Question 1

A number is selected at random from the integers 1 to 30 inclusive. What is the probability that the selected number is divisible by 3 or by 7 (or by both)?

Accepted Answer

Introduction / Context: This question uses basic number theory and probability. You must choose a random integer between 1 and 30 and find the probability that it is divisible by 3 or by 7. Because some numbers may be divisible by both 3 and 7, you should apply the inclusion exclusion principle for counting to avoid double counting overlap. Given Data / Assumptions: The sample space is the set of integers from 1 to 30 inclusive. Total number of possible outcomes = 30. We are interested in numbers divisible by 3 or 7. A number divisible by both 3 and 7 is divisible by 21. Concept / Approach: Let A be the event that the chosen number is divisible by 3, and let B be the event that it is divisible by 7. We use: P(A or B) = P(A) + P(B) - P(A and B). We first count the numbers divisible by 3, then those divisible by 7, then subtract the numbers divisible by both 3 and 7 (that is, divisible by 21) to avoid double counting. Step-by-Step Solution: Step 1: Count numbers between 1 and 30 that are divisible by 3. They are 3, 6, 9, ..., 30. The largest multiple of 3 less than or equal to 30 is 30 = 3 * 10. So, there are 10 multiples of 3 in this range. Step 2: Count numbers between 1 and 30 that are divisible by 7. They are 7, 14, 21, 28. The largest multiple of 7 less than or equal to 30 is 28 = 7 * 4. So, there are 4 multiples of 7 in this range. Step 3: Count numbers divisible by both 3 and 7, that is, by 21. Between 1 and 30, the multiples of 21 are just 21 itself, so there is 1 such number. Step 4: Use the inclusion exclusion principle to count numbers divisible by 3 or 7. Count = 10 + 4 - 1 = 13. Step 5: Total numbers in the sample space = 30. Step 6: Probability that a randomly selected number is divisible by 3 or 7 = 13 / 30. Verification / Alternative check: You can verify by listing these numbers explicitly: 3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30. Counting them one by one gives exactly 13 numbers. Because the sample space has 30 equally likely outcomes, the probability 13/30 is consistent with this direct enumeration. Why Other Options Are Wrong: Options 11/30 and 7/30 are too small and result from undercounting the qualifying numbers or failing to use inclusion exclusion correctly. Option 2/15 equals 4/30, which is far lower than the correct value and suggests that only multiples of 7 were counted. Option 1/3 equals 10/30, which is the probability of divisibility by 3 alone and ignores the additional multiples of 7 that are not multiples of 3. Common Pitfalls: The most common error is to simply add the probabilities of divisibility by 3 and by 7 without subtracting the overlap, leading to 10/30 + 4/30 = 14/30. This overcounts the multiples of 21. Another mistake is to forget that the range is inclusive of both 1 and 30. Always check end points and remember to subtract the intersection when using the rule for P(A or B) in situations where the sets can overlap. Final Answer: The probability that the selected number is divisible by 3 or by 7 is 13/30.

Question 2

A fair coin is tossed 11 times in succession. What is the probability that only the first two tosses result in heads and all of the remaining nine tosses result in tails?

Accepted Answer

Introduction / Context: This question involves repeated independent trials of a fair coin and a very specific pattern of outcomes. You toss a fair coin 11 times and want the probability that the first two tosses are heads while every one of the remaining nine tosses is tails. Problems like this test your ability to work with independent events and to translate a pattern into a product of probabilities. Given Data / Assumptions: A fair coin is tossed 11 times. Each toss is independent of the others. Probability of head on a single toss = 1/2. Probability of tail on a single toss = 1/2. We require the exact pattern: H, H, T, T, T, T, T, T, T, T, T. Concept / Approach: Because all tosses are independent, the probability of a particular sequence of heads and tails is the product of the probabilities for each toss. For each head, we multiply by 1/2, and for each tail, we also multiply by 1/2. Since the pattern is completely specified (no variations allowed), we do not multiply by any combinatorial factor such as a binomial coefficient. Step-by-Step Solution: Step 1: Probability that the first toss is a head = 1/2. Step 2: Probability that the second toss is a head = 1/2. Step 3: For each of the remaining nine tosses, the probability of tail on a single toss = 1/2. Step 4: Probability that all nine remaining tosses are tails = (1/2)^9. Step 5: Because tosses are independent, multiply all probabilities. Overall probability = (1/2) * (1/2) * (1/2)^9. Step 6: Combine exponents: (1/2) * (1/2) * (1/2)^9 = (1/2)^(1 + 1 + 9) = (1/2)^11. Verification / Alternative check: The total number of possible sequences of heads and tails in 11 tosses is 2^11, since each toss has 2 outcomes and they are independent. For this specific event, only one of those 2^11 sequences satisfies the condition of heads only on the first two tosses and tails everywhere else. Thus the probability of this single sequence among all equally likely sequences is 1 / 2^11, which is exactly (1/2)^11. This is another way to see the same result. Why Other Options Are Wrong: Option (1/2)^10 would correspond to a sequence involving 10 specified outcomes instead of 11 and ignores one toss. Option (1/2)^9 and (1/2)^8 also incorrectly count fewer tosses in the exponent. Option 11*(1/2)^11 would be correct for the probability of exactly two heads occurring anywhere in the 11 tosses, but here the positions of the heads are fixed to the first two tosses, so no combinatorial factor of 11 choose 2 is needed. Hence, only (1/2)^11 matches the exact event description. Common Pitfalls: Students often confuse the probability of exactly k heads in n tosses with the probability of a particular arrangement of outcomes. For a specific pattern, you use simple multiplication of probabilities with no extra combinatorial factor. Another pitfall is to misread only the first two tosses will yield heads and think that later tosses could also be heads, which would completely change the event. Always make sure you understand whether the pattern is exact or just a count of successes with no regard to order. Final Answer: The probability that only the first two tosses yield heads and all remaining tosses are tails is (1/2)^11.

Question 3

One card is drawn at random from a standard pack of 52 playing cards. What is the probability that the card drawn is neither a spade nor a king?

