Question

question: With probability = .25 , two switches are selected without replacement from box A, & with probability = .75 , two switches are selected without replacement from box B.

box A contains 4 defective switches & 6 good switches (prob = .)

box B contains 8 defective switches & 2 good switches (prob = .)

Determine the probability that the 2nd chosen switch is defective, given that the 1st chosen switch is defective.

Answer 1

Here we want to find  the probability that the 2nd chosen switch is defective, given that the 1st chosen switch is defective.

There are two ways of selecting  the 2nd chosen switch is defective, given that the 1st chosen switch is defective from each boxes.

i) P(switches are selected without replacement from box A and second switch is defective given that the 1st chosen switch is defective) = 0.25*(3/9) = (1/4)*(1/3) = 1/12 = 0.0833

ii) P(switches are selected without replacement from box B and second switch is defective given that the 1st chosen switch is defective) = 0.75*(7/9) = (3/4)*(7/9) = 7/12 = 0.5833

Add both these probabilities then we get the final result = (1/12) + (7/12) = 8/12 = 2/3 = 0.6667 ( this is the final answer).