Question

The probability that a couple will have first 3 daughters then 1 son is

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{3}{4}$
• D. $\frac{1}{16}$

Answer 1

The correct option is D $\frac{1}{16}$

Probability of 1st child being daughter = $\frac{1}{2}$
2nd child being daughter = $\frac{1}{2}$
3rd child being daughter = $\frac{1}{2}$
4th child being son = $\frac{1}{2}$

$\therefore$ Probability of having 3 daughter and then 1 son
= $\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}$
$=\frac{1}{16}$