First slide

Theorems of probability

Question

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

Moderate

A

$(12 \times 5^{3}) / 7^{5}$
B

$(10 \times 6^{2}) / 7^{5}$
C

$\frac{2}{5}$
D

$(10 \times 6^{3}) / 7^{5}$

Solution

A person can have his/her birthday on any one of the seven days of the week. So 5 persons can have their birthdays in 7⁵ ways. Out of 5, three persons can have their birthdays on days other than Sundays in 6³ways and other 2 on Sundays. Hence, the required probability is
$\frac{^{5} C_{2} \times 6^{3}}{7^{5}} = \frac{10 \times 6^{3}}{7^{5}}$
$(Note that 2 persons can be selected out of 5 {in}^{5} C_{2} ways.)$

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