SOLUTION: Suppose that we have a fuse box containing 63 fuses, of which 12 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first

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Question 1182142: Suppose that we have a fuse box containing 63 fuses, of which 12 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first, what is the probability that the second fuse selected is good given that the first one was defective? Round your answer to 4 decimal places.
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!

They ask about the conditional probability


    P(the 2nd fuse is good | 1st fuse was defective).      (1)



This conditional probability  (1)  is the ratio  of  P(2nd is good and 1st is defective) to P(1st is defective).



The numerator   of this ratio  P(both are defective) is  %2817%2F51%29%2A%28%2851-17%29%2F50%29 = %2817%2F51%29%2A%2834%2F50%29  ( ! 34 fuses are good, initially )


The denominator of this ratio  P(1st is defective) is  17%2F51.


THEREFORE, the final probability is  %28%2817%2F51%29%2A%2834%2F50%29%29%2F%28%2817%2F51%29%29 = 34%2F50 = 68%2F100 = 0.68.    ANSWER

Solved.


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The answer becomes absolutely  OBVIOUS  and the solution can be done  MENTALLY,
if you work in the  REDUCED  space of events.


In the reduced space of events,  after selecting and removing the first defective fuse,
you have only  50  fuses,  of which  51-17 = 34  are good.

Then the probability to get the good fuse at the second selection is,  OBVIOUSLY,   34%2F50 = 68%2F100 = 0.68,
which is the same answer as we got in the full space of event above.