A batch of 500 containers for frozen orange juice contain 5 that are defective. 2 are selected at random without replacement.

A=event that 1st is defective, B=event that 2nd is defective, C=event that 3rd is defective.

a) What is the probability that the 2nd one selected is defective given that 1st one was defective?

b) What is the probability that both are defective?

c) What is probability that both are acceptable?

d) Three containers are selected without replacement. What is the probability that the 3rd one selected is defective given 1st & 2nd are defective?

Question

Sloane

Physics

21 August, 03:50

A batch of 500 containers for frozen orange juice contain 5 that are defective. 2 are selected at random without replacement.

A=event that 1st is defective, B=event that 2nd is defective, C=event that 3rd is defective.

a) What is the probability that the 2nd one selected is defective given that 1st one was defective?

b) What is the probability that both are defective?

c) What is probability that both are acceptable?

d) Three containers are selected without replacement. What is the probability that the 3rd one selected is defective given 1st & 2nd are defective?

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Answers (1)

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Misty · Answer 1

Given:

500 containers; 5 of these are defective.

A = event that 1st is defective = 5/500

B = event that 2nd is defective = 4/499

C = event that 3rd is defective = 3/498

a) 4/499

b) 5/500 * 4/499 = 20/249,500 = 1/12,475

c) 495/500 * 494/500 = 244,530 / 250,000 = 24,453/25,000

d) 5/500 * 4/499 * 3/498 = 60/124,251,000 = 1/2,070,850