Question 1182142
<pre>

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  {{{(17/51)*((51-17)/50)}}} = {{{(17/51)*(34/50)}}}  ( ! 34 fuses are good, initially )


The denominator of this ratio  P(1st is defective) is  {{{17/51}}}.


THEREFORE, the final probability is  {{{((17/51)*(34/50))/((17/51))}}} = {{{34/50}}} = {{{68/100}}} = 0.68.    <U>ANSWER</U>
</pre>

Solved.



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The answer becomes absolutely &nbsp;OBVIOUS &nbsp;and the solution can be done &nbsp;MENTALLY,

if you work in the &nbsp;REDUCED &nbsp;space of events.



In the reduced space of events, &nbsp;after selecting and removing the first defective fuse,

you have only &nbsp;50 &nbsp;fuses, &nbsp;of which &nbsp;51-17 = 34 &nbsp;are good.


Then the probability to get the good fuse at the second selection is, &nbsp;OBVIOUSLY,  &nbsp;&nbsp;{{{34/50}}} = {{{68/100}}} = 0.68,

which is the same answer as we got in the full space of event above.