Question

The probability that in a family of four children, there will be atleast one boy is

• A. $\frac{1}{16}$
• B. $\frac{3}{16}$
• C. $\frac{13}{16}$
• D. $\frac{15}{16}$

Answer 1

The correct option is D $\frac{15}{16}$

Total number of possibility = $2^4$

each child can be boy or girl

$\Rightarrow 2$ possibilities

$\Rightarrow$ for 4 children, Total = $2\times 2\times 2\times 2$

$=16$

If the family have no boy

$\Rightarrow$ all girls

$\Rightarrow$ Total weigh to have

all girls = 1

$\Rightarrow$ P (No boy) = $\frac{1}{16}$

$\Rightarrow P$ (atleast one boy) = 1-P (No boy)

$=1-\frac{1}{16}$

$=\frac{15}{16}$ (D)