Question 1

A problem is given independently to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. Assuming their attempts are independent, what is the probability that the problem will be solved by at least one of them?

Accepted Answer

Introduction / Context:This is a standard probability problem about independent events and the complement rule. Three students attempt the same problem independently, each with a given probability of solving it. Instead of directly calculating all cases in which the problem is solved, the more efficient approach is to calculate the probability that none of them solves it and then subtract that value from 1. This is a common technique in probability questions involving "at least one" events.Given Data / Assumptions: Student 1 solves the problem with probability 1/2. Student 2 solves the problem with probability 1/3. Student 3 solves the problem with probability 1/4. Their attempts are independent of each other. Concept / Approach:The event "the problem is solved" is the complement of the event "none of the students solves the problem". For independent events, the probability that none of them succeeds is the product of the probabilities of each student failing. Once that value is found, the required probability is 1 minus this product. Using the complement avoids having to sum multiple overlapping cases where one, two or three students succeed.Step-by-Step Solution:Let S1, S2 and S3 be the events that student 1, 2 and 3 solve the problem respectively.Given: P(S1) = 1/2, P(S2) = 1/3, P(S3) = 1/4.Therefore, probabilities that they do not solve the problem are:P(not S1) = 1 - 1/2 = 1/2.P(not S2) = 1 - 1/3 = 2/3.P(not S3) = 1 - 1/4 = 3/4.The probability that none of them solves the problem is the product of these, since the attempts are independent:P(none solves) = (1/2) * (2/3) * (3/4).Multiply: (1/2) * (2/3) = 1/3, and (1/3) * (3/4) = 1/4.So P(none solves) = 1/4.The required probability that at least one solves the problem is:P(at least one solves) = 1 - P(none solves) = 1 - 1/4 = 3/4.Verification / Alternative check:We could also expand cases where exactly one, exactly two or all three solve the problem, but that is more cumbersome.The complement method works because the only way the problem is not solved is when all three fail simultaneously.Since our final probability is between 0 and 1 and matches the intuitive idea that the combined chance should be high but not certain, it is reasonable and confirmed by the arithmetic.Why Other Options Are Wrong:A value of 1/4 corresponds to the probability that none solves it, not that at least one solves. 1/2 and 2/3 underestimate the combined success probability for three students. The option 7/12 arises from adding individual probabilities without proper inclusion exclusion and is not correct for independent events. Only 3/4 correctly represents 1 minus the probability that all three fail.Common Pitfalls:Many students mistakenly add the individual probabilities directly, writing 1/2 + 1/3 + 1/4, which can exceed 1 and does not reflect independence or overlap. Others forget to use the complement rule and attempt to compute probabilities for each combination of successes and failures, increasing the chance of counting errors. Remembering that "at least one" is best handled via 1 minus "none" is very helpful in such questions.Final Answer:The probability that the problem is solved by at least one student is 3/4.

Question 2

Two cards are drawn at random from a standard deck of 52 playing cards without replacement. What is the probability that either both cards are black or both cards are queens?

Accepted Answer

Introduction / Context:This problem tests understanding of compound events in probability and the inclusion exclusion principle. We have two types of favourable outcomes: both cards are black, or both cards are queens. These events overlap when both cards are black queens, so counting must be done carefully to avoid double counting. A standard 52 card deck with equal suits and ranks is assumed.Given Data / Assumptions: A standard deck has 52 cards: 26 black (spades and clubs) and 26 red (hearts and diamonds). There are 4 queens in total, of which 2 are black (queen of spades and queen of clubs). Two cards are drawn at random without replacement. Each unordered pair of cards is equally likely. Concept / Approach:We consider two events: A = both cards are black, B = both cards are queens. The required probability is P(A or B). For overlapping events, P(A or B) = P(A) + P(B) - P(A and B). We compute each probability using combinations: total possible pairs from 52 cards and favourable pairs for each event. Subtracting the intersection term avoids double counting pairs that satisfy both conditions at once.Step-by-Step Solution:Total number of ways to choose 2 cards from 52 = C(52,2).C(52,2) = 52 * 51 / 2 = 1326.Event A: both cards are black.Number of black cards = 26, so ways to choose 2 black cards = C(26,2) = 26 * 25 / 2 = 325.Event B: both cards are queens.Number of queens = 4, so ways to choose 2 queens = C(4,2) = 4 * 3 / 2 = 6.Intersection A and B: both cards are black queens.Number of black queens = 2, so ways to choose both = C(2,2) = 1.Total favourable pairs = 325 + 6 - 1 = 330.Required probability = 330 / 1326.Simplify by dividing numerator and denominator by 6: 330 / 6 = 55 and 1326 / 6 = 221.Thus probability = 55/221.Verification / Alternative check:We can cross check by computing P(A) and P(B) separately and subtracting P(A and B).P(A) = 325 / 1326, P(B) = 6 / 1326, P(A and B) = 1 / 1326.Then P(A or B) = P(A) + P(B) - P(A and B) = (325 + 6 - 1) / 1326 = 330 / 1326 = 55/221.This agrees with the direct favourable counting result.Why Other Options Are Wrong:52/221 and 55/190 are incorrect simplifications and do not correspond to the correct counting. 19/221 and 5/52 underestimate the probability, likely due to missing some favourable cases or incorrect handling of the overlap. Only 55/221 matches the careful inclusion exclusion calculation.Common Pitfalls:A common mistake is to add the counts for both black and both queens without subtracting the intersection, thereby double counting the pair of black queens. Some students also confuse ordered pairs with unordered combinations, leading to incorrect denominators. Remember that in card problems, combinations are usually used because order does not matter when drawing multiple cards at once.Final Answer:The probability that either both cards are black or both are queens is 55/221.

