Assume that the birth of a boy or girl to a couple to be equally likely ,mutually exclusive,independent of the other children in the family .for a couple having $6$ children ,the probability that their "three oldest are boys" is

\frac{20}{64}

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\frac{1}{64}

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\frac{2}{64}

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\frac{8}{64}

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Solution

The correct option is D $\frac{8}{64}$
$p r o b a b i l i t y o f s o n = \frac{1}{2}$
$p r o b a b i l i t y o f g i r l = \frac{1}{2}$
$f r o m 6 children probability of 3 oldest are boys is$
$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

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Assume that the birth of a boy or girl to a couple to be equally likely ,mutually exclusive,independent of the other children in the family .for a couple having 6 children ,the probability that their "three oldest are boys" is

The correct option is D 864 probabilityofson=12 probabilityofgirl=12 from 6 children probability of 3 oldest are boys is 12×12×12=18

Assume that the birth of a boy or girl to a couple to be equally likely ,mutually exclusive,independent of the other children in the family .for a couple having $6$ children ,the probability that their "three oldest are boys" is

The correct option is D $\frac{8}{64}$
$p r o b a b i l i t y o f s o n = \frac{1}{2}$
$p r o b a b i l i t y o f g i r l = \frac{1}{2}$
$f r o m 6 children probability of 3 oldest are boys is$
$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$