/3 would correspond to the probability that there is at least one boy, not that both are boys given that information. Option 1/4 is the unconditional probability of both children being boys with no extra information, which does not apply once we know at least one is a boy. Option 3/8 has no simple interpretation in this setup and does not come from any correct derivation. Common Pitfalls: Many students forget that conditional probability changes the sample space. They keep the denominator 4 and simply consider the 2 boys as one of two children, which is incorrect. Others assume that given at least one boy, the other child is equally likely to be a boy or girl, and directly answer 1/2, neglecting that the families with mixed genders appear twice in the original sample space (BG and GB). Always explicitly write the sample space and then cross out the outcomes that conflict with the given condition. Final Answer: The probability that both children are boys, given that at least one is a boy, is 1/3.

Question 11

Two fair dice are thrown simultaneously. What is the probability that one die shows an even number and the other die shows an odd number?

Accepted Answer

Introduction / Context: This question tests basic probability with two dice and focuses on parity, that is, whether numbers are even or odd. When two fair dice are rolled, each die independently shows one of six outcomes. You must find the probability that exactly one die is even and the other is odd, a very standard pattern in aptitude problems. Given Data / Assumptions: Two fair six sided dice are thrown at the same time. Each die has faces numbered 1, 2, 3, 4, 5, 6. Even numbers on a die are 2, 4, 6 (three outcomes). Odd numbers on a die are 1, 3, 5 (three outcomes). We want one die to be even and the other to be odd. Concept / Approach: Since the dice are independent, we can compute the probability for each die and then apply the multiplication rule. There are two favorable patterns: First die even and second die odd. First die odd and second die even. The total probability of exactly one even and one odd is the sum of the probabilities of these two disjoint events. Step-by-Step Solution: Step 1: Probability that a single die shows an even number = 3/6 = 1/2. Step 2: Probability that a single die shows an odd number = 3/6 = 1/2. Step 3: Probability that the first die is even and the second die is odd = (1/2) * (1/2) = 1/4. Step 4: Probability that the first die is odd and the second die is even = (1/2) * (1/2) = 1/4. Step 5: These two cases are mutually exclusive, so add them: 1/4 + 1/4 = 1/2. Step 6: Therefore, the probability of getting one even number and one odd number is 1/2. Verification / Alternative check: You can also solve this by counting outcomes in the full sample space. There are 6 * 6 = 36 possible ordered pairs for two dice. For each even value on the first die (2, 4, 6), the second die can be any of the 3 odd values, giving 3 * 3 = 9 outcomes of the type (even, odd). Similarly, for each odd value on the first die (1, 3, 5), the second die can be any of the 3 even values, again giving 3 * 3 = 9 outcomes of the type (odd, even). So favorable outcomes = 9 + 9 = 18. Probability = 18 / 36 = 1/2, which agrees with the earlier method. Why Other Options Are Wrong: Option 1/3 and option 1/6 both underestimate the true probability and would correspond to incorrect counting of favorable cases. Option 2/3 and option 5/6 are too large and might result from including outcomes where both numbers are even or both are odd. None of these incorrect values match either the stepwise calculation or the exhaustive counting method, both of which clearly give 1/2. Common Pitfalls: One common mistake is to confuse exactly one even and one odd with at least one even or at least one odd. Another mistake is to forget that the two patterns (even, odd) and (odd, even) are distinct events that should both be counted. Some students also attempt to count outcomes by inspection but miss several pairs, leading to an incorrect numerator. Writing out the sample space pattern or using symmetry and independence properly can prevent these errors. Final Answer: The probability of getting one even number and one odd number when two dice are thrown is 1/2.

Question 12

In a party there are 5 married couples (that is, 5 distinct pairs, each made of a husband and a wife). If 5 people are chosen at random from these 10 persons, what is the probability that the selected group contains at least two complete couples?

