A batch of 471 containers for frozen orange juice contains 7 that are defective.

Question

A batch of 471 containers for frozen orange juice contains 7 that are defective. Two are selected, at random, without replacement from the batch. a) What is the probability that the second one selected is defective given that the first one was defective? b) What is the probability that both are defective? c) What is the probability that both are acceptable? Three containers are selected, at random, without replacement, from the batch. d) What is the probability that the third one selected is defective given that the first and second one selected were defective? e) What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? f) What is the probability that all three are defective?

Explanation / Answer

a)

probability that the second one selected is defective given that the first one was defective

=P( there remain 6 defective out of 470 after first one is defective) =6/470=3/235=0.012766

b)

probability that both are defective =P( first one defective*second one defective) =(7/471)*(6/470)=0.00019

c)

probability that both are acceptable =((471-7)/471)*((470-7))/470)=0.970466

d)probability that the third one selected is defective given that the first and second one selected were defective

=P( there remains 5 defective when first and second were out of 469)=5/469=0.010661

e)probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay =P( there remain 6 defective out of 469) =6/469=0.01279

f)probability that all three are defective =(7/471)*(6/470)*(5/469)=0.00000202

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