Question

The probability that in a family of $5$ members, exactly two members have birthday on Sunday is

• A. $\frac{(12\times 5^3)}{7^5}$
• B. $\frac{(10\times 6^2)}{7^5}$
• C. $\frac{(10\times 5^3)}{7^5}$
• D. $\frac{(10\times 6^3)}{7^5}$

Answer 1

The correct option is D $\frac{(10\times 6^3)}{7^5}$

A person can have his/her birthday on any one of the seven days of the week. So $5$ perons can have birthdays in $7^5$ ways.
Out of $5,$ three persons can have their birthdays on days other than sundays in $6^3$ ways and other $2$ on sundays. Hence, the required probability is
$\frac{{}^5{C}_2\times 6^3}{7^5}=\frac{10\times 6^3}{7^5}$

$($Note that $2$ persons can be selected out of $5$ in ${}^5{C}_2$ ways.$)$