Chapter 8.6, Problem 70E

a.

To determine

To explain: why the following given pattern gives the probability that the n people have distinct birthdays.

Answer to Problem 70E

The probability of no two people from the group of n people sharing the same birthday is

  $P(E)=\frac{n(E)}{n(S)}=\frac{365·364·\mathrm{363.....}\left[365-(n-1)\right]}{{365}^n}.$

Explanation of Solution

Given information: Consider a group of n people, the given pattern is,

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

Calculation:

The pattern gives the probability that n people have distinct birthdays using the standard formula of $P(E)=\frac{n(E)}{n(S)}$ . The sample space is the total number of possible combinations of birthdays amongst the n people. Since each person could be born on any of the 365 days of the year, this number is ${356}^n$ . In order for all these people to have different birthdays, the first person could be born on any of the 365 days of the year, the second on any of the 365 days of the year on which the first was not born, the third on any of the 365 days the first two were not born on and so on. Therefore the probability of no two people from the group of n people sharing the same birthday is

  $P(E)=\frac{n(E)}{n(S)}=\frac{365·364·\mathrm{363.....}\left[365-(n-1)\right]}{{365}^n}.$

b.

To determine

To write: an expression for the probability that four people ( n =4) have distinct birthdays using the pattern in part (a).

Answer to Problem 70E

  $n=4;\frac{365}{365}·\frac{364}{365}·\frac{363}{365}·\frac{362}{365}=\frac{365·364·363·362}{{365}^4}$

Explanation of Solution

Given information: Consider a group of n people, the given pattern is,

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

Calculation:

To find the probability that amongst 4 people , no 2 share the same birthday, plug n= 4 into the given pattern.

  $n=4;\frac{365}{365}·\frac{364}{365}·\frac{363}{365}·\frac{362}{365}=\frac{365·364·363·362}{{365}^4}$

c.

To determine

To verify: that $P_n$ probability that the n people have distinct birthdays can be obtained recursively by

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

Answer to Problem 70E

  $P_n$ Probability that the n people have distinct birthdays can be obtained recursively by

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

Explanation of Solution

Given information: Given $P_n$ probability that the n people have distinct birthdays can be obtained recursively by

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

Calculation:

Here, use the probability of 4 people having different birthdays as an example of how the recursive method provided is a valid way to find this probability. Using the recursive definition,

  $\begin{array}{c}{P}_4=\frac{365-(4-1)}{356}{P}_{4-1}\\ =\frac{362}{356}{P}_3\\ =\frac{362}{356}·\frac{365-(3-1)}{356}{P}_{3-1}\\ =\frac{363·362}{{356}^2}{P}_2\\ =\frac{363·362}{{356}^2}·\frac{365-(2-1)}{356}{P}_{2-1}\\ =\frac{364·363·362}{{356}^3}{P}_1\\ =\frac{364·363·362}{{356}^3}·\frac{365}{365}\\ =\frac{365·364·363·362}{{356}^4}.\end{array}$

Same result using this recursive definition as the one obtained in part (b).

Hence, Verified.

d.

To determine

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

Answer to Problem 70E

  $Q_n=1-P_n$ .

Explanation of Solution

Given information: $Q_n=1-P_n$ gives the probability that at least two people in a group of n people have the same birthday.

Calculation:

  $P_n$ gives the probability that in a group of n people , none share the same birthday. The complement of the event is $Q_n$ , is the event that at least two people share the same birthday. Therefore, by the law of complementation, $Q_n=1-P_n.$

e.

To determine

To complete: the given table using the result of parts (c) and (d).

Answer to Problem 70E

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: Given table is,

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

Calculation:

When constructing the probability of this table, it can help to recognize that:

  $\begin{array}{l}{P}_n=\frac{365·364·\mathrm{.....}\left[365-(n-1)\right]}{{365}^n}=\frac{365!}{{365}^n(365-n)!}=\frac{{}_{365}{P}_n}{{365}^n}.\\ Q_n=1-P_n\end{array}$

The completed table shown below,

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

Answer to Problem 70E

23 people in a group so that the probability of at least two of them having the same birthday is greater than $\frac{1}{2}$ , explain.

Explanation of Solution

Given information:

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

Calculation:

It is clear from the above table, 23 people must be in a group for the probability of at least two them having the same birthday to be greater than $\frac{1}{2}.$