Problem 744

In a family with 9 children, excluding multiple births, what is the probability of having 8 boys and 1 girl, in any order? As assume that a boy is as likely as a girl at each birth.




Probability of an Arbitrary Event under an Equally likely Assumption

P(E) =\frac{n(E)}{n(S)}

Let the sample space S be of all possible permutations of girls and boys for 9 children.

There are 2 possibilities for each of 9 children. Thus using the multiplication principle, the total number of permutation will be 29.

Therefore, the number of element is S is

n(s)  =29

= 512

Let the event E be the set of all element that correspond to the outcome “having 8 boys and 1 girl”. Since the could be 1st, 2nd , or 3rd, there are 9 ways that 8 boys and 1 girl occur with 9 children. Therefore the number of element in E is

n(E) = 9

Thus , the probability of having 8 boys and 1 girl is

P(E) = \frac{n(E)}{n(S)}

= \frac{9}{512}


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