Chapter 13.4, Problem 54E

a.

To determine

To find the probability that any two students in the classroom having the same birthday.

Answer to Problem 54E

  $0.346$ .

Explanation of Solution

Given:

There are total $18$ students and total $366$ birthdays are possible.

Calculation:

P ( $2$ students with the same birthday) $=1-$ P (no student with the same birthday)

  $\begin{array}{l}=1-\left(\frac{366}{366}\cdot \frac{365}{366}\cdot \frac{364}{366}\cdot \mathrm{.......................}\cdot \frac{349}{366}\right)\\ =1-\frac{P(366,18)}{{366}^{18}}\\ =0.346\end{array}$

b.

To determine

To write the inequality that can be used to determine the given probability.

Answer to Problem 54E

  $1-\frac{P(366,n)}{{366}^n}>\frac{1}{2}$ .

Explanation of Solution

Given:

Any two students having the same birth day is greater than $\frac{1}{2}$ .

Calculation:

P ( $2$ students with the same birthday)

  $=1-\frac{P(366,n)}{{366}^n}$

Now, writing the inequality as follows-

  $=1-\frac{P(366,n)}{{366}^n}>\frac{1}{2}$

c.

To determine

To find out whether there are enough students if not how many more would be required.

Answer to Problem 54E

  $\frac{4}{52}+\frac{13-1}{48}$

Explanation of Solution

Given:

The probability in part (A) to be greater than $\frac{1}{2}$ .

Calculation:

If the number of students are $18$ the P ( $2$ students with the same birthday)

  $=1-\frac{P(366,18)}{{366}^{18}}=0.346$

.If the number of students are $23$ the P ( $2$ students with the same birthday)

  $=1-\frac{P(366,18)}{{366}^{18}}\approx 0.51$

Therefore the more number of students needed are $23-18=5$ .