Mr. A has six children and at least one child | vedclass

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Question

(A) The total number of ways to have $6$ children is $2^6 = 64$.Since at least one child is a girl,we exclude the case where all children are boys.The

Vedclass · Accepted Answer

$\frac{20}{63}$

Vedclass · Answer

(A) The total number of ways to have $6$ children is $2^6 = 64$.Since at least one child is a girl,we exclude the case where all children are boys.The number of favorable outcomes where at least one child is a girl is $64 - 1 = 63$.The number of ways to have $3$ boys and $3$ girls is given by the binomial coefficient $\binom{6}{3} = \frac{6!}{3!3!} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.Therefore,the required probability is $\frac{20}{63}$.

Mr. $A$ has six children and at least one child is a girl. Then,the probability that Mr. $A$ has $3$ boys and $3$ girls is:

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