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

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