Question: In a group of n people (n ≤ 7), what is the probability that at least 2 of them have the same birth…

In a group of n people (n ≤ 7), what is the probability that at least 2 of them have the same birth day of the week? (Assume that any birth day of the week is as likely as any other.)

Let E be the event at least 2 people have the same birth day of the week. Choose the correct value of P(E) below.

A. P(E) = 1 - \frac{7!}{7^n(7 - n)!}

B. P(E) = 1 - \frac{7^n(7 - n)!}{7!}

C. P(E) = 1 - \frac{n!(7 - n)!}{7}

D. P(E) = \frac{7n!}{(7 - n)!} + 1

Solution

To find the probability that at least 2 people have the same birthday in a group of \( n \) people (where \( n \leq 7 \)), we can use the complement probability approach. First, calculate the probability that all \( n \) people have different birth days of the week. The total number of ways to assign birth days to \( n \) people is \( 7^n \). Now calculate the number of ways to assign different birth days to each of the \( n \) people: The first person can have any of the 7 days, the second person can only have any of the remaining 6 days, the third person can only have any of the remaining 5 days, and so on. The number of ways to assign different birth days to \( n \) people is given by: \[ \begin{aligned} 7 \times 6 \times 5 \times \ldots \times (7-n+1) = \frac{7!}{(7-n)!} \end{aligned} \] The probability that all \( n \) people have different birth days is: \[ P(\text{all different}) = \frac{\frac{7!}{(7-n)!}}{7^n} \] The probability that at least 2 people have the same birth day is the complement: \[ P(E) = 1 - P(\text{all different}) = 1 - \frac{\frac{7!}{(7-n)!}}{7^n} \] This matches option A: \[ P(E) = 1 - \frac{7!}{7^n (7-n)!} \] Thus, the correct answer is A.