Question 3

A man and his wife apply independently for two vacancies on the same post. The probability that the man is selected is 1/7, and the probability that the wife is selected is 1/5. What is the probability that exactly one of them is selected?

Accepted Answer

Introduction / Context:This question is about independent events and the probability that exactly one of two candidates is selected. It appears frequently in aptitude tests and recruitment exam practice, where candidates must understand how to combine probabilities of mutually exclusive scenarios and use complement events correctly. The key is to recognize that "exactly one" means one selected and the other rejected, which can happen in two distinct ways.Given Data / Assumptions: Probability that the man is selected = 1/7. Probability that the wife is selected = 1/5. Their selections are assumed independent. We are interested in the probability that exactly one of them is selected. Concept / Approach:There are two mutually exclusive ways for exactly one of them to be selected: (1) the man is selected and the wife is not; (2) the wife is selected and the man is not. Because the events are independent, the probability of a specific combination is the product of the individual selection or rejection probabilities. The total probability for exactly one selection is the sum of the probabilities of these two mutually exclusive cases.Step-by-Step Solution:Let M be the event that the man is selected and W be the event that the wife is selected.Given: P(M) = 1/7, so P(not M) = 1 - 1/7 = 6/7.Given: P(W) = 1/5, so P(not W) = 1 - 1/5 = 4/5.Exactly one of them is selected means either (M and not W) or (not M and W).Because the events are independent, we can multiply probabilities within each case.P(M and not W) = P(M) * P(not W) = (1/7) * (4/5) = 4/35.P(not M and W) = P(not M) * P(W) = (6/7) * (1/5) = 6/35.Total probability for exactly one selection = 4/35 + 6/35 = 10/35.Simplify 10/35 by dividing numerator and denominator by 5 to get 2/7.Verification / Alternative check:We can check consistency by also considering all possible outcomes.Possible outcomes: both selected, neither selected, exactly one selected.If we compute these separately and sum them, the total must be 1.However, for the purposes of this problem we only need the exactly one case, and the calculation gives 2/7, which is between 0 and 1 and seems reasonable given the relatively small individual probabilities.Why Other Options Are Wrong:1/7 corresponds to one portion of the exact one case, but omits the scenario where the wife is selected and the man is not. 3/4 and 4/5 are too large to be plausible given the small individual selection probabilities. 1/5 is simply the wife's selection probability and does not reflect the condition about the man's non selection.Common Pitfalls:Students sometimes confuse "exactly one" with "at least one" and compute 1 minus the probability that neither is selected. That approach would produce a different value. Another common mistake is forgetting that both ways of having exactly one selected must be counted. Always list the mutually exclusive cases explicitly when dealing with "exactly one" type problems.Final Answer:The probability that exactly one of them is selected is 2/7.

Question 4

In a lottery, there are 10 prize tickets and 25 blank tickets. One ticket is drawn at random. What is the probability that the ticket drawn is a prize ticket?

Accepted Answer

Introduction / Context:This is a simple probability problem dealing with a finite set of equally likely outcomes. The lottery contains a certain number of prize tickets and a certain number of blank tickets. The question asks for the probability of drawing a prize ticket in a single random draw. It tests basic understanding of probability as the ratio of favourable outcomes to total possible outcomes.Given Data / Assumptions: Number of prize tickets = 10. Number of blank tickets = 25. Total tickets in the lottery = prize tickets + blank tickets. One ticket is drawn at random, and each ticket is equally likely to be chosen. Concept / Approach:The theoretical probability of an event is defined as the number of favourable outcomes divided by the total number of equally likely outcomes. Here, the favourable outcomes are the prize tickets. The total outcomes are all tickets, both prize and blank. Therefore, the required probability is simply the ratio 10 divided by the total number of tickets. Simplifying this fraction gives the final answer.Step-by-Step Solution:Total number of tickets = number of prize tickets + number of blank tickets.Total tickets = 10 + 25 = 35.Number of favourable tickets (prize tickets) = 10.Required probability = favourable / total = 10 / 35.Simplify 10/35 by dividing numerator and denominator by 5.10/35 = 2/7.Thus, the probability of drawing a prize ticket is 2/7.Verification / Alternative check:As a quick check, note that 2/7 is less than 1/2, which makes sense because prizes are fewer than blanks (10 versus 25).Also, the probability of drawing a blank is 25/35 = 5/7.2/7 + 5/7 = 1, which confirms that the probabilities of all possible outcomes sum to 1.Why Other Options Are Wrong:5/7 is the probability of drawing a blank ticket, not a prize. 1/5 and 1/2 do not reflect the actual ratio 10/35. 3/7 would imply 15 prize tickets, which is not the case here. Only 2/7 correctly represents 10 favourable outcomes out of 35 total tickets.Common Pitfalls:Students sometimes mistakenly use only the number of blanks or only the number of prizes as the denominator. It is important to always include all possible outcomes in the denominator. Another error is failing to simplify the fraction; while 10/35 is correct, exams usually expect the simplest form 2/7. Always check if the fraction can be reduced.Final Answer:The probability that the ticket drawn is a prize ticket is 2/7.

Question 5

Tickets numbered from 1 to 20 are mixed thoroughly and then one ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or a multiple of 5?