Accepted Answer

Introduction / Context: This question tests your ability to use complements and the inclusion exclusion principle with a standard deck of cards. You must find the probability that a randomly drawn card is neither a spade nor a king. Rather than counting directly, it is much easier to find the probability of the card being a spade or a king and then subtract that from 1. Given Data / Assumptions: A standard deck has 52 cards. There are 4 suits: spades, hearts, diamonds, clubs. Each suit has 13 cards. There are 4 kings in the deck, one in each suit. We need the probability that a drawn card is not a spade and not a king. Concept / Approach: Let A be the event that the card is a spade and B be the event that the card is a king. We first find P(A or B), the probability that the card is either a spade or a king (or both). Then we use: P(neither A nor B) = 1 - P(A or B). To get P(A or B), we apply the formula: P(A or B) = P(A) + P(B) - P(A and B). Step-by-Step Solution: Step 1: Number of spade cards = 13, so P(A) = 13 / 52. Step 2: Number of kings in the deck = 4, so P(B) = 4 / 52. Step 3: One card, the king of spades, is both a spade and a king, so P(A and B) = 1 / 52. Step 4: Compute P(A or B) using inclusion exclusion. P(A or B) = 13/52 + 4/52 - 1/52 = 16/52. Step 5: Simplify 16/52 by dividing numerator and denominator by 4 to get 4/13. Step 6: Now compute P(neither A nor B). P(neither spade nor king) = 1 - 4/13. Step 7: Convert 1 to 13/13 and subtract: 13/13 - 4/13 = 9/13. Verification / Alternative check: You can also verify by direct counting. First, count how many cards you must exclude: all spades (13 cards) and any remaining kings that are not spades (3 more cards). That is 13 + 3 = 16 cards that are either spades or non spade kings. The remaining cards in the deck are 52 - 16 = 36. Therefore, there are 36 cards that are neither spades nor kings. The probability is 36 / 52, which simplifies to 9 / 13. This matches the complement based calculation. Why Other Options Are Wrong: Option 4/13 is the probability of drawing a spade or a king, not the probability of avoiding both. Option 1/2 suggests that exactly half the deck is excluded, which is not accurate. Option 3/13 and option 1/13 are too small and imply that almost all cards are spades or kings, which is clearly not the case. None of these match the correct count of 36 safe cards out of 52 total. Common Pitfalls: One frequent mistake is to add P(spade) and P(king) and forget to subtract P(king of spades), thereby double counting that card. Another mistake is to attempt to directly count the cards that are neither spade nor king without systematically subtracting the excluded cards. Using complements and inclusion exclusion reduces the chance of such errors and is a powerful method for many exam style probability problems. Final Answer: The probability that the card drawn is neither a spade nor a king is 9/13.

Question 4

For two events A and B, the probability of event A is 1/4, the probability of event B is 1/2, and the probability that both A and B occur together is 7/50. What is the probability that neither event A nor event B occurs?

Accepted Answer

Introduction / Context: This problem uses the basic rules of probability for two events, including the formula for the probability of a union and its complement. You are given probabilities of events A, B, and their intersection, and you must compute the probability that neither event occurs. This is standard in quantitative aptitude and helps reinforce the relationship between unions, intersections, and complements. Given Data / Assumptions: P(A) = 1/4. P(B) = 1/2. P(A and B) = 7/50. We need P(neither A nor B), which is the probability that event A does not occur and event B does not occur. Concept / Approach: First, find the probability that at least one of A or B occurs, which is P(A or B). For any two events: P(A or B) = P(A) + P(B) - P(A and B). Then use the complement rule. The event neither A nor B is the complement of A or B. Thus: P(neither A nor B) = 1 - P(A or B). Step-by-Step Solution: Step 1: Write down the formula for P(A or B). P(A or B) = P(A) + P(B) - P(A and B). Step 2: Substitute the given values: P(A) = 1/4, P(B) = 1/2, P(A and B) = 7/50. Step 3: Convert all fractions to a common denominator, for example 100. 1/4 = 25/100, 1/2 = 50/100, and 7/50 = 14/100. Step 4: Compute P(A or B) using these equivalent fractions. P(A or B) = 25/100 + 50/100 - 14/100 = (25 + 50 - 14) / 100 = 61/100. Step 5: Use the complement rule to find P(neither A nor B). P(neither A nor B) = 1 - P(A or B) = 1 - 61/100. Step 6: Convert 1 to 100/100 and subtract to get 100/100 - 61/100 = 39/100. Verification / Alternative check: You can interpret the probabilities as percentages of a large population or of many repeated trials. For example, imagine 100 trials. On average, A occurs in 25 trials, B occurs in 50 trials, and both occur together in 14 trials. By inclusion exclusion, A or B occurs in 25 + 50 - 14 = 61 trials. Therefore, in the remaining 39 trials, neither A nor B occurs, giving a proportion of 39/100. This reasoning directly matches the computed probability. Why Other Options Are Wrong: Option 61/100 is actually P(A or B), not the probability of neither event occurring. Option 17/100 and option 25/100 are smaller than the correct complement and would arise from subtracting incorrectly or misusing the intersection term. Option 49/100 does not align with either 61/100 or its complement and does not come from a consistent application of the union formula and complement rule. Only 39/100 fits the given information correctly. Common Pitfalls: Students sometimes forget to subtract P(A and B) when computing P(A or B), which double counts the overlap and leads to a union probability greater than 1 in extreme cases. Another mistake is to confuse P(neither A nor B) with P(A complement) * P(B complement), which is only valid if A and B are independent and P(A and B) equals P(A) * P(B). Because we are explicitly given P(A and B), we should use that data instead of assuming independence. Always follow the union formula and complement rule step by step to avoid these errors. Final Answer: The probability that neither A nor B occurs is 39/100.

Question 5

From a well shuffled pack of 52 playing cards, two cards are drawn together at random. What is the probability that both of the drawn cards are kings?