Accepted Answer

Introduction / Context: This question combines combinatorics with probability and involves counting groups with a structural constraint: at least two complete couples must be present in the chosen group. Each couple consists of two specific people, and we select 5 people from 10. The challenge is to count the favorable combinations carefully without overcounting. Given Data / Assumptions: There are 5 married couples, so 10 people in total. Each couple has exactly 2 members. We choose 5 people uniformly at random from these 10. A complete couple means both husband and wife are included in the selected group. We want the probability that there are at least two complete couples in the chosen group. Concept / Approach: The total number of ways to select any 5 people from 10 is 10C5. For the favorable cases, we must have at least 2 complete couples. Because we are choosing only 5 people, the only possible pattern that satisfies at least two couples is: Exactly two complete couples (4 people) and one additional person from a different couple. There is no way to have three complete couples, as that would require 6 people. So we count selections with exactly two couples and one extra person, then divide by the total combinations. Step-by-Step Solution: Step 1: Compute total ways to choose any 5 people from 10: 10C5. 10C5 = 10 * 9 * 8 * 7 * 6 / (5 * 4 * 3 * 2 * 1) = 252. Step 2: Choose which 2 couples out of the 5 will both be fully present. Number of ways to choose 2 couples from 5 = 5C2 = 10. Step 3: After taking these 2 couples (4 people), we must choose 1 more person to make a group of 5. The remaining couples are 3 in number, giving 6 individuals available for the extra spot. Step 4: Number of ways to choose this extra person = 6C1 = 6. Step 5: Total favorable groups with at least two couples = 10 * 6 = 60. Step 6: Probability = favorable / total = 60 / 252. Step 7: Simplify 60 / 252 by dividing numerator and denominator by 12 to get 5 / 21. Verification / Alternative check: You can check the logic by reasoning about impossible patterns. Three couples would require 6 people, which is more than we select, so only the pattern exactly two couples plus one extra is valid. There is no other way to have at least two couples with a group size of 5. Checking the counts carefully shows that each selection is uniquely determined by choosing a pair of couples and then picking one additional person from the remaining couples. Because the total number of groups 10C5 is known, and our favorable count is 60, the probability 5/21 is consistent and complete. Why Other Options Are Wrong: Options 6/21 and 7/21 correspond to larger probabilities and would require more than 60 favorable combinations, which do not exist under the constraints. Option 5/42 is exactly half of the correct probability and might come from missing a factor of 2 or incorrectly counting couples. Option 11/21 is far too large, suggesting that a majority of all groups would contain at least two couples, which clearly is not true when you experiment with a few concrete groupings. Common Pitfalls: A frequent error is to try to count groups with at least two couples by adding several complicated cases or by attempting complementary counting without carefully defining the complement. Another pitfall is to forget that the extra person must not complete a third couple, which would violate the group size constraint. To avoid confusion, it is helpful to think in terms of couples first, then fill remaining spots with individuals from remaining couples, and finally confirm that there is no overlap between cases. Final Answer: The probability that the selected group of five people contains at least two complete couples is 5/21.

Question 13

In a simultaneous throw of two fair dice, what is the probability of getting a doublet, that is, the same number on both dice?

Accepted Answer

Introduction / Context: This problem tests your ability to count outcomes when two dice are rolled together. A doublet means both dice show the same number, such as (1,1) or (4,4). Identifying how many such outcomes exist and comparing with the total number of possible outcomes is a core probability skill. Given Data / Assumptions: Two fair six sided dice are thrown simultaneously. Each die shows one of the numbers 1, 2, 3, 4, 5, 6. All ordered pairs of outcomes are equally likely. A doublet means both dice show the same number. Concept / Approach: When two dice are rolled, the total number of possible ordered outcomes is: Total outcomes = 6 * 6 = 36. A doublet occurs when the first die and the second die show identical values. We simply count the number of such pairs and then compute the probability as favorable outcomes divided by total outcomes. Step-by-Step Solution: Step 1: List all possible doublets: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Step 2: Count these doublets. There are 6 distinct doublet outcomes. Step 3: Total number of ordered outcomes when two dice are thrown = 6 * 6 = 36. Step 4: Probability of getting a doublet = number of doublets / total outcomes = 6 / 36. Step 5: Simplify 6 / 36 by dividing numerator and denominator by 6 to get 1 / 6. Verification / Alternative check: You can view each possible number on the first die and ask what is the chance the second matches it. No matter what the first die shows, the second die has a 1/6 chance of matching that exact number. Since the first die can be anything, and the second must match it, the overall probability of a doublet is 1/6. This reasoning matches the explicit counting of 6 favorable outcomes out of 36 total outcomes. Why Other Options Are Wrong: Option 1/3 would suggest that 12 out of 36 outcomes are doublets, which is too many because there are only 6 equal pairs. Option 5/36 treats only 5 outcomes as favorable, which would ignore one of the six possible equal pairs. Option 2/3 is far too large and implies that doublets are very common, which goes against direct outcome counting. Option 1/12 underestimates the true probability and does not align with any correct combination of favorable and total outcomes here. Common Pitfalls: Students sometimes confuse ordered pairs with unordered outcomes, or they mistakenly think that some doublets are more likely than others. Another common issue is miscounting the total sample space by assuming 11 or some other wrong number of outcomes. Always remember that two dice give 6 possibilities for each die, and thus 36 ordered outcome pairs in total. For doublets, simply match equal numbers and count those pairs carefully. Final Answer: The probability of getting a doublet when two dice are thrown is 1/6.

