Problem 743

In a family with 8 children, excluding multiple births, what is the probability of having 8 girls?

Assume that a girl is as likely as a boy at each birth.




Let the sample space S be of all possible permutations of girls and boys for 8 children.

There are 2 possibilities for each of 8 children. Thus using the multiplication principle, the total number of permutation will be 28. Therefore, the number of element is S is

n(S) = 28 = 256

Let  the event E be the set of all element that correspond to the outcome “having  8 girls”.

Since there is only one way that 8 girls out of 8 children can occur, there is only 1 element in the E.

n(E) = 1

Thus, the probability of having 8 girls is

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

= \frac{1}{25}


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