Accepted Answer

Introduction / Context: This problem checks understanding of basic probability with playing cards and the use of combinations when two cards are drawn together without replacement. The focus is on counting how many ways we can obtain two kings from a standard deck and comparing this to the total number of possible pairs of cards. Given Data / Assumptions: There is a standard deck with 52 cards. The deck contains 4 kings in total. Two cards are drawn together without replacement. All unordered pairs of cards are equally likely. We want the probability that both selected cards are kings. Concept / Approach: When several objects are drawn together and order does not matter, combinations are used. The probability of an event is equal to the number of favourable combinations divided by the total number of possible combinations. Here, favourable outcomes are all 2 card combinations that consist of kings only, and the total outcomes are all 2 card combinations from the full deck of 52 cards. Step-by-Step Solution: Total number of ways to choose any 2 cards from 52 is C(52, 2). C(52, 2) = 52 * 51 / 2 = 1326 possible pairs. Number of kings in the deck is 4. Number of ways to choose 2 kings from these 4 is C(4, 2). C(4, 2) = 4 * 3 / 2 = 6 favourable pairs. Required probability = favourable pairs / total pairs = 6 / 1326. Simplify 6 / 1326 by dividing numerator and denominator by 6 to get 1 / 221. Verification / Alternative check: An alternative method is to compute the probability sequentially. The probability that the first card is a king is 4 / 52. Once this happens, 3 kings remain out of 51 cards, so the probability that the second card is also a king is 3 / 51. The combined probability is (4 / 52) * (3 / 51) which equals 12 / 2652. After simplification this again reduces to 1 / 221, confirming the combination based result. Why Other Options Are Wrong: 1/15 is far too large and would correspond to many more favourable outcomes than actually possible. 1/26 would mean 51 favourable pairs, which does not match the count of 6 ways to choose two kings. 2/221 is larger than the correct probability and does not result from any valid combination count. Common Pitfalls: Learners often confuse combinations and permutations by accidentally counting the pair king of hearts followed by king of spades as different from king of spades followed by king of hearts, even though the question treats both as the same pair. Another frequent error is to divide by 52 * 51 instead of C(52, 2), which corresponds to treating order as important when it is not. Always match the counting method to the wording of the question, especially when cards are drawn together. Final Answer: The probability that both cards drawn are kings is 1/221.

Question 6

One card is drawn at random from a standard pack of 52 playing cards. What is the probability that the card drawn is neither a spade nor a Jack?

Accepted Answer

Introduction / Context: This problem checks basic set based probability with playing cards. Instead of asking for one particular suit or rank, it asks for cards that are not spades and also not Jacks. To solve this, we use the idea of excluding unwanted cards from the total and then computing the probability for the remaining cards. Given Data / Assumptions: A standard deck contains 52 cards. There are 4 suits, so there are 13 spades. There are 4 Jacks in the deck, one from each suit. The Jack of spades is counted both as a spade and as a Jack. One card is drawn at random and all cards are equally likely. We want the card to be neither a spade nor a Jack. Concept / Approach: We use the complement principle through set operations. First we count how many cards are in the union of the set of spades and the set of Jacks. These are the cards that we do not want. Then we subtract this count from the total number of cards to get the allowed cards. Finally, the probability is the number of allowed cards divided by 52. Step-by-Step Solution: Number of spades = 13. Number of Jacks = 4. Jack of spades is common to both sets, so it is counted twice above. Cards that are spades or Jacks = 13 + 4 - 1 = 16. Cards that are neither spades nor Jacks = 52 - 16 = 36. Required probability = 36 / 52. Simplify 36 / 52 by dividing numerator and denominator by 4 to get 9 / 13. Verification / Alternative check: We can also think directly about suits and ranks. Non spade suits are hearts, diamonds and clubs, which together have 39 cards. From these, we must remove the remaining 3 Jacks: Jack of hearts, Jack of diamonds and Jack of clubs. So allowed cards = 39 minus 3 = 36. Dividing by 52 again gives the same probability 9 / 13, which verifies the previous calculation. Why Other Options Are Wrong: 4/13 corresponds to 16 favourable cards, which actually equals the number that are spades or Jacks, not those excluded from that union. 10/13 and 8/13 are larger than 9/13 and would require more allowed cards than the 36 that really remain after removing all spades and all Jacks. Common Pitfalls: A common mistake is to subtract 13 and 4 directly from 52 without correcting for the double counted Jack of spades, which gives 52 minus 17 instead of 52 minus 16. Another error is to treat spades and Jacks as independent, which they are not, since suits and ranks combine in specific cards. Always remember to adjust for overlap when applying the union formula for sets. Final Answer: The probability that the card drawn is neither a spade nor a Jack is 9/13.

Question 7

Three unbiased coins are tossed simultaneously. What is the probability of getting at most two heads?

Accepted Answer

Introduction / Context: This question focuses on probability for repeated coin tosses. The phrase at most two heads means any outcome where the number of heads is zero, one or two, but not three. Because the coins are unbiased, every possible sequence of heads and tails is equally likely, and we can use either direct counting or the complement rule to find the required probability. Given Data / Assumptions: There are three fair coins. Each coin has two possible outcomes: head or tail. All 2^3 = 8 outcomes are equally likely. Event of interest: number of heads is less than or equal to two. Concept / Approach: The easiest way is to use the complement rule. The only way to have more than two heads with three coins is to have exactly three heads. So at most two heads is the complement of exactly three heads. We find the probability of three heads and subtract it from one. This saves time compared to listing all the zero, one and two head cases separately, although we can list them later as a check. Step-by-Step Solution: Total number of outcomes = 2^3 = 8. The only outcome with three heads is HHH, so there is exactly 1 outcome with three heads. Probability of exactly three heads = 1 / 8. Probability of at most two heads = 1 - probability of three heads. So probability of at most two heads = 1 - 1 / 8 = 7 / 8. Verification / Alternative check: We can list all outcomes explicitly: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Outcomes with at most two heads are HHT, HTH, THH, HTT, THT, TTH, TTT, which is 7 outcomes. Dividing by the total 8 outcomes gives 7 / 8, which matches the complement method and confirms the result. Why Other Options Are Wrong: 3/4 would correspond to only 6 favourable outcomes out of 8, but we have 7. 2/3 would require about 5.33 favourable outcomes, which does not make sense for a discrete sample of 8 equally likely cases. 5/8 would mean only 5 favourable outcomes, again not matching the correct count. Common Pitfalls: Learners sometimes misread at most two heads as exactly two heads. That would give only 3 outcomes and a probability of 3 / 8, which is incorrect. Others forget that three heads is only a single outcome and assume it has a larger probability. Carefully distinguishing between complement events and counting all possibilities avoids these errors. Final Answer: The probability of getting at most two heads when three fair coins are tossed is 7/8.

