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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
$\frac{1}{{6}}$
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