Accepted Answer

Introduction / Context: This question tests basic probability with simple counting and the idea of numbers that are multiples of 3 or 5. It also illustrates how to handle overlap when a number satisfies both conditions at the same time, using the addition principle for events that are not mutually exclusive. Given Data / Assumptions: Tickets are numbered from 1 to 20. Exactly one ticket is drawn at random. Each ticket is equally likely to be drawn. We are interested in numbers that are multiples of 3 or multiples of 5. Concept / Approach: We use classical probability: P(event) = favourable outcomes / total outcomes. Here the event is that the ticket number is a multiple of 3 or 5. We count multiples of 3, multiples of 5, and then subtract the overlap (multiples of both 3 and 5, that is multiples of 15) to avoid double counting. Step-by-Step Solution: Total tickets = 20, so total outcomes = 20. Multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18 (6 numbers). Multiples of 5 between 1 and 20: 5, 10, 15, 20 (4 numbers). Number 15 is common to both sets, so it is counted twice above. Favourable outcomes = 6 + 4 - 1 = 9 (inclusion exclusion principle). Probability = favourable / total = 9 / 20. Verification / Alternative check: We can also list all numbers from 1 to 20 and highlight those divisible by 3 or 5. Counting them directly again gives 9 numbers. Since 9 and 20 have no common factor other than 1, the fraction 9/20 is already in simplest form, which confirms our result. Why Other Options Are Wrong: 1/2: This would require 10 favourable numbers, but there are only 9. 3/5: This equals 12/20, which is larger than 9/20 and overestimates the probability. 2/5: This equals 8/20, which is slightly less than the correct count of 9 favourable numbers. Common Pitfalls: Students often forget to subtract the overlap when using the idea of multiples of 3 or 5. Counting 6 multiples of 3 and 4 multiples of 5 and simply adding them gives 10, which is wrong because 15 is then counted twice. Always check for numbers that satisfy both conditions when using the addition principle for probability problems. Final Answer: Therefore, the required probability that the drawn ticket shows a number that is a multiple of 3 or 5 is 9/20.

Question 6

Two fair dice are tossed simultaneously. What is the probability that the total score shown on the two upper faces is a prime number?

Accepted Answer

Introduction / Context: This question checks familiarity with basic probability for two dice and understanding of prime numbers. The key idea is to list all possible sums when two dice are thrown and then count how many of those sums are prime numbers, while remembering that some sums occur more frequently than others. Given Data / Assumptions: Two standard fair six sided dice are thrown at the same time. Each die shows an integer from 1 to 6. All 36 ordered pairs of outcomes are equally likely. Prime numbers to consider among possible sums are 2, 3, 5, 7 and 11. Concept / Approach: For two dice, there are 6 * 6 = 36 total possible ordered outcomes. For each possible sum, we count how many pairs (first die, second die) produce that sum. The probability that the sum is prime is the number of favourable outcome pairs divided by 36. We do not treat all sums as equally likely; we treat all ordered pairs as equally likely. Step-by-Step Solution: Total number of outcomes = 6 * 6 = 36. Sum = 2: only (1, 1) so 1 way. Sum = 3: (1, 2), (2, 1) so 2 ways. Sum = 5: (1, 4), (2, 3), (3, 2), (4, 1) so 4 ways. Sum = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) so 6 ways. Sum = 11: (5, 6), (6, 5) so 2 ways. Total favourable outcomes = 1 + 2 + 4 + 6 + 2 = 15. Probability = 15 / 36 = 5 / 12 after simplification. Verification / Alternative check: We can verify by writing the full 6 by 6 table of sums, circling those that are 2, 3, 5, 7 or 11 and counting circles. This again produces 15 favourable pairs out of 36. Dividing numerator and denominator of 15/36 by 3 gives 5/12, confirming the final fraction is simplified. Why Other Options Are Wrong: 1/6: This is too small and would correspond to only 6 favourable cases. 1/2: This would require 18 favourable outcomes, which is more than the actual 15. 7/9: This is greater than 3/4 and suggests 28 favourable outcomes, which is impossible here. Common Pitfalls: Learners sometimes count prime sums but then mistakenly divide by 11 possible sums instead of 36 outcome pairs. The sample space in dice problems is the set of ordered pairs, not merely the set of distinct sums, which have different frequencies. Forgetting that, leads to incorrect probabilities. Final Answer: Hence, the probability that the total score is a prime number is 5/12.

Question 7

A leap year is selected at random. What is the probability that this leap year will contain exactly 53 Mondays?

Accepted Answer

Introduction / Context: This question connects basic probability with calendar facts. It tests whether you know how many days are in a leap year and how weekdays repeat. The focus is on working out how often a specific day, Monday, occurs in a leap year and expressing that as a probability when the starting weekday is not fixed. Given Data / Assumptions: A leap year has 366 days. 366 days is equal to 52 full weeks plus 2 extra days. Each day of the week is equally likely to be the first day of the leap year. We want the leap year to have exactly 53 Mondays. Concept / Approach: In any leap year, every day of the week occurs at least 52 times because 52 weeks cover 364 days. The remaining 2 days fall on two consecutive weekdays. Therefore, exactly two weekdays will occur 53 times. We need to count how many choices for the first day of the year make Monday one of the two weekdays that occur 53 times, and then divide by the total possible starting weekdays (7). Step-by-Step Solution: 366 days = 52 weeks (364 days) + 2 extra days. Every weekday appears at least 52 times. The extra 2 days are consecutive weekdays starting from the weekday of 1 January. If 1 January is Monday, then Monday and Tuesday occur 53 times. If 1 January is Sunday, then Sunday and Monday occur 53 times. So Monday appears 53 times if 1 January is Monday or Sunday (2 cases). Total possible starting weekdays = 7. Required probability = 2 / 7. Verification / Alternative check: We can systematically consider all seven possible starting weekdays: Sunday to Saturday. For each, list the two weekdays that occur 53 times (starting day and next day). Only when the starting day is Sunday or Monday does Monday appear among these two. That again gives 2 favourable cases out of 7 possibilities. Why Other Options Are Wrong: 1/7: This would mean only one starting weekday leads to 53 Mondays, which is not correct. 3/7: This implies three possible starting days, but we found only two. 1: This would mean every leap year has 53 Mondays, which is clearly false. Common Pitfalls: Some learners divide by 366 directly or guess without using the week structure. Others forget that in a leap year there are exactly two weekdays with 53 occurrences and the rest have 52. The right way is to track the starting weekday and see how the additional two days distribute. Final Answer: Therefore, the probability that a randomly chosen leap year has exactly 53 Mondays is 2/7.

