Quantitative Aptitude — Probability

Total possible outcomes when rolling a fair six-sided die are {1, 2, 3, 4, 5, 6}, so there are 6 total outcomes. Favorable outcomes (odd numbers) are {1, 3, 5}, so there are 3 favorable outcomes. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(getting an odd number) = 3/6 = 1/2.

When two fair coins are tossed simultaneously, the total possible outcomes are HH, HT, TH, TT. So, there are 4 total outcomes. Favorable outcomes (getting exactly one head) are HT, TH. So, there are 2 favorable outcomes. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(getting exactly one head) = 2/4 = 1/2.

Total number of balls in the bag = Number of red balls + Number of blue balls + Number of green balls = 5 + 3 + 2 = 10 balls. Number of blue balls = 3. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(drawing a blue ball) = 3/10.

Total number of cards in a well-shuffled deck = 52. Number of Kings in a deck = 4. Number of Queens in a deck = 4. Number of favorable outcomes (drawing a King or a Queen) = Number of Kings + Number of Queens = 4 + 4 = 8. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(drawing a King or a Queen) = 8/52 = 2/13.

When two fair six-sided dice are rolled, the total number of possible outcomes is 6 * 6 = 36. Favorable outcomes where the sum of the numbers is 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 such outcomes. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(sum is 7) = 6/36 = 1/6.

Let P(A) be the probability that A solves the problem = 2/3. Let P(B) be the probability that B solves the problem = 3/5. Since A and B try independently, the probability that A does NOT solve the problem, P(A') = 1 - P(A) = 1 - 2/3 = 1/3. The probability that B does NOT solve the problem, P(B') = 1 - P(B) = 1 - 3/5 = 2/5. The probability that the problem is NOT solved by either A or B (since they are independent events) is P(A' and B') = P(A') * P(B') = (1/3) * (2/5) = 2/15. The probability that the problem IS solved is the complement of the problem not being solved: P(problem solved) = 1 - P(problem not solved) = 1 - 2/15 = 13/15.

Total number of marbles in the box = 4 (red) + 6 (blue) = 10 marbles. Probability of drawing the first red marble: P(1st Red) = (Number of red marbles) / (Total marbles) = 4/10. Since the first marble is drawn without replacement, there are now 3 red marbles left and a total of 9 marbles. Probability of drawing the second red marble (given the first was red): P(2nd Red | 1st Red) = (Number of remaining red marbles) / (Total remaining marbles) = 3/9. The probability that both marbles are red is the product of these probabilities: P(both red) = P(1st Red) * P(2nd Red | 1st Red) = (4/10) * (3/9) = (2/5) * (1/3) = 2/15.

When three fair coins are tossed simultaneously, the total number of possible outcomes is 2^3 = 8. These outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The event 'at least one tail' is the complement of the event 'no tails' (which means all heads). The only outcome with 'no tails' is HHH. So, the probability of getting no tails, P(no tails) = 1/8. The probability of getting at least one tail is 1 minus the probability of getting no tails: P(at least one tail) = 1 - P(no tails) = 1 - 1/8 = 7/8.

Total number of people in the group = 5 men + 4 women = 9 people. Total ways to form a committee of 3 members from 9 people (using combinations formula C(n,k) = n! / (k! * (n-k)!)): Total ways = C(9,3) = (9 * 8 * 7) / (3 * 2 * 1) = 3 * 4 * 7 = 84. Ways to choose exactly 2 men from 5 men = C(5,2) = (5 * 4) / (2 * 1) = 10. Ways to choose exactly 1 woman from 4 women = C(4,1) = 4. Number of favorable ways (committee with exactly 2 men and 1 woman) = C(5,2) * C(4,1) = 10 * 4 = 40. Using the formula for probability, P(Event) = (Number of Favorable Outcomes) / (Total Number of Outcomes): P(committee with 2 men and 1 woman) = 40/84. Simplifying the fraction: 40/84 = (4 * 10) / (4 * 21) = 10/21.

Let P(A), P(B), P(C) be the probabilities that an item is produced by machine A, B, or C respectively. P(A) = 20% = 0.20 P(B) = 30% = 0.30 P(C) = 50% = 0.50 Let P(D|A), P(D|B), P(D|C) be the probabilities that an item is defective given it was produced by A, B, or C respectively. P(D|A) = 5% = 0.05 P(D|B) = 2% = 0.02 P(D|C) = 3% = 0.03 We need to find P(A|D), the probability that the item was produced by machine A given it is defective. We use Bayes' Theorem: P(A|D) = [P(D|A) * P(A)] / P(D) First, calculate the total probability of an item being defective, P(D), using the Law of Total Probability: P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) P(D) = (0.05 * 0.20) + (0.02 * 0.30) + (0.03 * 0.50) P(D) = 0.010 + 0.006 + 0.015 P(D) = 0.031 Now, substitute P(D) back into Bayes' Theorem to find P(A|D): P(A|D) = (0.010) / (0.031) = 10/31.