Question 8

Tickets numbered from 1 to 20 are mixed thoroughly and then one ticket is drawn at random. What is the probability that the number on the drawn ticket is a multiple of 4 or a multiple of 15?

Accepted Answer

Introduction / Context: This question uses simple probability with numbers on tickets. The event is defined in terms of divisibility by 4 or 15. To solve it correctly, we must count how many numbers in the range from 1 to 20 satisfy the divisibility condition and then divide by the total number of tickets. Given Data / Assumptions: Tickets are numbered from 1 to 20 inclusive. One ticket is drawn at random. Each ticket has an equal chance of being chosen. Favourable tickets are those whose numbers are multiples of 4 or multiples of 15. Concept / Approach: We use the classical probability formula: probability equals number of favourable outcomes divided by total number of outcomes. For a condition of the form multiple of 4 or 15, we apply the union principle. Count multiples of 4, count multiples of 15, then subtract any common multiples to avoid double counting. Since the range is small, we can list multiples explicitly. Step-by-Step Solution: Total tickets = 20, so total outcomes = 20. Multiples of 4 between 1 and 20: 4, 8, 12, 16, 20. That is 5 numbers. Multiples of 15 between 1 and 20: 15 only. That is 1 number. Least common multiple of 4 and 15 is 60, which is greater than 20, so there is no overlap. Total favourable numbers = 5 + 1 = 6. Required probability = 6 / 20. Simplify 6 / 20 by dividing numerator and denominator by 2 to get 3 / 10. Verification / Alternative check: We can check by writing all numbers from 1 to 20 and marking those divisible by 4 or 15. The marked numbers are 4, 8, 12, 15, 16 and 20. Counting them gives 6 favourable tickets. Because there are 20 tickets total, the fraction 6 / 20 simplifies to 3 / 10, which confirms the previous calculation is correct. Why Other Options Are Wrong: 1/4 would require exactly 5 favourable numbers, but there are 6. 2/5 corresponds to 8 favourable outcomes, which is too large for this range. 1/2 implies that 10 out of 20 numbers meet the condition, which clearly is not the case. Common Pitfalls: Some learners forget to check for common multiples of 4 and 15 and either subtract incorrectly or think there is an overlap when there is not. Others may miscount multiples of 4 by skipping 20 or including 0, which is outside the given range. Listing the numbers in the range and verifying each condition is a reliable method to avoid such mistakes. Final Answer: The probability that the drawn ticket shows a number that is a multiple of 4 or 15 is 3/10.

Question 9

A manager will approve the leave request of exactly one employee for a particular day. Five employees Roshan, Mahesh, Sripad, Laxmipriya and Shreyan submit leave applications for the Diwali day. Assuming all five have equal chance, what is the proba…

Accepted Answer

Introduction / Context: This question is a direct application of basic probability in a simple selection situation. A manager must choose exactly one person from a group of five employees who all apply for leave on the same day. We are asked to find the probability that a specific employee, Laxmipriya, is the one whose request is approved. Given Data / Assumptions: There are 5 employees: Roshan, Mahesh, Sripad, Laxmipriya and Shreyan. The manager approves leave for exactly one of these five employees. Each employee is assumed to have an equal chance of being selected. Event of interest: the manager selects Laxmipriya for leave. Concept / Approach: This is a uniform probability model. When all outcomes are equally likely, the probability of a particular individual being chosen is 1 divided by the total number of individuals. There is no complex counting or conditional reasoning here, just a simple ratio of one favourable case to the total number of possible cases. Step-by-Step Solution: Total number of possible approvals = 5, one for each employee. Only one of these outcomes is favourable, namely the outcome where Laxmipriya is selected. Therefore the number of favourable outcomes = 1. Required probability = favourable outcomes / total outcomes = 1 / 5. Verification / Alternative check: We can think of the selection as picking one name from a box containing five slips, each with a different name. Because every slip is equally likely to be picked, each person has a probability of 1 / 5 of being chosen. This matches the calculation and shows the reasoning is consistent with everyday intuition about fairness. Why Other Options Are Wrong: 4/5 would suggest that Laxmipriya is almost certain to be chosen, which contradicts the equal chance assumption. 1/4 would correspond to a situation with four employees instead of five. 1/2 would mean there are only two equally likely outcomes, which is not true here. Common Pitfalls: A common mistake is to misinterpret the phrase accepts only one employees leave request as implying some extra condition that changes the probabilities. In fact, unless additional information is given about preferences or priorities, the natural assumption is that all employees are equally likely to be chosen. Staying close to this assumption keeps the reasoning straightforward. Final Answer: The probability that the manager approves the leave request of Laxmipriya is 1/5.

Question 10

Tickets are numbered from 1 to 18 and mixed together. One ticket is drawn at random. What is the probability that the number on the drawn ticket is a multiple of 2 or a multiple of 3?

Accepted Answer

Introduction / Context: This question uses basic probability with divisibility conditions on integers. The tickets are numbered from 1 to 18, and we want the probability that a randomly selected ticket shows a number that is divisible by 2 or divisible by 3. The task is to count such numbers and divide by the total number of tickets. Given Data / Assumptions: Tickets are numbered 1, 2, 3, up to 18. One ticket is drawn at random from these 18 tickets. Each ticket has equal chance of being selected. The favourable numbers are those that are multiples of 2 or multiples of 3. Concept / Approach: Use the union principle for counting. Count all multiples of 2, count all multiples of 3, and then subtract the numbers that are multiples of both 2 and 3, because those numbers have been counted twice. Once the count of favourable numbers is known, probability equals favourable count divided by total count, which here is 18. Step-by-Step Solution: Total tickets = 18. Multiples of 2 from 1 to 18: 2, 4, 6, 8, 10, 12, 14, 16, 18, which is 9 numbers. Multiples of 3 from 1 to 18: 3, 6, 9, 12, 15, 18, which is 6 numbers. Multiples of both 2 and 3 are multiples of 6: 6, 12, 18, which is 3 numbers. Total favourable numbers = 9 + 6 - 3 = 12. Required probability = 12 / 18. Simplify 12 / 18 by dividing numerator and denominator by 6 to get 2 / 3. Verification / Alternative check: We can check by writing all numbers from 1 to 18 and marking those that are multiples of 2 or 3. The marked numbers will be 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18, which is exactly 12 numbers. Since there are 18 numbers in total, the probability is 12 / 18, which simplifies to 2 / 3, confirming the result. Why Other Options Are Wrong: 1/3 would correspond to only 6 favourable numbers, which is too small. 3/5 equals 10 out of 18, but we found 12 favourable numbers. 5/6 would require 15 favourable numbers, which is more than actually satisfy the condition. Common Pitfalls: Learners sometimes forget to subtract the overlap, that is the multiples of 6, and obtain 15 instead of 12 favourable numbers. Others may miss one or two numbers when listing multiples by hand. Using the inclusion exclusion idea and checking the list carefully helps avoid such counting errors. Final Answer: The probability that the number on the ticket is a multiple of 2 or 3 is 2/3.