Question 8

A bag contains 4 white balls, 5 red balls and 6 blue balls. Three balls are drawn at random without replacement. What is the probability that all the three balls drawn are red?

Accepted Answer

Introduction / Context: This problem is a classic example of probability without replacement and uses combinations. It asks for the probability that all three chosen balls are red when drawing from a bag containing balls of three different colours. The focus is on counting combinations correctly and understanding that the order of drawing does not matter in this context. Given Data / Assumptions: White balls = 4. Red balls = 5. Blue balls = 6. Total balls = 4 + 5 + 6 = 15. Three balls are drawn at random without replacement. Event of interest: all three drawn balls are red. Concept / Approach: Since order does not matter, we use combinations. The total number of possible selections of 3 balls is C(15, 3). The number of favourable selections, where every ball is red, is C(5, 3), because we must choose all three from the 5 red balls. The required probability is then given by favourable combinations divided by total combinations. Step-by-Step Solution: Total balls = 15, so total combinations = C(15, 3). C(15, 3) = 15 * 14 * 13 / (3 * 2 * 1) = 455. Red balls = 5, so favourable combinations = C(5, 3). C(5, 3) = 5 * 4 * 3 / (3 * 2 * 1) = 10. Required probability = 10 / 455. Simplify 10 / 455 by dividing numerator and denominator by 5: 10 / 455 = 2 / 91. Verification / Alternative check: We can also compute stepwise probability without combinations. First draw red: 5/15. Second red: 4/14. Third red: 3/13. Multiply: (5/15) * (4/14) * (3/13) = (5 * 4 * 3) / (15 * 14 * 13) = 60 / 2730 = 2 / 91 after simplification. This matches the combination method answer. Why Other Options Are Wrong: 1/22 and 3/22 are larger than 2/91 and do not match the correct counting. 2/77 corresponds to 2 * 455 / (77 * 455), which does not arise from any correct favourable count. Common Pitfalls: Students sometimes mix permutations and combinations, or forget that the total number of balls is 15, not 5. Another error is computing probabilities additively rather than multiplicatively for successive draws. Always check whether order matters and whether replacement occurs. Final Answer: So, the probability that all three drawn balls are red is 2/91.

Question 9

Two fair dice are thrown together. What is the probability that the sum of the numbers appearing on the two faces is divisible by 4 or divisible by 6?

Accepted Answer

Introduction / Context: This question tests understanding of probability on two dice where the event is expressed in terms of the sum being divisible by given numbers. It requires identifying all possible sums of two dice and determining for which of those sums divisibility by 4 or 6 holds, including handling overlap correctly. Given Data / Assumptions: Two fair six sided dice are thrown. Each ordered pair of outcomes (1 to 6 on each die) is equally likely. We are interested in sums divisible by 4 or by 6. Total number of possible outcomes is 6 * 6 = 36. Concept / Approach: First list all possible sums of two dice, which range from 2 to 12. Identify sums divisible by 4 or 6. For each such sum, count the number of ordered pairs giving that sum. Because we work at the level of pairs, some sums have more combinations than others. The desired probability is the number of favourable pairs divided by 36. Step-by-Step Solution: Possible sums: 2 to 12. Sums divisible by 4: 4, 8, 12. Sums divisible by 6: 6, 12. List combinations: Sum 4: (1, 3), (2, 2), (3, 1) so 3 ways. Sum 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) so 5 ways. Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) so 5 ways. Sum 12: (6, 6) so 1 way. Total favourable outcomes = 3 + 5 + 5 + 1 = 14. Probability = 14 / 36 = 7 / 18 after simplification. Verification / Alternative check: Instead of listing pairs, we can think in terms of the frequency table of sums. The counts 3, 5, 5, and 1 for sums 4, 6, 8, and 12 are standard values for two dice and sum to 14. Dividing 14 by 36 again gives 7/18, confirming the answer. Why Other Options Are Wrong: 14/35 reduces to 2/5, which would correspond to 14 or 15 favourable out of 35, not out of 36. 8/18 simplifies to 4/9 and corresponds to 16 favourable outcomes, which is not the case. 7/35 is far too small and would match 7 favourable outcomes, which is incorrect. Common Pitfalls: Common mistakes include treating all sums from 2 to 12 as equally likely, ignoring their different frequencies, or double counting sums like 12 which satisfy both divisibility conditions. The safest method is always to count ordered pairs when working with two dice problems. Final Answer: Thus, the probability that the sum is divisible by 4 or by 6 is 7/18.

Question 10

One letter is chosen at random from the word ASSISTANT and another letter is chosen at random from the word STATISTICS. What is the probability that the two chosen letters are the same?

