SOLUTION: Suppose that we have a fuse box containing 51 fuses, of which 17 are defective. Two fuses are selected at random and removed from the box in succession assuming that the first fuse

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose that we have a fuse box containing 51 fuses, of which 17 are defective. Two fuses are selected at random and removed from the box in succession assuming that the first fuse      Log On


   

Question 1182139: Suppose that we have a fuse box containing 51 fuses, of which 17 are defective. Two fuses are selected at random and removed from the box in succession assuming that the first fuse has been replaced before selecting the second one. What is the probability that the second fuse selected is defective given that the first one was defective? Round your answer to 4 decimal places.
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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Suppose that we have a fuse box containing 51 fuses, of which 17 are defective.
Two fuses are selected at random and removed from the box in succession
assuming that the first fuse has been replaced before selecting the second one.
What is the probability that the second fuse selected is defective
given that the first one was defective? Round your answer to 4 decimal places.
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They ask about the conditional probability

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



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



The numerator   of this ratio  P(both are defective) is  %2817%2F51%29%2A%2816%2F50%29.


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


THEREFORE, the final probability is  %28%2817%2F51%29%2A%2816%2F50%29%29%2F%28%2817%2F51%29%29 = 16%2F50 = 32%2F100 = 0.32.    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  16  are defective.

Then the probability to get the defective fuse at the second selection is,  OBVIOUSLY,   16%2F50 = 32%2F100 = 0.32,
which is the same answer as we got in the full space of event above.