   Chapter 9.3, Problem 32ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# Assuming that all years have 365 days and all birth-days occur with equal probability, how large must n be so that in any randomly chosen group of n people, the probability that two or more have the same birthday is at least ½?(This is called the birthday problem. Many people find the answer surprising.)

To determine

The size of n such that so that in any randomly chosen group of n people, the probability that two or more have the same birthday is atleast 1/2.

Explanation

Given information:

Assuming that all years have 365 days arid all birthdays occur with equal probability,

Concept used:

1st person birthday can happen any of 365 days.

2nd person birthday can happen any of 365 days.

3rd person birthday can happen any of 365 days.

And so on.

Calculation:

There are 365 possible birth days for each person.

Then, by multiplication rule, the total number of ways is 365n.

If no two persons share the same birth day, there are 365 possible days for first person, 364 for

2nd person, and 363 for 3rd person, and 365n+1 for nth person.

Then, by multiplication rule, the total number of ways is 365(3651).....(365n+1).

By difference rule, the total number of ways birth day could be associated with n peoples such that at least two people share the same birth day is 365n365364......(365n+1).

The probability that at least two of the n people share the same birthday is as follows:

P(n, all have different b'days)=365×364×363×......×365n+1 365n= 365Pn 365nP( More than 2 people have same birthday)=1 365Pn 365n

Here 365Pn denotes the permutations.

Compute the value of n such that the probability that two or more than 2 people have the same birthday is at least 12.

1 365Pn 365n12 365Pn 365n12 365Pn 365n12365! 365n( 365n)!12365n(365n)!2×365!

Substitute n=1,2,3..... in inequality 365n(365n)!2×365! to compute the value of n.

For n=1, the inequality becomes.

365(3651)!2×365!365(364)!2×365!365!2×365!

This inequality is not valid

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