Accepted Answer

Introduction / Context: This question tests probability with unequal letter frequencies in two different words. The idea is to select one letter from each word and find the probability that both letters match. Because letters occur with different frequencies, we must account for these frequencies when calculating probabilities. Given Data / Assumptions: Word 1: ASSISTANT (9 letters). Word 2: STATISTICS (10 letters). One letter is chosen uniformly at random from each word. We want the letters to be identical. Concept / Approach: The probability that both letters are the same is the sum over all letters that appear in both words of [probability of choosing that letter from ASSISTANT] * [probability of choosing that letter from STATISTICS]. We must first count how many times each relevant letter occurs in each word, then use these counts over 9 and 10 respectively. Step-by-Step Solution: In ASSISTANT: A = 2, S = 3, I = 1, T = 2, N = 1, total = 9. In STATISTICS: S = 3, T = 3, A = 1, I = 2, C = 1, total = 10. Relevant common letters: A, S, I, T. P(both A) = (2/9) * (1/10) = 2/90. P(both S) = (3/9) * (3/10) = 9/90. P(both I) = (1/9) * (2/10) = 2/90. P(both T) = (2/9) * (3/10) = 6/90. Total probability = (2 + 9 + 2 + 6) / 90 = 19 / 90. Verification / Alternative check: We can compute each letter contribution with fractions directly: 2/9 * 1/10, 3/9 * 3/10, 1/9 * 2/10, 2/9 * 3/10. Adding them with a common denominator 90 again gives 19/90. Because the denominator already reflects independent choices, there is no further simplification needed beyond 19/90. Why Other Options Are Wrong: 35/96 and 19/96 use an incorrect denominator and do not match the detailed count. None of these is incorrect because we have found an exact match 19/90 in the options. Common Pitfalls: A common error is to assume each distinct letter has equal probability, rather than weighting by frequency. Another mistake is to add probabilities of selecting letters without multiplying for the two independent choices. Always multiply probabilities when two selections are independent and then sum over possible matching letters. Final Answer: Hence, the probability that the two selected letters are the same is 19/90.

Question 11

In a class there are 15 boys and 10 girls. Three students are selected at random. What is the probability that the selected group consists of exactly 1 girl and 2 boys?

Accepted Answer

Introduction / Context: This question involves combinations and mixed groups from two categories, boys and girls. It asks for the probability that a randomly chosen group of three students has exactly one girl and two boys, so we must count favourable and total combinations carefully and then form the probability fraction. Given Data / Assumptions: Boys = 15. Girls = 10. Total students = 25. We select 3 students at random without replacement. Event of interest: exactly 1 girl and 2 boys. Concept / Approach: Order does not matter in forming a group, so we use combinations. The total number of possible groups of 3 from 25 students is C(25, 3). The favourable groups must have 1 girl chosen from 10 and 2 boys chosen from 15. The probability is favourable combinations divided by total combinations. Step-by-Step Solution: Total groups of 3 from 25 = C(25, 3). C(25, 3) = 25 * 24 * 23 / (3 * 2 * 1) = 2300. Ways to choose 1 girl from 10 = C(10, 1) = 10. Ways to choose 2 boys from 15 = C(15, 2). C(15, 2) = 15 * 14 / 2 = 105. Favourable groups = 10 * 105 = 1050. Probability = 1050 / 2300. Simplify by dividing numerator and denominator by 50: 1050 / 2300 = 21 / 46. Verification / Alternative check: We can also use a stepwise probability approach. First pick a girl then two boys or in any order, but account for permutations. However, summing over all orders will return the same result once we simplify, which acts as a check on the combination method. The combination method is simpler and less error prone here. Why Other Options Are Wrong: 1/5 and 3/25 are much smaller than the correct fraction and do not correspond to any natural counting in this context. 1/50 is far too small and would suggest very few favourable groups, which contradicts the large number of ways to choose 2 boys and 1 girl. Common Pitfalls: Learners sometimes confuse exactly 1 girl with at least 1 girl and accidentally include cases of 2 or 3 girls. Another mistake is to use permutations instead of combinations or to forget to multiply the choices for boys and girls. Keeping track of the structure "choose girls then boys" helps avoid these errors. Final Answer: Therefore, the probability of selecting exactly 1 girl and 2 boys is 21/46.

Question 12

A basket contains 10 apples and 20 oranges. Of these, 3 apples and 5 oranges are defective. If two fruits are chosen at random without replacement, what is the probability that either both fruits are oranges or both fruits are non defective?

Accepted Answer

Introduction / Context: This question combines classification into defective and non defective fruits with the probability of drawing fruits of specific types. We are asked to find the probability of the union of two events: both fruits are oranges, or both fruits are non defective. Because these events overlap, we must use the addition rule with subtraction of the intersection. Given Data / Assumptions: Apples = 10, of which defective apples = 3, so non defective apples = 7. Oranges = 20, of which defective oranges = 5, so non defective oranges = 15. Total fruits = 10 + 20 = 30. Two fruits are drawn without replacement. Event A: both fruits are oranges. Event B: both fruits are non defective. Concept / Approach: We use combinations to count pairs of fruits. The total number of ways to choose 2 fruits from 30 is C(30, 2). We compute P(A), P(B), and P(A and B). The required probability P(A or B) equals P(A) + P(B) minus P(A and B). Intersection A and B means both fruits are non defective oranges. Step-by-Step Solution: Total pairs = C(30, 2) = 30 * 29 / 2 = 435. Event A: both oranges. Oranges = 20, so C(20, 2) = 20 * 19 / 2 = 190. Event B: both non defective. Non defective fruits = 7 + 15 = 22, so C(22, 2) = 22 * 21 / 2 = 231. A and B: both non defective oranges. Non defective oranges = 15, so C(15, 2) = 15 * 14 / 2 = 105. Favourable pairs = 190 + 231 - 105 = 316. Required probability = 316 / 435. Verification / Alternative check: We could also calculate probabilities directly: P(A) = (20/30) * (19/29), P(B) = (22/30) * (21/29), and P(A and B) = (15/30) * (14/29). Expanding and simplifying with a common denominator 30 * 29 gives the same result 316/435 when combined as P(A) + P(B) minus P(A and B). This confirms the combination method. Why Other Options Are Wrong: 136/345 and 158/435 are smaller, corresponding to fewer favourable outcomes and do not match the correct counting. 17/87 is not based on any natural pair count from 435 and does not simplify to 316/435. Common Pitfalls: Students often forget to subtract the intersection when using P(A or B) and end up double counting cases where both fruits are non defective oranges. Another mistake is to treat the two events as mutually exclusive, which they are not. Careful classification avoids such errors. Final Answer: Thus, the required probability that both fruits are oranges or both are non defective is 316/435.

