Unit 3 — Probability · Statistics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 91 questions on this lesson, each with an explanation.
If B is an event of the sample space S where P(B) > 0, then P(S | B) = ……
Correct answer: (D) 1
Explanation: Step 1: P(S | B) = P(S ∩ B) ÷ P(B), and S ∩ B = B Step 2: P(S | B) = P(B) ÷ P(B) = 1 ∴ P(S | B) = 1
A regular coin is tossed twice. The probability of getting two heads given that the first toss shows a head equals ……
Correct answer: (B) 1/2
Explanation: Step 1: The condition reduces the sample space to B = {(H, H), (H, T)} Step 2: Two heads = {(H, H)}, one of the 2 outcomes of B, so P = (1/4) ÷ (2/4) = 1/2 ∴ The probability = 1/2 (1/4 is a trap: it is P(two heads) without the condition)
A regular die is rolled once. The probability of getting the number 3 given that the number that appeared is odd equals ……
Correct answer: (B) 1/3
Explanation: Step 1: The condition reduces the sample space to B = {1, 3, 5}, and A = {3}, so A ∩ B = {3} Step 2: P(A | B) = P(A ∩ B) ÷ P(B) = (1/6) ÷ (3/6) = 1/3 ∴ The probability = 1/3 (1/6 is a trap: it is P(3) without the condition)
A card is drawn at random from 30 identical cards numbered from 1 to 30. The probability that the number is prime given that it is odd equals ……
Correct answer: (A) 3/5
Explanation: Step 1: There are 15 odd numbers; the prime ones are 3, 5, 7, 11, 13, 17, 19, 23, 29, that is 9 numbers Step 2: P(prime | odd) = (9/30) ÷ (15/30) = 9/15 = 3/5 ∴ The probability = 3/5 (2/3 = 10/15 is a trap: 1 is not a prime number; 3/10 = 9/30 ignores the condition)
120 people applied for a job. Qualified: married males 30, single males 15, married females 20, single females 15. Unqualified: married males 10, single males 12, married females 8, single females 10. An applicant is chosen at random. The probability that the applicant is qualified given that he (she) is married equals ……
Correct answer: (C) 25/34
Explanation: Step 1: Married applicants = 30 + 20 + 10 + 8 = 68, of whom 30 + 20 = 50 are qualified Step 2: P(qualified | married) = (50/120) ÷ (68/120) = 50/68 = 25/34 ∴ The probability = 25/34 (5/8 = 50/80 is a trap: it is P(married | qualified), the reversed condition; 5/12 = 50/120 is the probability of both)
The probability that a customer of a shop buys online is 0.8, and the probability that he pays by bank card if he buys online is 0.35. A customer is chosen at random. The probability that he buys online and does not pay by bank card equals ……
Correct answer: (B) 0.52
Explanation: Step 1: P(no card | online) = 1 − P(card | online) = 1 − 0.35 = 0.65 Step 2: P(online ∩ no card) = P(no card | online) × P(online) = 0.65 × 0.8 = 0.52 ∴ The probability = 0.52 (0.28 = 0.8 × 0.35 is a trap: it is the probability of buying online and paying by card)
A box contains 10 lamps, 3 of which are defective. Three lamps are drawn one after another without replacement. The probability that all three lamps are good equals ……
Correct answer: (A) 7/24
Explanation: Step 1: Each draw is conditioned on the draws before it: P = 7/10 × 6/9 × 5/8 Step 2: P = 210/720 = 7/24 ∴ The probability = 7/24 (343/1000 = (7/10)³ treats the drawing as with replacement; 17/24 is the probability of at least one defective lamp)
A factory produces 60% of its products on the first machine and 40% on the second machine. The probability that a piece is defective if it is made on the first machine is 0.05, and if it is made on the second machine is 0.10. A piece is chosen at random from the production. The probability that it is defective equals ……
Correct answer: (C) 0.07
Explanation: Step 1: From the tree diagram: P(first ∩ defective) = 0.6 × 0.05 = 0.03 and P(second ∩ defective) = 0.4 × 0.10 = 0.04 Step 2: P(defective) = 0.03 + 0.04 = 0.07 ∴ The probability = 0.07 (0.075 is the plain average of the two rates; 0.03 keeps only the first machine's branch)
An employee goes to work by one of two roads: the first road with probability 0.6 and the second with probability 0.4. If he takes the first road, the probability that he is late is 0.1, and if he takes the second, it is 0.25. One day the employee was late. The probability that he took the first road equals ……
Correct answer: (C) 0.375
Explanation: Step 1: P(first ∩ late) = 0.6 × 0.1 = 0.06 and P(second ∩ late) = 0.4 × 0.25 = 0.1, so P(late) = 0.16 Step 2: P(first | late) = P(first ∩ late) ÷ P(late) = 0.06 ÷ 0.16 = 0.375 ∴ The probability = 0.375 (0.625 is P(second | late); 0.16 is P(late) itself)
If A and B are two events of a sample space S where P(A | B) = 0.5, P(B | A) = 0.25 and P(A ∪ B) = 0.5, then P(A) = ……
Correct answer: (B) 0.4
Explanation: Step 1: Let P(A ∩ B) = m, then P(B) = m ÷ 0.5 = 2m and P(A) = m ÷ 0.25 = 4m Step 2: P(A ∪ B) = 4m + 2m − m = 5m = 0.5, so m = 0.1 ∴ P(A) = 4 × 0.1 = 0.4 (0.2 is P(B); 0.1 is P(A ∩ B))
The first 10 questions in every subject are free, each with an explanation.