Question 14

From a group of 5 boys and 4 girls (9 people in total), a group of 5 persons is formed at random. What is the probability that the selected group contains at least two girls?

Accepted Answer

Introduction / Context: This question combines combinations and probability in a selection problem involving boys and girls. You are asked to find the probability that a randomly chosen group of five people contains at least two girls. Such problems are standard in aptitude tests and train you to think in terms of counting favorable combinations and using complements when convenient. Given Data / Assumptions: Number of boys = 5. Number of girls = 4. Total people = 9. We select 5 people at random from these 9. We want groups that contain at least two girls. Concept / Approach: The total number of ways to select any 5 people from 9 is 9C5. Instead of counting all cases with at least two girls directly, it is simpler to use the complement rule: At least two girls = Total groups - (groups with zero girls + groups with exactly one girl). We compute each of these counts using combinations and then convert the result into a probability. Step-by-Step Solution: Step 1: Total number of groups of 5 from 9 people = 9C5. 9C5 = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126. Step 2: Count groups with zero girls (all boys). To have zero girls, we must choose all 5 people from the 5 boys: this is 5C5 = 1 group. Step 3: Count groups with exactly one girl. Choose 1 girl from 4: 4C1 = 4. Choose the remaining 4 members from the 5 boys: 5C4 = 5. So groups with exactly one girl = 4 * 5 = 20. Step 4: Total groups with fewer than two girls = 1 + 20 = 21. Step 5: Groups with at least two girls = total groups minus these = 126 - 21 = 105. Step 6: Probability of at least two girls = 105 / 126. Step 7: Simplify 105 / 126 by dividing numerator and denominator by 21 to get 5 / 6. Verification / Alternative check: You can cross check by explicitly listing the possible numbers of girls in the group: 2, 3, or 4. Then compute: Exactly two girls: 4C2 * 5C3. Exactly three girls: 4C3 * 5C2. Exactly four girls: 4C4 * 5C1. Adding these three values and dividing by 9C5 also gives 5/6. This second method confirms that the complementary counting approach is consistent and accurate. Why Other Options Are Wrong: Option 2/3, option 3/5, and option 11/18 are all less than the true probability and arise from various partial counts or missing cases. Option 7/9 is larger than 5/6 and would imply that almost all groups meet the condition, which is not correct. None of these values agree with the complete combination counting or the complement approach, while 5/6 does. Common Pitfalls: A frequent mistake is to misinterpret at least two girls as exactly two girls, ignoring cases with three or four girls. Another common error is to forget to subtract both zero girls and one girl cases when using the complement method. Also, when working quickly, students sometimes calculate 9C5 incorrectly, which then affects every subsequent ratio. Careful use of combinations and explicit consideration of all relevant cases can prevent these errors. Final Answer: The probability that a randomly chosen group of five people contains at least two girls is 5/6.

Question 15

Two fair dice are rolled simultaneously. What is the probability that the sum of the numbers appearing on the two dice is exactly 9?

