Question 1150805
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            The problem can be solved by various ways.



<U>Solution 1</U>.  &nbsp;&nbsp;Make and use the sample space directly and explicitly.


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The sample space is 

    BBB  BBG  BGB  BGG  GBB  GBG  GGB  GGG

(by coding B = boy,  G = girl).


All outcomes are equally likely with the probability of  {{{(1/2)^3}}} = {{{1/8}}}  each.


The number of outcomes with a boy as a second child is 4  (BBB  BBG  GBB  GBG).

Of them,  the number of outcomes with only one boy in the family is 1 (GBG).


Hence, the probability under the question is  P = {{{1/4}}}.
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<U>Solution 2</U>.  &nbsp;&nbsp;Logical analysis.


<pre>
If the boy (B) is in the 2-nd  of the 3 positions, then for the 1-st and for 3-rd position only  G  is possible.

Each G at the 1-st position and each G at the 3-rd position goes with the probability of {{{1/2}}}.

Hence, the probability to have  G  at  the 1-st and 3-rd positions, under the condition, that the 2-nd position is B, is  {{{(1/2)^2}}} = {{{1/4}}}.
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In both solutions, you have the same <U>ANSWER</U> :


<pre>
          the probability under the question is  {{{1/4}}}.
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Solved, answered, and explained.