Probability — Practice

Question 1

If P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2, find P(A|B) and P(B|A).

Accepted Answer

Divide the joint probability by each event's probability in turn, since the conditioning event supplies the denominator. Answer: P(A|B) = 0.5 and P(B|A) = 0.4.

Question 2

If A and B are independent with P(A) = 0.4 and P(B) = 0.5, find P(A or B).

Accepted Answer

Independence gives P(A and B) = 0.20, and the addition rule then supplies the union. Answer: P(A or B) = 0.7.

Question 3

Two cards are drawn without replacement from a well shuffled deck of 52. Find the probability that both are kings.

Accepted Answer

The first card is a king with probability 4/52, and given that, only 3 kings remain among 51 cards. Answer: (4/52)(3/51) = 1/221.

Question 4

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4. Find the probability that the problem is solved.

Accepted Answer

The problem is unsolved only if all three fail, which happens with probability (1/2)(2/3)(3/4) = 1/4. Answer: P(solved) = 3/4.

Question 5

Bag A has 3 red and 5 black balls; bag B has 4 red and 4 black. A bag is chosen at random and a ball drawn. Find the probability that it is red.

Accepted Answer

Weight each bag's red probability by the 1/2 chance of selecting that bag and add. Answer: (1/2)(3/8) + (1/2)(4/8) = 7/16.

Question 6

Two machines produce 40% and 60% of output, with defect rates 3% and 2%. An item is found defective. Find the probability it came from the first machine.

Accepted Answer

The numerator is the first machine's term and the denominator is the sum of both, since the machines partition the output. Answer: (0.4)(0.03)/[(0.4)(0.03)+(0.6)(0.02)] = 1/2.

Question 7

A doctor knows a patient has one of three diseases with probabilities 0.5, 0.3 and 0.2. A test is positive with probabilities 0.8, 0.6 and 0.4 respectively. Given a positive test, find the probability of the first disease.

Accepted Answer

The three diseases partition the possibilities, so apply Bayes with the first disease's term on top. Answer: (0.5)(0.8)/[(0.5)(0.8)+(0.3)(0.6)+(0.2)(0.4)] = 0.40/0.66 = 20/33.

Question 8

A family has three children. Given that at least one is a boy, find the probability that all three are boys.

Accepted Answer

Of the 8 equally likely birth orders, 7 contain at least one boy, and only BBB has all three. Answer: 1/7.

Practice — Probability

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