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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.

AWhat is the probability that the 2nd one selected is defective given that 1st one was defective?
BWhat is the probability that both are defective?
CWhat is probability that both are acceptable?
DThree containers are selected without replacement. What is the probability that the 3rd one selected is defective given 1st & 2nd are defective?

Verified Answer :

taskmasters

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

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