Question

The probability that among $7$ persons, no $2$ were born on the same day of a week is

• A. $\frac{2}{7}$
• B. $\frac{7!}{7}$
• C. $\frac{7!}{7^7}$
• D. $\frac{2}{7^7}$

Answer 1

Answer: (C)

$\frac{7!}{7^7}$

Total number of persons $7$.

For the sample space $n\left(S\right)$ the number of possible ways $7$ persons have birthday in a week is $7^7$.

$\therefore n\left(S\right)=7^7$

The number of ways all have birthdays in different days of a week, $n\left(B\right)$, is the number of ways of selecting $7$ different days for $7$ persons.

$\therefore n\left(B\right)={}^7{P}_7=\frac{7!}{(7-7)!}=\frac{7!}{0!}=7!$

Thus the probability that no $2$ were born on the same day of a week is $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{7!}{7^7}$