Question 11

Two cards are drawn at random from a well shuffled pack of 52 playing cards. What is the probability that either both cards are red or both cards are queens?

Accepted Answer

Introduction / Context: This question combines two events for card selection. We draw two cards and want the probability that the pair is either both red or both queens. Because two queens can also both be red, we must handle the overlap between these events carefully using the union rule in probability. Given Data / Assumptions: A standard deck has 52 cards. There are 26 red cards, consisting of hearts and diamonds. There are 4 queens in total. Among these queens, 2 are red queens and 2 are black queens. Two cards are drawn together without replacement, and all unordered pairs are equally likely. Event A: both cards are red. Event B: both cards are queens. Concept / Approach: We use combinations to count outcomes. The total number of possible pairs is C(52, 2). To find the probability of A or B, we use P(A or B) = P(A) + P(B) minus P(A and B). For counting P(A and B), we note that both cards must be red queens, so both must be chosen from the two red queens available. Step-by-Step Solution: Total pairs = C(52, 2) = 52 * 51 / 2 = 1326. Number of red cards = 26. Pairs with both red cards = C(26, 2) = 26 * 25 / 2 = 325. Number of queens = 4, so pairs with both queens = C(4, 2) = 4 * 3 / 2 = 6. Red queens are 2 in number, so pairs that are both red and both queens = C(2, 2) = 1. Favourable pairs = 325 + 6 - 1 = 330. Required probability = 330 / 1326. Simplify 330 / 1326 by dividing numerator and denominator by 6 to get 55 / 221. Verification / Alternative check: We can also compute probabilities directly. P(both red) = (26/52) * (25/51). P(both queens) = (4/52) * (3/51). P(both red queens) = (2/52) * (1/51). Then P(A or B) = P(both red) + P(both queens) minus P(both red queens). Substituting the fractions and simplifying leads again to 55 / 221, so the combination approach is confirmed. Why Other Options Are Wrong: 17/112 and 33/221 do not correspond to the correct count of 330 favourable pairs out of 1326. 5/26 is larger than 55/221 and does not arise from any natural combination count for this situation. Common Pitfalls: A frequent error is to add the probabilities of both red and both queens without subtracting the overlap where the pair consists of two red queens. This double counts that case and produces a probability that is too large. Another mistake is to forget that order does not matter and use 52 * 51 as the denominator instead of C(52, 2). Careful use of the union rule and combinations prevents these issues. Final Answer: The probability that either both cards are red or both cards are queens is 55/221.

Question 12

A bag contains 6 black balls and 8 white balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is white?

Accepted Answer

Introduction / Context: This is a basic probability question involving a single draw from a collection of balls of two colours. The task is to compute the chance that the ball drawn at random from the bag is white. Problems of this type give good practice with the fundamental formula for probability based on equally likely outcomes. Given Data / Assumptions: Number of black balls = 6. Number of white balls = 8. Total number of balls in the bag = 6 + 8 = 14. One ball is drawn at random, and each ball is equally likely to be selected. The event of interest is that the selected ball is white. Concept / Approach: In a simple uniform sample space, the probability of an event is equal to the ratio of favourable outcomes to total possible outcomes. Here, each ball is considered one outcome. Favourable outcomes correspond to the white balls. We do not need combinations because only one ball is drawn, and order is not an issue. Step-by-Step Solution: Compute total number of balls: 6 + 8 = 14. Number of favourable balls (white) = 8. Probability of drawing a white ball = favourable outcomes / total outcomes. So required probability = 8 / 14. Simplify 8 / 14 by dividing numerator and denominator by 2 to obtain 4 / 7. Verification / Alternative check: We can also compute the probability of drawing a black ball and then use the complement. Probability of black ball = 6 / 14 = 3 / 7. Therefore probability of white ball = 1 minus 3 / 7 = 4 / 7. This second method arrives at the same fraction and gives a quick consistency check for the calculation. Why Other Options Are Wrong: 3/7 is the probability of drawing a black ball, not a white ball. 1/8 would correspond to 1 white ball out of 8 total balls, which does not match the situation. 3/4 suggests that three quarters of all balls are white, which would require 10.5 white balls out of 14, which is impossible. Common Pitfalls: The most common error is mixing up the counts of black and white balls or forgetting to add them correctly when finding the total. Another mistake is to leave the fraction unsimplified, although in probability questions it is usually better to present the answer in simplest form. Always identify the number of favourable cases clearly before forming the ratio. Final Answer: The probability that the ball drawn from the bag is white is 4/7.

Question 13

If the odds in favour of an event are given as p:q, what is the probability that the event will occur?