Accepted Answer

Introduction / Context: Here you must work with the sample space of two dice and identify which ordered pairs give a specific total, in this case a sum of 9. Problems involving sums of dice outcomes are very common in probability sections and help you practise systematic counting of favorable outcomes. Given Data / Assumptions: Two fair six sided dice are rolled together. Each die shows one of 1, 2, 3, 4, 5, 6. All 36 ordered pairs of outcomes are equally likely. We need the probability that the sum of the two results is 9. Concept / Approach: We must identify all ordered pairs (a, b) with a from 1 to 6 and b from 1 to 6 such that a + b = 9. Once we have counted all such pairs, the probability will be: Probability = (Number of pairs with sum 9) / 36. This direct counting strategy is straightforward and avoids confusion about symmetry. Step-by-Step Solution: Step 1: List all pairs of numbers between 1 and 6 that add to 9. Step 2: The valid pairs are: (3, 6), (4, 5), (5, 4), and (6, 3). Step 3: Count these pairs. There are 4 such ordered outcomes. Step 4: Total possible ordered outcomes when rolling two dice = 6 * 6 = 36. Step 5: Probability that the sum is 9 = number of favorable outcomes / total outcomes = 4 / 36. Step 6: Simplify 4 / 36 by dividing numerator and denominator by 4 to get 1 / 9. Verification / Alternative check: You can also reason by symmetry. The smallest possible sum of two dice is 2 and the largest is 12. For sums in the middle, such as 7, 8, and 9, there are several combinations, but as the target sum moves away from 7 the number of combinations decreases. Specifically, 7 has the most combinations, and 8 and 9 have equal numbers of combinations to each other. By listing systematically, you confirm that there are 4 combinations for 9, so the probability 4/36 = 1/9 is consistent with this pattern. Why Other Options Are Wrong: Option 1/6 would require 6 favorable outcomes, which is not true here. Option 2/9 and 4/9 are larger than the correct probability and would imply many more pairs summing to 9 than actually exist. Option 5/18 is also too large and does not correspond to any correct counting pattern. Only 1/9 matches the actual ratio of favorable to total outcomes based on the list of valid pairs. Common Pitfalls: Students sometimes forget that the order of the dice matters when counting outcomes, so they may count (3,6) and (6,3) as a single outcome instead of two. Others may mistakenly include pairs that sum to 8 or 10 when working quickly. To avoid such mistakes, write down the pairs explicitly and check each sum. Remember that 36 is the total number of ordered outcomes for two fair six sided dice, and keep this constant in your denominator for such problems. Final Answer: The probability that the sum of the numbers on the two dice is 9 is 1/9.

Question 16

Four fair dice are thrown simultaneously. What is the probability that exactly two of them show the same face (forming one pair) and the remaining two dice show two different faces, both different from the pair and from each other?

Accepted Answer

Introduction / Context: This question is a more advanced dice probability problem. You are asked for the probability that when four dice are rolled, the pattern of faces is exactly one pair and two distinct singletons, with all three values different from one another. This means the outcome looks like a,a,b,c with a, b, and c distinct. Counting such patterns correctly requires careful combinatorial reasoning. Given Data / Assumptions: Four fair six sided dice are thrown at the same time. Each die has faces 1, 2, 3, 4, 5, 6. All 6^4 = 1296 ordered outcomes are equally likely. We want outcomes with exactly one pair and two other distinct values. Thus the multiset of values must be of the form {a, a, b, c} with a, b, c all different. Concept / Approach: To count favorable outcomes: First choose the value that will appear twice (the pair). Then choose the two distinct values that will appear once each. Finally, arrange these four dice so that the multiset a, a, b, c is assigned to positions in all possible ways. After counting favorable ordered outcomes, we divide by the total 6^4 to obtain the probability. This is a standard counting pattern in problems about dice and repeated values. Step-by-Step Solution: Step 1: Choose the value for the pair. There are 6 possible face values, so 6 choices for a. Step 2: From the remaining 5 distinct face values, choose 2 values for the singletons b and c. Number of ways = 5C2 = 10. Step 3: For a fixed choice of a, b, and c, we must count the permutations of the multiset {a, a, b, c} on four dice. Step 4: The number of distinct permutations of a multiset with 4 items where one value repeats twice is 4! / 2!. Step 5: 4! / 2! = 24 / 2 = 12 ordered arrangements. Step 6: Therefore, the total number of favorable outcomes = 6 * 10 * 12 = 720. Step 7: The total number of ordered outcomes when four dice are rolled is 6^4 = 1296. Step 8: Probability = favorable / total = 720 / 1296. Step 9: Simplify 720 / 1296 by dividing numerator and denominator by 72 to obtain 10 / 18, and then by 2 to get 5 / 9. Verification / Alternative check: To verify, consider that all possible value patterns for four dice can be categorized by the type of repetition: four of a kind, three of a kind plus a distinct singleton, two disjoint pairs, exactly one pair and two distinct singletons, and all four distinct. You can compute the counts for all categories using standard formulas and confirm that the sum is 1296. The count 720 for the exactly one pair and two distinct singles category fits consistently in this breakdown, and the resulting probability 5/9 is therefore trustworthy. Why Other Options Are Wrong: Option 4/9 would correspond to 4 * 1296 / 9 = 576 favorable outcomes, which is too small. Option 7/18 is approximately 0.388 and does not arise from any natural counting split for this pattern. Option 1/3 and option 2/9 are both significantly smaller probabilities and would imply far fewer favorable cases than we actually count. None of these incorrect options match the careful enumeration of 6 choices for the pair value, 10 ways to pick the two singletons, and 12 permutations of each multiset. Common Pitfalls: A common mistake is to forget that b and c must be distinct from each other and from a, which leads to counting patterns like a, a, a, b or a, a, b, b that do not satisfy the condition. Another frequent error is to assume the dice are indistinguishable and use combinations instead of permutations, which ignores the fact that each die is a separate position. Some students also double count permutations when dealing with repeated values. Always fix the value counts first, then count permutations with the correct division for identical items, and finally multiply by the number of ways to choose the distinct values themselves. Final Answer: The probability that exactly one pair appears and the other two dice show different faces is 5/9.

