GO Classes DPP | GATE DA | Probability | Counting Methods

Answer 1

Let $A$ be the event that at least two people in the group have the same birthday.

It is easier to first find the probability of the complement event, namely that all $n$ people have different birthdays.

The first person can have any birthday, so the probability is $1$.

The second person must have a different birthday from the first, so the probability is $\dfrac{364}{365}$.

The third person must have a birthday different from the first two, so the probability is $\dfrac{363}{365}$.

Continuing in this way, we get

$P(A^c) = \dfrac{365}{365} \cdot \dfrac{364}{365} \cdot \dfrac{363}{365} \cdot \ldots \cdot \dfrac{365-n+1}{365}$

$\Rightarrow P(A^c) = \dfrac{365 \cdot 364 \cdot \ldots \cdot (365-n+1)}{365^n}$

By the complement rule,

$P(A) = 1 - P(A^c)$

$\therefore P(A) = 1 - \dfrac{365 \cdot 364 \cdot \ldots \cdot (365-n+1)}{365^n}$

Hence, the correct answer is A.

Answer 2

Answer: A

Answer 3

Problem Statement:

Assume that birthdays are uniformly distributed over $365$ days of the year and ignore leap years. In a randomly chosen group of $n$ people, what is the probability that at least two people have the same birthday?

Solution:

It is easier to solve this by finding the probability of the complementary event, which is that no two people share the same birthday (i.e., all $n$ people have distinct birthdays).

The total number of possible birthday combinations for $n$ people is $365^n$.

For all $n$ people to have distinct birthdays:

• The $1^{\text{st}}$ person can have a birthday on any of the $365$ days.
• The $2^{\text{nd}}$ person must have a birthday on one of the remaining $364$ days.
• The $3^{\text{rd}}$ person must have a birthday on one of the remaining $363$ days.
• $\dots$
• The $n^{\text{th}}$ person must have a birthday on one of the remaining $(365 - (n - 1)) = (365 - n + 1)$ days.

The number of favorable ways all $n$ people can have distinct birthdays is:

$$365 \times 364 \times \dots \times (365 - n + 1)$$

The probability that all $n$ people have distinct birthdays is:

$$P(\text{All distinct}) = \frac{365 \cdot 364 \cdot \dots \cdot (365 - n + 1)}{365^n}$$

Using the rule of complement, the probability that at least two people share the same birthday is:

\begin{align*}

P(\text{At least two share}) &= 1 - P(\text{All distinct}) \\

&= 1 - \frac{365 \cdot 364 \cdot \dots \cdot (365 - n + 1)}{365^n}

\end{align*}

Correct Option: A. $1 - \frac{365 \cdot 364 \cdot \dots \cdot (365 - n + 1)}{365^n}$