Mr. A has 6 children and at least one child i | vedclass

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(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 (wh

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 (which is $1$ case).Thus,the total number of possible outcomes is $2^6 - 1 = 64 - 1 = 63$.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 $6$ children and at least one child is a girl. What is the probability that Mr. $A$ has $3$ boys and $3$ girls?

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