Accepted Answer

Introduction / Context: This conceptual question links the idea of odds with the usual definition of probability. In many competitive exams and real life situations, information is sometimes expressed as odds in favour of an event rather than as a simple probability. Understanding how to convert between odds and probability is therefore an important skill. Given Data / Assumptions: The odds in favour of a certain event are given as p:q. This means that for every p favourable cases, there are q unfavourable cases. All cases under consideration are assumed to be equally likely. We want the probability that the event occurs, expressed in terms of p and q. Concept / Approach: Odds in favour of an event describe the ratio of favourable outcomes to unfavourable outcomes. If the odds are p:q, then there are p favourable outcomes and q unfavourable outcomes. The total number of possible outcomes is p plus q. Probability, by definition, is the ratio of favourable outcomes to total outcomes. So we simply divide the number of favourable cases by the sum of favourable and unfavourable cases. Step-by-Step Solution: Odds in favour p:q means favourable outcomes = p and unfavourable outcomes = q. Total outcomes under consideration = p + q. Probability of the event occurring = favourable outcomes / total outcomes. So probability = p / (p + q). Verification / Alternative check: To verify, consider a simple numerical example. Suppose the odds in favour of an event are 3:2. This implies 3 favourable and 2 unfavourable cases, giving total outcomes equal to 5. The probability of the event is then 3 / 5, which matches the formula p / (p + q) when p equals 3 and q equals 2. Trying other values, such as p equal to 1 and q equal to 4, gives probability 1 / 5, which also fits the formula. Why Other Options Are Wrong: p/q represents the odds ratio itself, not the probability. It compares favourable to unfavourable cases but does not include the total number of outcomes. q/(p+q) actually gives the probability that the event does not occur, because it uses the count of unfavourable outcomes over the total. q/p is simply the odds against the event, which is the reciprocal of the odds in favour, not the required probability. Common Pitfalls: Learners sometimes confuse odds and probability and attempt to use p/q directly as a probability. This leads to values greater than 1 when p is larger than q, which cannot be a valid probability. Always remember that probability must lie between 0 and 1 and that odds must be converted by dividing the favourable part by the total of favourable plus unfavourable cases. Final Answer: If the odds in favour of an event are p:q, then the probability that the event occurs is p/(p+q).

Question 14

A three digit number is formed at random using the digits 1, 2, 3, 4 and 5 without repetition. What is the probability that the number formed is divisible by 5?

Accepted Answer

Introduction / Context: This question involves forming three digit numbers from a given set of digits without repetition. The focus is on divisibility by 5, which depends entirely on the last digit of the number. By understanding how many numbers can be formed in total and how many of those end with the required digit, we can find the probability efficiently. Given Data / Assumptions: Available digits are 1, 2, 3, 4 and 5. Each digit can be used at most once in the number. A three digit number is formed, so the first digit cannot be zero, but there is no zero in the set anyway. We are interested in numbers that are divisible by 5. Concept / Approach: A number is divisible by 5 if and only if its units digit is 0 or 5. Since 0 is not in the list of digits, any valid number divisible by 5 must end with 5. That means the last digit is fixed as 5. The first and second digits are then arranged using the remaining four digits. We use permutations because the order of digits in a number matters. Finally, the probability is favourable permutations divided by total permutations. Step-by-Step Solution: Total digits available = 5. Total three digit numbers without repetition = 5P3 = 5 * 4 * 3 = 60. For divisibility by 5, the last digit must be 5. Fix the units digit as 5. Remaining digits for the first two places are 1, 2, 3 and 4, which is 4 digits. Number of ways to arrange the first two digits from these 4 digits = 4P2 = 4 * 3 = 12. So favourable numbers divisible by 5 = 12. Required probability = favourable / total = 12 / 60 = 1 / 5. Verification / Alternative check: We can double check by noticing that in all three digit numbers formed without repetition from five digits, each of the five digits is equally likely to appear in the last place. Therefore, the probability that the last digit is 5 is simply 1 out of 5. Since divisibility by 5 depends only on the last digit here, this reasoning again gives probability 1 / 5, confirming our detailed calculation. Why Other Options Are Wrong: 1/4 would be correct only if there were four possible digits that could make the number divisible by 5, which is not true. 1/2 and 2/5 are too large and would imply that a very high proportion of all possible numbers are divisible by 5, which contradicts the exact count of 12 favourable numbers out of 60. Common Pitfalls: A common mistake is to forget that digits are used without repetition and to treat the number of possibilities as 5^3 instead of 5P3. Another error is to think that any occurrence of the digit 5 in the number makes it divisible by 5, whereas only the last digit matters. Focusing on the units place rule for divisibility by 5 is the key to solving this problem correctly. Final Answer: The probability that the three digit number formed is divisible by 5 is 1/5.

Question 15

The probabilities that drivers A, B and C will each drive home safely after consuming liquor are 2/5, 3/7 and 3/4 respectively. Assuming their driving events are independent, what is the probability that all three drivers drive home safely?

Accepted Answer

Introduction / Context: This question illustrates how to combine independent probabilities for several people performing the same type of action. Each driver has a given probability of reaching home safely after consuming liquor, and the problem asks for the chance that all three do so. This is a straightforward application of the multiplication rule for independent events. Given Data / Assumptions: Probability that driver A drives home safely = 2/5. Probability that driver B drives home safely = 3/7. Probability that driver C drives home safely = 3/4. The driving results of A, B and C are independent of each other. Event of interest: all three drivers reach home safely. Concept / Approach: When events are independent, the probability that they all occur together is the product of their individual probabilities. Here, we multiply the probabilities for A, B and C each driving safely. There is no need to consider any order, because the events occur simultaneously and independently, and the multiplication rule captures this. Step-by-Step Solution: Probability that A is safe = 2/5. Probability that B is safe = 3/7. Probability that C is safe = 3/4. Required probability that all three are safe = (2/5) * (3/7) * (3/4). First multiply numerators: 2 * 3 * 3 = 18. Multiply denominators: 5 * 7 * 4 = 140. So probability = 18 / 140. Simplify 18 / 140 by dividing numerator and denominator by 2 to get 9 / 70. Verification / Alternative check: We can confirm the result by simplifying at an earlier step. For example, write (2/5) * (3/7) * (3/4) and cancel a factor of 2 from 2 and 4 to get (1/5) * (3/7) * (3/2). Then multiply 1 * 3 * 3 = 9 and 5 * 7 * 2 = 70, giving 9 / 70 again. This double check shows that our simplification is consistent and that no arithmetic error has been made. Why Other Options Are Wrong: 3/70 and 4/70 are smaller fractions that would imply a much lower probability than is justified by the given numbers. 1/50 does not correspond to any natural simplification of the product of 2/5, 3/7 and 3/4. Common Pitfalls: A frequent misunderstanding is to add probabilities instead of multiplying them when dealing with independent events. Another mistake is to use incorrect denominators by multiplying only some of the terms or by failing to simplify properly. Carefully multiplying all three fractions and simplifying step by step avoids these issues and leads to the correct result. Final Answer: The probability that all three drivers will drive home safely is 9/70.