Question 17

A box contains 5 defective bulbs and 15 non defective bulbs, for a total of 20 bulbs. Two bulbs are chosen at random without replacement. What is the probability that both bulbs chosen are non defective?

Accepted Answer

Introduction / Context: This question tests probability with selections from a finite population containing defective and non defective items. You are required to find the probability that both chosen bulbs are non defective when the choice is made without replacement, which means each draw changes the composition of the box. Given Data / Assumptions: Number of defective bulbs = 5. Number of non defective bulbs = 15. Total bulbs = 5 + 15 = 20. Two bulbs are drawn at random without replacement. We want the probability that both bulbs are non defective. Concept / Approach: The probability can be computed either stepwise using conditional probabilities or using combinations. With combinations, we compare the number of ways to select 2 non defective bulbs to the number of ways to select any 2 bulbs from the total. The formula is: Probability = (Number of ways to choose 2 non defective bulbs) / (Number of ways to choose any 2 bulbs). Step-by-Step Solution: Step 1: Total number of ways to choose any 2 bulbs from 20 = 20C2. 20C2 = 20 * 19 / (2 * 1) = 190. Step 2: Total number of non defective bulbs = 15. Step 3: Number of ways to choose 2 non defective bulbs = 15C2. 15C2 = 15 * 14 / (2 * 1) = 105. Step 4: Probability that both bulbs are non defective = 105 / 190. Step 5: Simplify 105 / 190 by dividing numerator and denominator by 5 to get 21 / 38. Verification / Alternative check: You can also use a stepwise approach. The probability that the first bulb is non defective is 15/20 = 3/4. After one non defective bulb is taken, the remaining non defective bulbs are 14 and the total bulbs are 19. The probability that the second bulb is also non defective is 14/19. Therefore, probability both are non defective is: (3/4) * (14/19) = 42 / 76 = 21 / 38 after simplification. This matches the combination method exactly. Why Other Options Are Wrong: Option 15/38 and option 19/38 are both smaller than 21/38 and do not result from any correct combination or stepwise multiplication. Option 3/19 is much too small, and option 5/19 does not match either of the systematic methods. Only 21/38 aligns with both the combination based and the conditional probability based calculations. Common Pitfalls: A common mistake is to treat the draws as independent with replacement and mistakenly use (15/20)^2 with no adjustment, which actually does equal 9/16 and is different from the correct result. Another error is to use 15C2 in the numerator but incorrectly put 20^2 or 20C1 * 19C1 in the denominator without proper reasoning. Always confirm whether the selection is with or without replacement and then be consistent in your use of combinations or sequential probabilities, not mixing concepts from different models. Final Answer: The probability that both bulbs drawn are non defective is 21/38.

Question 18

The letters of the word "CASTIGATION" are arranged at random in all possible distinct ways. What is the probability that all the vowels in the word occupy only the even positions of the arrangement?

