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Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day of the year. Consider a group of 10 people, of which you are not a member. The sample space consists of all possible sequences of 10 birthdays (onefor each person).

a. Define the probability function P for S.
b. Consider the following events:
1. "someone in the group shares your birthday";
2. "some two people in the group share a birthday";
3. "some three people in the group share a birthday". Carefully describe the subset of S that corresponds to each event.

c. Find an exact formula for P(A). What is the smallest m such that P(A) > .5?
d. Justify why m is greater than 365 without doing any computation. (We 3 are looking for 2 a short answer giving a heuristic sense of why this is so.)
e. Find an exact formula for P(B).

Sagot :

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

[tex]S=365^n[/tex]

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

[tex]P(x)=\frac{1}{365^{n}}[/tex]

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

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