Chapter 8.6, Problem 73E

(a)

To determine

To explain: the reason the following pattern given the probability that the $n$ people have distinct birthdays.

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Calculation:

As consider successive people with distinct birthday, the probabilities must decrease to take into account the birth dates used. Because the birth dates of people are independent events, multiply the respective probabilities of distinct birthdays.

(b)

To determine

To write: an expression for the probability that four people have distinct birthdays using the pattern in part (a).

Answer to Problem 73E

  $P_n=\frac{365-\left(n-1\right)}{365}{P}_{n-1}$

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Calculation:

  $\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}\cdot \frac{362}{365}$

As consider successive people with distinct birthdays, the probabilities must decrease to take into account birth dates already used. Because the birth dates of people are independent events, multiply the respective probabilities of distinct birthdays.

The expression is $P_n=\frac{365-\left(n-1\right)}{365}{P}_{n-1}$ .

(c)

To determine

To verify: the probability can be obtained recursively by $P_1=1$ and $P_n=\frac{365-\left(n-1\right)}{365}{P}_{n-1}$ .

Answer to Problem 73E

Probability can be obtained recursively by $P_1=1\text{and }{P}_n=\frac{365-\left(n-1\right)}{36}{P}_{n-1}$ .

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Calculation:

The probabilities that the $n$ people have distinct birth days

  $\begin{array}{l}n=1;\frac{365}{365}\\ n=2;\frac{365}{365}\cdot \frac{364}{365}\\ n=3;\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}\end{array}$

And so on.

By, this phenomenon concludes that the probabilities can be obtained recursively by.

  $P_1=1\text{and }{P}_n=\frac{365-\left(n-1\right)}{36}{P}_{n-1}$ .

Probability can be obtained recursively by $P_1=1\text{and }{P}_n=\frac{365-\left(n-1\right)}{36}{P}_{n-1}$ .

(d)

To determine

To explain: the reason $Q_n=1-P_n$ gives the probability that at least two people in group of $n$ people have the same birthday.

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Calculation:

  $Q_n$ is the probability that the birthdays are not distinct, which is equivalent to at least two people having the same birthday.

(e)

To determine

To complete: the table.

Answer to Problem 73E

$n$	10	15	20	23	30	40	50
$P_n$	0.88	0.75	0.59	0.49	0.29	0.11	0.03
$Q_n$	0.12	0.25	0.41	0.51	0.71	0.89	0.97

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Given table:

$n$	10	15	20	23	30	40	50
$P_n$
$Q_n$

Calculation:

From (c)

Probabilities can be obtained recursively by $P_1=1and{P}_n=\frac{365-\left(n-1\right)}{365}{P}_{n-1}$

From all these have,

$n$	10	15	20	23	30	40	50
$P_n$	0.88	0.75	0.59	0.49	0.29	0.11	0.03
$Q_n$	0.12	0.25	0.41	0.51	0.71	0.89	0.97

(f)

To determine

To find: the number of people must be in a group so that the probability of at least two of them having the same birthday is greater than $\frac{1}{2}$ .

Answer to Problem 73E

That 23 people must be in a group so that the probability of at least two them having same birthday is greater than half

Explanation of Solution

Given information:

Consider a group of $n$ people,

  $\begin{array}{l}n=2:\frac{365}{365}\cdot \frac{364}{365}=\frac{365\cdot 364}{{365}^2}\\ n=3:\frac{365}{365}\cdot \frac{364}{365}\cdot \frac{363}{365}=\frac{365\cdot 364\cdot 363}{{365}^3}\end{array}$

Calculation:

From the table in (e), notice that 23 people must be in a group so that the probability of at least two them having same birthday is greater than half.