Question 1155751
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First consider the probability that with four children the first is a girl and the last is a boy:<br>
P(GxxB) = (1/2)(2/2)(2/2)(1/2) = 4/16<br>
Now consider the requirement that there be at least one boy and at least one girl.  That eliminates two possible cases -- all boys or all girls -- so the denominator of the probability fraction is 16-2 = 14.<br>
Of the two cases that were thrown out, neither has a girl first and a boy last, so the numerator of the probability fraction is still 4.<br>
ANSWER: 4/14, or 2/7<br>
Note there are other formal ways of calculating that probability; however, for a small problem like this, it is easiest simply to list all 16 possible sequences of 4 boys or girls and find how many satisfy the given conditions.  14 of the 16 satisfy the condition that there is at least one boy and one girl; and 4 of those 14 have a girl first and a boy last.<br>