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Two boxes contain 20 light bulbs each. The material used for the bulbs in one of the two boxes was faulty so that one out of four bulbs go off as soon as you use them. The other box doesn't contain any fault bulbs. One box is selected at random and two bulbs are selected (without replacement) from it and tested. None of these two bulbs go off. What is the probability they are from the box containing the bulbs made from the faulty material?

P(B1) = 0.5         P(B2) = 0.5
P(F | B1) = 0.25    P(F | B2) = 0
P(F^C | B1) = 0.75  P(F^C | B2) = 1

So far I used Bayes' theorem :

P(B1 | F^C) = (P(B1)*(P(F^C | B1)) / ( P(B1)*P(F^C | B1) + P(B2)*P(F^C | B2) )

(0.5*0.75) / ((0.5*0.75)+(0.5*1)) = 3/7

However this is only selecting one bulb, I don't know how to then select bulb two as the box has already been chosen?

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  • $\begingroup$ What does $P(B1) = ... = 1$ mean? $\endgroup$ – gunes Nov 20 '18 at 19:19
  • $\begingroup$ Sorry, the probabiltity that a bulb is chosen from box 1! $\endgroup$ – Catherine Nov 20 '18 at 20:12
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Box 1:

P(bulb 1 works|box 1) = 15/20 (=0.75, what you had)

P(bulb 2 works|box 1) = 14/19 (since one of the fifteen good bulbs have been selected, so there are 14 good bulbs left in the remaining 19 bulbs)

Box 2:

P(bulb 1 works|box 2) = 1 (what you had)

P(bulb 2 works|box 2) = 1 (since all bulbs in box 2 work)

Assume the bulb performances in each box are independent (knowing one bulb works or not doesn't change the probability that another bulb works), then for box 1 (and vice versa for box 2):

P(both bulb work|box 1) = P(bulb 1 works|box 1) * P(bulb 2 works|box 1)

You can continue with your Bayes rule from then on.

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