Q.7.21 - A First Course in Probability

Question

For a group of 100 people, compute 

(a) the expected number of days of the year that are birthdays of exactly 3 people;

(b) the expected number of distinct birthdays. 

Step-by-Step Solution

Answer

According to the condition 

a) since we need to pick a gathering of $3$ individuals out of $100$ them. The number of days in the year that fulfill this condition is $N=\sum _{j=1}^{365}{I}_j$ Hence, the normal worth is

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot \left(\begin{array}{c}100\\ 3\end{array}\right){\left(\frac{1}{365}\right)}^3{\left(\frac{364}{365}\right)}^{97}$

b)The number of days in the year that fulfill this condition is $N=\sum _{j=1}^{365}{I}_j$

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot (1-{\left(\frac{364}{365}\right)}^{100})$

Step 1: Given Information (part a)

The expected number of days of the year that are birthdays of exactly $3$ people; 

Step 2: Explanation (part a)

Define indicator random variables Ij that marks if that day is the birthday of exactly three people or not. Observe that

$P(I_j=1)=\left(\begin{array}{c}100\\ 3\end{array}\right){\left(\frac{1}{365}\right)}^3{\left(\frac{364}{365}\right)}^{97}$

since we have to choose a group of $3$ people out of $100$ them. The number of days in the year that satisfy this condition is  $N=\sum _{j=1}^{365}{I}_j$. Hence, the expected value is

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot \left(\begin{array}{c}100\\ 3\end{array}\right){\left(\frac{1}{365}\right)}^3{\left(\frac{364}{365}\right)}^{97}$

Step 3: Final Answer (part a)

The expected number of days of the year that satisfy the condition is 

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot \left(\begin{array}{c}100\\ 3\end{array}\right){\left(\frac{1}{365}\right)}^3{\left(\frac{364}{365}\right)}^{97}$

Step 4: Given Information (part b)

The expected number of distinct birthdays.

Step 5: Explanation (part b)

Define indicator random variables Ij that marks if there exists a person that has a birthday on that day or not. We have that

$P(I_j=1)=1-{\left(\frac{364}{365}\right)}^{100}$

The number of days in the year that fulfill this condition is  $N=\sum _{j=1}^{365}{I}_j$

Hence, the expected value of a distinct birthday is

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot (1-{\left(\frac{364}{365}\right)}^{100})$

Step 6: Final Answer (part b)

The expected number of distinct birthdays that satisfy the condition is

$E\left(N\right)=\sum _{j}E\left(I_j\right)=365\cdot (1-{\left(\frac{364}{365}\right)}^{100})$