Question 16

One card is drawn at random from a standard pack of 52 playing cards. What is the probability that the card drawn is either a diamond or a king?

Accepted Answer

Introduction / Context: This card probability question involves a union of two events based on suit and rank. We want the probability that a single card randomly drawn from a standard deck is either a diamond or a king. Because one specific card belongs to both groups, we must take overlap into account when counting favourable outcomes. Given Data / Assumptions: A standard deck has 52 cards. There are 4 suits, so diamonds have 13 cards. There are 4 kings in the deck, one in each suit. The king of diamonds is both a king and a diamond. One card is drawn at random with all cards equally likely. Event A: card is a diamond. Event B: card is a king. Concept / Approach: To find the probability that the card is a diamond or a king, we use the union formula: P(A or B) = P(A) + P(B) minus P(A and B). In counting terms, that means we add the number of diamonds and the number of kings, then subtract one for the overlap, which is the king of diamonds. Finally, we divide by the total number of cards in the deck. Step-by-Step Solution: Number of diamonds = 13. Number of kings = 4. The king of diamonds is common to both sets. Favourable cards that are diamonds or kings = 13 + 4 - 1 = 16. Total cards in the deck = 52. Required probability = 16 / 52. Simplify 16 / 52 by dividing numerator and denominator by 4 to get 4 / 13. Verification / Alternative check: Compute probabilities directly. P(diamond) = 13 / 52. P(king) = 4 / 52. P(king of diamonds) = 1 / 52. So P(diamond or king) = 13 / 52 + 4 / 52 minus 1 / 52 = 16 / 52. This simplifies to 4 / 13, which matches the earlier combination based reasoning and confirms the answer. Why Other Options Are Wrong: 5/12 would require about 21.67 favourable cards, which is not an integer count and does not match 16. 5/13 implies 20 favourable cards, again inconsistent with the count of 16. 3/14 is much smaller and would represent only approximately 11.14 favourable cards, which is not correct. Common Pitfalls: Learners often forget to subtract the overlapping card that belongs to both categories and simply add 13 and 4 to get 17 favourable cards. This overcounts the king of diamonds and leads to an incorrect probability. Always check whether an item can satisfy both conditions when applying the union formula in probability questions. Final Answer: The probability that the drawn card is a diamond or a king is 4/13.

Question 17

Consider 5 independent Bernoulli trials, each with probability of success p. If the probability of at least one failure in these five trials is greater than or equal to 31/32, then p lies in which interval?

Accepted Answer

Introduction / Context: This problem deals with Bernoulli trials and connects the probability of at least one failure with the underlying probability of success. It requires translating a condition on the probability of a compound event into an inequality involving p and then interpreting the solution as an interval. This is a standard application of the complement rule and basic algebra. Given Data / Assumptions: There are 5 independent Bernoulli trials. Each trial has probability of success equal to p. Therefore probability of failure on a single trial is 1 minus p. The probability that there is at least one failure in the 5 trials is greater than or equal to 31/32. We need to determine the range of values of p that satisfy this condition. Concept / Approach: The key idea is to use the complement of at least one failure. The event at least one failure is the complement of the event no failures at all. No failures means all trials are successful. Because the trials are independent, the probability that all 5 are successful is p^5. Therefore, the probability of at least one failure is 1 minus p^5. We set this greater than or equal to 31/32 and solve the resulting inequality for p. Step-by-Step Solution: Let F be the event that there is at least one failure. P(F) = 1 - P(no failure). No failure means all 5 trials are successful, so P(no failure) = p^5. Given condition: P(F) greater than or equal to 31/32. So 1 - p^5 greater than or equal to 31/32. Rearrange: p^5 less than or equal to 1 - 31/32 = 1/32. Thus p^5 less than or equal to 1/32. Note that 32 = 2^5, so 1/32 = 1 / 2^5 = (1/2)^5. Therefore p^5 less than or equal to (1/2)^5 implies p less than or equal to 1/2, because p is between 0 and 1. So the allowed interval for p is 0 less than or equal to p less than or equal to 1/2. Verification / Alternative check: Check boundary values. If p equals 1/2, then p^5 equals (1/2)^5 which is 1/32. Then P(F) = 1 minus 1/32 = 31/32, which satisfies the given condition exactly. If p is smaller than 1/2, then p^5 is smaller than 1/32, so 1 minus p^5 is greater than 31/32, which still satisfies the condition. If p is greater than 1/2, then p^5 becomes larger than 1/32 and 1 minus p^5 becomes less than 31/32, which no longer meets the requirement. This confirms that the interval [0, 1/2] is correct. Why Other Options Are Wrong: (0, 1) is too wide because values greater than 1/2 do not satisfy the inequality. [1/2, 1] includes values where p is greater than 1/2, which lead to a probability of at least one failure less than 31/32. (1/2, 1] is even worse because it excludes the valid boundary value p equal to 1/2 and contains only values that violate the condition. Common Pitfalls: Learners sometimes confuse at least one failure with exactly one failure, which would require a binomial expansion instead of using a simple complement. Another common error is to forget that probabilities lie between 0 and 1 and to take an incorrect root when solving p^5 less than or equal to a fraction. Always apply the complement rule first for at least one type questions and then solve the resulting inequality carefully with attention to the allowed range of p. Final Answer: The probability condition is satisfied when p lies in the interval [0, 1/2].

Question 18

There are 10 letters and 10 corresponding envelopes with 10 different addresses. Each letter is placed randomly into an envelope. What is the probability that exactly 9 letters are placed in their correct envelopes?