Question 13

Two cards are drawn together at random from a standard pack of 52 playing cards. What is the probability that exactly one card is a spade and the other card is a heart?

Accepted Answer

Introduction / Context: This question is about selecting two cards from a full deck and asking for a specific combination of suits: one spade and one heart. It tests knowledge of combinations and understanding that we must count all ways to get one card from each suit, without caring about the order in which the cards are drawn. Given Data / Assumptions: A standard deck has 52 cards. There are 13 spades and 13 hearts in the deck. Two cards are drawn simultaneously, that is, without order and without replacement. Event of interest: one card is a spade and the other is a heart. Concept / Approach: The total number of ways to choose 2 cards from 52 is C(52, 2). To have one spade and one heart in the same pair, we must choose one card out of 13 spades and one card out of 13 hearts, so favourable outcomes are 13 * 13. The required probability is the ratio of favourable pairs to total pairs. Step-by-Step Solution: Total number of unordered pairs = C(52, 2) = 52 * 51 / 2 = 1326. Number of spades = 13, number of hearts = 13. Favourable ways to choose one spade and one heart = 13 * 13 = 169. Required probability = 169 / 1326. Simplify 169 / 1326 by dividing numerator and denominator by 13: 169 / 1326 = 13 / 102. Verification / Alternative check: We can also consider ordered draws: pick any card first and then the second, and then divide by total ordered outcomes. For example, P(spade then heart) = (13/52) * (13/51), and P(heart then spade) = (13/52) * (13/51). Adding gives 2 * 13 * 13 / (52 * 51) which simplifies to 13/102, matching the combination method. Why Other Options Are Wrong: 3/20, 47/100 and 29/34 do not simplify to 13/102 and arise from incorrect favourable counts or denominators. 29/34 is extremely large and would imply that nearly every pair is one spade and one heart, which is clearly not true. Common Pitfalls: Some learners incorrectly treat suits as equally likely for each card and then multiply probabilities for independent suits without adjusting for the second draw. Others forget that the denominator must be based on 52 choose 2, not 52 squared, when working with unordered pairs. Clearly identifying favourable combinations greatly reduces such errors. Final Answer: Therefore, the probability that exactly one card is a spade and the other is a heart is 13/102.

Question 14

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

Accepted Answer

Introduction / Context: This is a straightforward single draw probability question, testing understanding of the basic formula for probability in a simple situation. The aim is to compute the probability of drawing a white ball from a bag that contains both black and white balls. Given Data / Assumptions: Black balls = 6. White balls = 8. Total balls in the bag = 6 + 8 = 14. One ball is drawn at random and every ball has an equal chance of being drawn. Event of interest: the drawn ball is white. Concept / Approach: In the classical definition, probability of an event equals number of favourable outcomes divided by total number of possible outcomes, assuming all outcomes are equally likely. Here, favourable outcomes are white balls and total outcomes are all balls in the bag. Step-by-Step Solution: Total balls = 6 + 8 = 14. Number of favourable outcomes (white balls) = 8. Probability of drawing a white ball = favourable / total = 8 / 14. Simplify 8 / 14 by dividing numerator and denominator by 2: 8 / 14 = 4 / 7. Verification / Alternative check: We can also compute the probability of drawing a black ball and then use complement. Probability of black = 6/14 = 3/7. Therefore probability of white = 1 - 3/7 = 4/7. This agrees with the direct counting method and confirms the result. Why Other Options Are Wrong: 3/7 is the probability of drawing a black ball, not a white ball. 1/8 would be correct only if there were 8 balls total with 1 white, which is not the case. 3/4 would require 3 out of every 4 balls being white, that is 10.5 white balls out of 14, which is impossible. Common Pitfalls: The most common error is miscounting the total number of balls or mixing up the counts of black and white. Another minor mistake is forgetting to simplify the final fraction. Always identify clearly what outcomes are favourable and then divide by the total number of outcomes. Final Answer: Hence, the probability that the drawn ball is white is 4/7.

Question 15

Three unbiased (fair) coins are tossed simultaneously. What is the probability of getting at least two heads?

Accepted Answer

Introduction / Context: This question uses simple probability of coin tosses. It asks for the probability of getting at least two heads when three fair coins are tossed together. The phrase at least two heads means either exactly two heads or exactly three heads, so we sum probabilities of these outcomes. Given Data / Assumptions: Three unbiased coins are tossed. Each coin has two outcomes: head (H) or tail (T). All 2^3 = 8 possible outcomes of three coin tosses are equally likely. Event of interest: at least two heads (that is, 2 or 3 heads). Concept / Approach: We can explicitly list all sample space elements for three coin tosses and count the favourable outcomes. Alternatively, we can count outcomes with 0 or 1 head and subtract from 1. Both methods should give the same probability. Here we will list all outcomes directly for clarity. Step-by-Step Solution: Total outcomes for three coins = 2^3 = 8. List them: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Outcomes with at least two heads: HHH, HHT, HTH, THH. Number of favourable outcomes = 4. Probability = favourable / total = 4 / 8 = 1 / 2. Verification / Alternative check: We can use the complement method. P(0 heads) = P(TTT) = 1/8. Outcomes with exactly one head: HTT, THT, TTH, so P(1 head) = 3/8. Hence P(at least two heads) = 1 - (1/8 + 3/8) = 1 - 4/8 = 4/8 = 1/2. This matches the direct counting method. Why Other Options Are Wrong: 1/4 would correspond to only 2 favourable outcomes, but there are 4. 3/4 would require 6 favourable outcomes, which is not the case here. 1/3 does not match any natural partition of the 8 outcomes. Common Pitfalls: Learners sometimes misinterpret at least two heads as exactly two heads and forget HHH. Others may forget to list all outcomes or miscount them. Systematic listing or using combinations (C(3,2) for exactly two heads and C(3,3) for three heads) helps avoid mistakes. Final Answer: Therefore, the probability of getting at least two heads is 1/2.