Accepted Answer

Introduction / Context: This is a challenging probability problem involving permutations with repeated letters and a positional restriction on vowels. You must consider all distinct arrangements of the letters in the word CASTIGATION and then determine the fraction of those arrangements in which every vowel is placed only in even positions. This problem tests your understanding of permutations with repetition and probability as a ratio of favorable to total arrangements. Given Data / Assumptions: The word is CASTIGATION, which has 11 letters. Letters and their frequencies: C (1), A (2), S (1), T (2), I (2), G (1), O (1), N (1). Vowels in the word: A, A, I, I, O (total 5 vowels). Consonants in the word: C, S, T, T, G, N (total 6 consonants). We count positions from left to right as 1 to 11; even positions are 2, 4, 6, 8, 10. We want every vowel to appear only in positions 2, 4, 6, 8, and 10, and no consonant in those positions. Concept / Approach: First, compute the total number of distinct permutations of the 11 letter multiset. Then count how many of these permutations satisfy the vowel position condition. Finally, express the probability as the ratio favorable / total. Because there are repeated letters (A, T, I appearing twice), we must divide by factorials for repeated letters when counting permutations. Step-by-Step Solution: Step 1: Compute the total number of distinct permutations of the letters in CASTIGATION. Total letters = 11. Repeated letters: A appears 2 times, T appears 2 times, I appears 2 times. Total permutations = 11! / (2! * 2! * 2!). Step 2: Identify positions for vowels and consonants. There are 5 even positions (2, 4, 6, 8, 10) and 6 odd positions (1, 3, 5, 7, 9, 11). We must place the 5 vowels A, A, I, I, O in the 5 even positions, and the 6 consonants C, S, T, T, G, N in the 6 odd positions. Step 3: Count favorable permutations of vowels among the 5 even positions. Number of arrangements of 5 vowels (A, A, I, I, O) = 5! / (2! * 2!). Step 4: Count favorable permutations of consonants among the 6 odd positions. Number of arrangements of 6 consonants (C, S, T, T, G, N) = 6! / 2!. Step 5: Multiply the vowel and consonant arrangements to get total favorable outcomes. Favorable permutations = [5! / (2! * 2!)] * [6! / 2!]. Step 6: Now form the probability ratio. Probability = { [5! / (2! * 2!)] * [6! / 2!] } / { 11! / (2! * 2! * 2!) }. Step 7: Simplify the expression by cancelling common factors of 2! and handling factorials carefully. Carrying out the arithmetic gives a simplified fraction of 1 / 462. Verification / Alternative check: To verify, compute numerical values stepwise. First, 5! / (2! * 2!) counts vowel arrangements and equals 30. Next, 6! / 2! counts consonant arrangements and equals 360. Thus favorable permutations = 30 * 360 = 10800. The total permutations 11! / (2! * 2! * 2!) equal 4989600. The ratio 10800 / 4989600 can be simplified by dividing numerator and denominator by 10800, which gives 1 / 462. This is a very small probability, which makes sense because requiring all vowels to sit only in specific even positions is a strong restriction. Why Other Options Are Wrong: Options 1/231, 1/154, 1/330, and 1/500 are different small fractions that do not match the exact count of favorable versus total permutations. They could arise from mistakes such as miscounting the number of positions, treating some repeated letters as distinct, or failing to separate the vowel and consonant arrangements. None of these alternative fractions simplify to the correct value of 1/462. Common Pitfalls: Frequent mistakes include ignoring repeated letters and using 11! as the total number of permutations without dividing by the appropriate factorials for duplicates. Another issue is misidentifying the number of vowels or the number of even positions in an 11 letter word. Some learners also forget to restrict consonants entirely to odd positions, which leads to overcounting. To avoid these errors, first list the frequency of each letter, then divide positions into vowel and consonant slots, and finally compute permutations for each group separately before taking their product. Final Answer: The probability that all vowels in CASTIGATION occupy only the even positions is 1/462.

Question 19

A single fair coin is tossed once. What is the probability that the outcome is a head?

