Respuesta :
Let the first person's birth month be x. Now, the second person must born in the same month (1/12 probability) to satisfy the given condition as both their births are independent events. So, required probability is 1/12 .
Now we have n persons and required to find if at least two have same birth month. So, we find the probability of none of them having same birth month and subtract this from 1.
Excluding leap years there are 365 different birthdays possible.So , any person can have any one of the 365 days of the year as a birthday.Same way the second person may also have any one of the 365 days of the year as a birthday and so on.
Hence in a group of N people , there are (365)^N possible combination of birthdays
The number of ways that all N people can have different birthdays is then
m/n=365 * 364 * 363*...*(365−N+1)/(365)^N
Therefore , the probability that no two birthdays coincide is given by mn
1-m/n=365 * 364 * 363*...*(365−N+1)/(365)^N
The answer is that the probability that in a group of n people chosen at random, there are at least two born in the same month of the year is (365) (364)(363)...(365−N+1)/(365)^N.
This is a question of advanced probability involving intersection of two independent events. So, the probability of two independent events ,
A and B occurring together is P(A).P(B) In this question also, we apply this formula.
Let the first person's birth month be m. Now, the second person must born in the same month (1/12 probability) as both their births are independent events and none of them affect each other. So, required probability is 1/12 .
Now we have n persons and required to find if at least two have same birth month. So, whenever we have to find the probability of 'at least' an easy and quicker way to do so is to subtract the probability of 'none' from the total probability i.e. 1.
So, here we find the probability of none of them having same birth month and subtract this from 1.
So , any person can have any one of the 365 days of the year as a birthday. Same way the second person may also have any one of the 365 days of the year as a birthday and so on.
(We are taking the general case of non leap years to simplify the problem)
Hence in a group of N people , there are (365)^N possible combination of birthdays are possible as each birth is independent of others.
The number of ways that all N people can have different birthdays is then
m/n=365 * 364 * 363*...*(365−N+1)/(365)^N
Therefore , the probability that no two birthdays coincide is given by
1-m/n=365 * 364 * 363*...*(365−N+1)/(365)^N.
To learn more about probability of independent events , visit link - brainly.com/question/11455301
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