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 a face card (that is, a Jack, Queen or King)?

Accepted Answer

Introduction / Context: This is a basic card probability question that focuses on counting specific types of cards, called face cards. It tests your knowledge of the composition of a standard 52 card deck and the ability to express the chance of drawing one of these special cards as a fraction. Given Data / Assumptions: Standard deck size = 52 cards. Four suits: hearts, diamonds, clubs and spades. Each suit has 3 face cards: Jack, Queen and King. Total face cards in the deck = 4 * 3 = 12. One card is drawn at random. Concept / Approach: Since all 52 cards are equally likely, the probability of drawing a face card is the number of face cards divided by the total number of cards. We simply count the favourable cards and use the classical probability formula P(event) = favourable outcomes / total outcomes. Step-by-Step Solution: Number of face cards per suit = 3 (Jack, Queen, King). Number of suits = 4. Total face cards = 3 * 4 = 12. Total cards in deck = 52. Probability of drawing a face card = 12 / 52. Simplify 12 / 52 by dividing numerator and denominator by 4: 12 / 52 = 3 / 13. Verification / Alternative check: We can also think in terms of suits. The probability that the card is a face card from hearts is 3/52, and similarly 3/52 for each of the other three suits. Summing these four equal probabilities gives 4 * 3/52 = 12/52 = 3/13, which confirms the result. Why Other Options Are Wrong: 1/13 corresponds to only 4 favourable cards, not the 12 face cards. 3/52 equals the probability that the card is a face card of a specific suit only, such as face card of hearts. 9/52 would imply 9 face cards total, which is not correct for a standard deck. Common Pitfalls: Sometimes learners confuse face cards with all picture cards including Ace, but Ace is not considered a face card in most probability questions unless explicitly stated. Counting all ranks from Jack to Ace would lead to an incorrect favourable count. Final Answer: Thus, the probability that the drawn card is a face card is 3/13.

Question 17

Eight married couples, that is 8 husbands and their 8 wives, attend a dance show called Nach Baliye. In a lucky draw, four persons are selected at random for a prize. What is the probability that at least one complete married couple is included amon…

Accepted Answer

Introduction / Context: This problem deals with selecting a group from married couples and asks for the probability that the chosen group of four contains at least one complete couple. It is often easier to solve such problems using the complement approach, first calculating the probability that no couple appears together and then subtracting this from 1. Given Data / Assumptions: There are 8 married couples, so total persons = 16. Four persons are selected at random without restriction. Each group of four persons is equally likely. We want at least one couple in the group, meaning one or more husband wife pairs. Concept / Approach: Let Event A be the event that at least one couple is selected. Instead of counting all groups that have one or more couples directly, we compute the probability of the complement event, B: no couple is selected. Then P(A) = 1 minus P(B). To enforce no couple, we ensure that at most one person from each couple is selected. Step-by-Step Solution: Total number of 4 person groups from 16 people = C(16, 4) = 1820. To count groups with no couple, first choose which 4 couples contribute one member: C(8, 4). C(8, 4) = 70. From each chosen couple, we can choose either the husband or the wife, so 2 choices per couple. Number of groups with no couple = 70 * 2^4 = 70 * 16 = 1120. Therefore, P(no couple) = 1120 / 1820. P(at least one couple) = 1 - 1120 / 1820 = (1820 - 1120) / 1820 = 700 / 1820. Simplify 700 / 1820: divide numerator and denominator by 140 to get 5 / 13, which equals 15 / 39. Verification / Alternative check: Instead of simplifying directly to 5/13, we notice that option 15/39 reduces to 5/13. The other given fractions, when simplified, do not match 5/13. This consistency check confirms that 15/39 represents the correct probability in the given options. Why Other Options Are Wrong: 8/39 is smaller and does not come from any natural count related to 700 favourable groups. 12/13 is very large and would mean almost every group contains a couple, which is not true. None of these is incorrect because there is a matching fraction 15/39 that equals 5/13. Common Pitfalls: A frequent mistake is to try to count groups with at least one couple directly, which involves overlapping cases with one, two or more couples and is complicated. Another error is to miscount the no couple cases by forgetting the factor 2^4 for selecting one person from each chosen couple. Using the complement method with clear combinatorial reasoning keeps the calculation manageable. Final Answer: Thus, the probability that at least one married couple is included among the four selected persons is 15/39, which simplifies to 5/13.

Question 18

Two letters are chosen at random without replacement from the word HOME. What is the probability that none of the chosen letters is a vowel?