Accepted Answer

Introduction / Context: This is one of the simplest probability questions, designed to test the basic idea of equally likely outcomes. You toss a fair coin once and must find the probability that the result is a head. Even though this feels intuitive, it is important to express it clearly in probability terms. Given Data / Assumptions: The coin is fair, meaning it is not biased. There are exactly two possible outcomes: head (H) and tail (T). Each outcome is equally likely. The coin is tossed once. Concept / Approach: When all outcomes are equally likely, probability is defined as: Probability = (Number of favorable outcomes) / (Total number of possible outcomes). Here, favorable outcomes are those in which the coin shows a head. The total possible outcomes are head and tail. Step-by-Step Solution: Step 1: List the sample space for one coin toss: {H, T}. Step 2: Total number of possible outcomes = 2. Step 3: The favorable outcome for getting a head is H. Step 4: Number of favorable outcomes = 1. Step 5: Probability of getting a head = 1 / 2. Verification / Alternative check: You can imagine performing a large number of coin tosses using a fair coin. Over time, the frequency of heads will get closer and closer to half of the total tosses. This long run frequency interpretation supports the theoretical result that the probability of a head is 1/2. Because the coin is assumed fair, there is no reason to favor one side over the other. Why Other Options Are Wrong: Option 1/3 and option 1/4 both imply that there are more possible outcomes than actually exist or that heads is significantly less likely than tails, which contradicts the fairness assumption. Option 2/3 and option 3/4 suggest that heads is more likely than tails, which is not true for a fair coin. None of these alternative fractions correctly represent the symmetric nature of the two outcomes of a coin toss. Common Pitfalls: Although this question is simple, it reveals misunderstandings when students mix up the total number of outcomes or misinterpret the term fair. In more complex problems, people sometimes forget that all elementary outcomes must be equally likely before applying the simple favorable over total formula. Establishing this basic idea with a coin toss builds a foundation for approaching more complicated probability questions correctly. Final Answer: The probability that the coin shows a head is 1/2.

Question 20

A box contains 100 balls numbered from 1 to 100. Three balls are selected at random with replacement. What is the probability that the sum of the three selected numbers is odd?

Accepted Answer

Introduction / Context: This problem combines basic probability with parity reasoning. You draw three numbered balls from 1 to 100 with replacement and must determine the probability that the sum of the three numbers is odd. Instead of tracking each possible total, it is much easier to use the parity (odd or even) of each draw and apply simple rules for sums of odd and even numbers. Given Data / Assumptions: The box contains the integers 1 through 100. There are 100 balls in total. Half of them, 50 numbers, are odd and the other 50 numbers are even. Three balls are selected with replacement, so each draw is independent. We want the probability that the sum of the three drawn numbers is odd. Concept / Approach: The parity of a sum depends on how many addends are odd. In particular: Even + even = even. Even + odd = odd. Odd + odd = even. For three numbers, the sum is odd exactly when there are an odd number of odd terms. Therefore we need either exactly one odd and two evens, or exactly three odds. We first compute the probability of drawing an odd or even on a single draw, then use independence to find the probability of each pattern, and finally add these probabilities. Step-by-Step Solution: Step 1: Count odd and even numbers between 1 and 100. There are 50 odd numbers and 50 even numbers. Step 2: Probability of an odd number on a single draw = 50 / 100 = 1 / 2. Step 3: Probability of an even number on a single draw = 50 / 100 = 1 / 2. Step 4: Case 1: exactly one odd and two evens. Number of ways to choose which draw is odd = 3C1 = 3. Probability of a specific pattern like odd, even, even = (1/2) * (1/2) * (1/2) = 1/8. Total probability for exactly one odd = 3 * 1/8 = 3/8. Step 5: Case 2: all three numbers are odd. Probability of three odds in a row = (1/2) * (1/2) * (1/2) = 1/8. Step 6: The sum is odd if either Case 1 or Case 2 occurs, so add the probabilities. Total probability = 3/8 + 1/8 = 4/8 = 1/2. Verification / Alternative check: You can also argue by symmetry. Each draw is equally likely to be odd or even, and there is no bias toward producing odd or even sums when three independent symmetric draws are added. For every triple that sums to an odd number, you can create a corresponding triple that sums to an even number by flipping the parity of exactly one draw. This pairing shows that exactly half of all possible parity patterns produce an odd sum, giving a probability of 1/2, which matches the explicit calculation above. Why Other Options Are Wrong: Options 1/3 and 1/4 are too small and would imply that odd sums are much less common than even sums. Option 3/4 and option 2/3 are too large and would mean that odd sums dominate, which contradicts the symmetry argument. None of these values match the careful case analysis that shows 3/8 + 1/8 = 1/2. Common Pitfalls: A common error is to think that the sum is equally likely to be odd or even for any number of draws because each number is equally likely to be odd or even, without checking the parity rule. This is true for three draws but not for every possible count of draws, so relying on memory alone can be risky. Another mistake is to miscount the number of odd or even numbers between 1 and 100, but since they alternate, the count of 50 odd and 50 even is straightforward. Always link the parity of the sum to the number of odd terms, and then apply independence carefully. Final Answer: The probability that the sum of the three selected numbers is odd is 1/2.

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