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14.Probability
normal
Mr. A has six children and atleast one child is a girl, then probability that Mr. A has $3$ boys and $3$ girls, is
A
$\frac{20}{63}$
B
$\frac{1}{3}$
C
$\frac{5}{11}$
D
$\frac{1}{32}$
Solution
$\frac{{\left( {\frac{{6!}}{{3!3!}}} \right)}}{{{2^6} – 1}} = \frac{{20}}{{63}}$
Standard 11
Mathematics