Accepted Answer

Introduction / Context: This question involves probability with letters from a short word. The aim is to draw two letters such that neither is a vowel. It is a simple combination problem since the order of drawing letters is not important, and the letters are drawn without replacement. Given Data / Assumptions: The word HOME has 4 letters: H, O, M and E. Vowels in HOME: O and E (2 vowels). Consonants in HOME: H and M (2 consonants). Two letters are chosen at random without replacement. Event of interest: both chosen letters are consonants. Concept / Approach: Order does not matter when simply choosing letters, so we use combinations. The probability of selecting two consonants is equal to the number of ways to choose 2 consonants divided by the total number of ways to choose any 2 letters from the 4 available letters. Step-by-Step Solution: Total number of letters = 4. Total ways to choose 2 letters = C(4, 2) = 4 * 3 / 2 = 6. Number of consonants = 2 (H and M). Ways to choose 2 consonants from 2 = C(2, 2) = 1. Probability that both letters are consonants = favourable / total = 1 / 6. Verification / Alternative check: We can list all possible pairs of letters: HO, HM, HE, OM, OE, ME. Only HM consists of two consonants. There are 6 pairs total and 1 favourable pair, so the probability is 1/6, confirming the combination approach. Why Other Options Are Wrong: 1/2, 1/3 and 1/4 would imply 3, 2, or 1.5 favourable outcomes, which do not match the actual single favourable pair. Thus any fraction other than 1/6 misrepresents the counting of consonant pairs. Common Pitfalls: A common mistake is to forget that HOME has two vowels and two consonants and to miscount consonants or vowels. Another error is to think order matters and use permutations, which leads to an incorrect denominator. Since the question does not care about order, combinations are the correct tool. Final Answer: Hence, the probability that neither of the chosen letters is a vowel is 1/6.

Question 19

In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both English and Hindi. If a student is selected at random, what is the probability that the student has offered English or Hindi (or both)?

Accepted Answer

Introduction / Context: This question tests the use of set theory within probability, specifically the inclusion exclusion principle for two sets, English and Hindi. We are given percentage data for students offering English, Hindi and both, and asked to find the probability that a randomly chosen student has offered at least one of the two subjects. Given Data / Assumptions: P(English) = 30% = 0.30. P(Hindi) = 20% = 0.20. P(English and Hindi) = 10% = 0.10. We assume percentages are based on the same total number of students. We want P(English or Hindi). Concept / Approach: For two events E and H, the probability of E or H is given by P(E or H) = P(E) + P(H) minus P(E and H). The minus term removes the double counting of students who offered both subjects. After computing the probability as a decimal, we convert it into a simplified fraction for the final answer. Step-by-Step Solution: P(English) = 0.30. P(Hindi) = 0.20. P(English and Hindi) = 0.10. P(English or Hindi) = P(English) + P(Hindi) - P(English and Hindi). P(English or Hindi) = 0.30 + 0.20 - 0.10 = 0.40. 0.40 as a fraction is 40 / 100 = 2 / 5. Verification / Alternative check: Imagine a class of 100 students. Then 30 offered English, 20 offered Hindi and 10 are counted in both subject groups. So total students who offered at least one of the two subjects = 30 + 20 - 10 = 40. The probability is therefore 40/100 = 2/5, consistent with the earlier calculation. Why Other Options Are Wrong: 1/2 represents 50%, which would require 50 students in 100 offering at least one subject, not 40. 3/4 (75%) and 4/5 (80%) are much larger than the computed 40%, so they are clearly incorrect. Common Pitfalls: A frequent mistake is to simply add 30% and 20% and conclude 50% without subtracting the 10% who are counted twice. Forgetting the overlap overestimates the probability. Always apply the inclusion exclusion formula for unions of sets. Final Answer: Thus, the probability that a randomly chosen student has offered English or Hindi is 2/5.

Question 20

A bag contains 2 red roses, 4 yellow roses and 6 pink roses. Two roses are drawn at random without replacement. What is the probability that the two roses drawn are not of the same colour?

Accepted Answer

Introduction / Context: This question is about drawing two items without replacement from a collection with three different colours. Instead of asking for matching colours, it asks for the probability that the two drawn roses are not of the same colour. This can be solved directly or via the complement approach using probability of same colour first. Given Data / Assumptions: Red roses = 2. Yellow roses = 4. Pink roses = 6. Total roses = 2 + 4 + 6 = 12. Two roses are drawn at random without replacement. Event of interest: the two roses are not of the same colour. Concept / Approach: It is usually easier to first find the probability that both roses are of the same colour, then subtract this from 1. Same colour means both red, both yellow or both pink. We compute this using combinations, then use complement: P(not same colour) = 1 minus P(same colour). Step-by-Step Solution: Total number of ways to choose 2 roses from 12 = C(12, 2) = 12 * 11 / 2 = 66. Ways to choose 2 red roses = C(2, 2) = 1. Ways to choose 2 yellow roses = C(4, 2) = 4 * 3 / 2 = 6. Ways to choose 2 pink roses = C(6, 2) = 6 * 5 / 2 = 15. Total same colour pairs = 1 + 6 + 15 = 22. P(same colour) = 22 / 66 = 1 / 3. P(not same colour) = 1 - 1 / 3 = 2 / 3. Verification / Alternative check: We could also compute P(not same colour) directly by summing probabilities of red yellow, red pink and yellow pink mixed pairs, using ordered draws and then converting to unordered results. The calculations are more tedious but lead to the same final fraction 2/3, confirming the complement approach is correct. Why Other Options Are Wrong: 1/6 and 14/33 do not match either P(same colour) or P(not same colour) and come from incorrect counting. 5/6 is much too large; it would mean same colour is very unlikely, which is not supported by the counts. Common Pitfalls: Some students compute the probability of same colour and forget to subtract from 1. Others miscalculate combinations, especially for the pink roses. Carefully writing C(n, 2) and computing it helps prevent numerical errors. Final Answer: So, the probability that the two roses drawn are of different colours is 2/3.

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