Accepted Answer

Introduction / Context: This is a classic problem from the topic of permutations and derangements, which studies how many ways we can place objects so that none or some of them sit in their natural position. Here we have 10 letters and 10 matching envelopes, and we want the probability that exactly 9 letters end up in the correct envelopes. This condition turns out to be impossible, and the explanation reveals why. Given Data / Assumptions: There are 10 distinct letters and 10 distinct envelopes. Each envelope has one correct letter that should be placed inside it. Letters are placed randomly, one in each envelope, so each permutation of the 10 letters is equally likely. We are interested in arrangements where exactly 9 letters are in the correct envelopes. Concept / Approach: If exactly 9 letters are in their correct envelopes, that means 9 positions match perfectly with their intended letters. The remaining one letter must then go into the remaining one envelope. However, that last envelope is precisely the correct envelope for that remaining letter, because all other envelopes are already correctly filled. This leads to a contradiction with the requirement that exactly 9 letters be correct and the remaining one be wrong. Step-by-Step Solution: Consider 10 letters labelled L1 to L10 and envelopes E1 to E10, where Li is supposed to go into Ei. If exactly 9 letters are correct, then there are 9 indices i where Li is in Ei. That means 9 envelopes are already filled with their correct letters, leaving exactly one empty envelope, say Ej. The only letter not yet placed is Lj, because all other letters L1 to L10 except Lj have already been used in their correct positions. The last remaining letter Lj must go into the only remaining envelope Ej. But Ej is the correct envelope for Lj, so Lj will also be correctly placed. This leads to all 10 letters being correct, not exactly 9, which contradicts the condition. Therefore there is no possible arrangement with exactly 9 correct letters. Number of favourable arrangements = 0. Probability = favourable arrangements / total arrangements = 0 / 10! = 0. Verification / Alternative check: We can verify this logic with a smaller example. If there are 2 letters and 2 envelopes and we want exactly one letter to be in the correct envelope, it is not possible. Either both letters are in their correct envelopes, or both are swapped, giving zero cases with exactly one correct placement. This simple example mirrors the same principle that prevents exactly 9 correct matches out of 10. Why Other Options Are Wrong: 1/10 or 1/9 suggest there is at least one valid arrangement with exactly 9 correct placements, which is not true. 1 would mean every arrangement has exactly 9 correct letters, which is clearly impossible. Common Pitfalls: Many learners initially think it is possible to have exactly one letter in the wrong envelope and imagine that last letter as being misplaced. The key is to remember that when all but one envelope are already correctly filled, the final letter has no choice but to go into its matching envelope. Recognizing this logical constraint is essential in derangement based probability questions. Final Answer: The probability that exactly 9 letters are placed into their correct envelopes is 0.

Question 19

One card is drawn at random from a standard pack of 52 playing 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 focusing on two specific cards in a standard deck. Rather than dealing with a whole suit or rank, it asks for the chance that a single draw yields one of two particular cards. This highlights the idea of counting individual favourable outcomes in a uniform sample space. Given Data / Assumptions: There is a standard deck of 52 distinct cards. We are interested in two specific cards: the queen of clubs and the king of hearts. One card is drawn at random from the deck. Every card has the same probability of being drawn. Concept / Approach: When each outcome in the sample space is equally likely, probability equals favourable outcomes divided by total outcomes. Here, each individual card is one outcome. The favourable outcomes are exactly the two mentioned cards. There is no overlap because a single card cannot be both the queen of clubs and the king of hearts. Step-by-Step Solution: Total number of cards in the deck = 52. Favourable cards = queen of clubs and king of hearts, so there are 2 favourable cards. Required probability = favourable outcomes / total outcomes = 2 / 52. Simplify 2 / 52 by dividing numerator and denominator by 2 to obtain 1 / 26. Verification / Alternative check: An alternative way to see this is to compute the probability of each card separately and then add them, because the events are mutually exclusive. Probability of queen of clubs = 1 / 52. Probability of king of hearts = 1 / 52. Since both cannot occur together on a single draw, the combined probability is 1 / 52 plus 1 / 52, which equals 2 / 52 or 1 / 26. This matches the combination reasoning. Why Other Options Are Wrong: 1/52 corresponds to the probability of drawing only one specific card, not one of two specific cards. 1/13 would mean 4 favourable cards, which would be correct for any one rank such as any king, but not just one king and one queen. 1/4 suggests 13 favourable cards, which is far too many for only two particular cards. Common Pitfalls: Some learners mistakenly multiply probabilities for the two cards instead of adding them. Multiplication is used for joint occurrence, but here a single card cannot be both the queen of clubs and the king of hearts at the same time. When events are mutually exclusive and we want either one or the other, probabilities should be added, not multiplied. Final Answer: The probability of drawing either the queen of clubs or the king of hearts is 1/26.

Question 20

An unbiased cubic die with faces numbered 1 to 6 is thrown once. What is the probability of getting a number that is a multiple of 3 or a multiple of 4?

Accepted Answer

Introduction / Context: This question tests understanding of probability for outcomes on a single fair die. The event is defined using divisibility conditions. We must identify which face values are multiples of 3 or multiples of 4, then express the count of those faces as a fraction of the total possible outcomes. Given Data / Assumptions: The die is fair and has six faces numbered 1, 2, 3, 4, 5 and 6. All six outcomes are equally likely. We require a result that is a multiple of 3 or a multiple of 4. Concept / Approach: We use the basic probability formula: probability equals number of favourable outcomes divided by total outcomes. The sample space has 6 outcomes. The favourable set is the union of multiples of 3 and multiples of 4 among the numbers 1 to 6. Since the set is small, we can list and check each number directly, and then compute the ratio. Step-by-Step Solution: Possible results when the die is thrown: 1, 2, 3, 4, 5, 6. Multiples of 3 in this set: 3 and 6. Multiples of 4 in this set: 4. Union of these values: 3, 4 and 6. Number of favourable outcomes = 3. Total number of possible outcomes = 6. Required probability = favourable / total = 3 / 6 = 1 / 2. Verification / Alternative check: We can check quickly by noting that half of the numbers from 1 to 6 satisfy the required condition. There are three favourable numbers out of six. Because the die is fair, each number appears with equal chance, so a probability of 1 / 2 makes sense intuitively and matches our computed ratio of 3 / 6. Why Other Options Are Wrong: 1/12 and 1/9 are too small and would suggest far fewer favourable outcomes than actually exist. 3/4 would require 4 or 5 favourable outcomes, but we only have 3 values that meet the divisibility rule. Common Pitfalls: Errors usually occur when candidates incorrectly list the multiples, for example by including 2 or 5, or by thinking that 1 is a multiple of every integer. Another mistake is to count 12 as a possible outcome, which is not on a standard six faced die. Careful listing of all face values and checking each against the divisibility conditions prevents these mistakes. Final Answer: The probability of getting a number that is a multiple of 3 or 4 is 1/2.

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