A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.

Alexus Deleon 2022-09-25 Answered
Intro to probability chapter 4 ex 31
A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.
Solution: Creating an indicator r.v. for each pair of people, we have that the expected number of pairs of people with the same birthday is (50C2 . 1/365) by linearity. Now create an indicator r.v. for each day of the year, taking the value 1 if at least two of the people were born that day (and 0 otherwise). Then the expected number of days on which at least two people were born is
365 ( 1 ( 364 / 365 ) 50 50 ( 1 / 365 ) ( 364 / 365 ) 49 )
Could someone explain how we got the answer ?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Brendon Melton
Answered 2022-09-26 Author has 5 answers
Step 1
(a) Number all possible pairs of people 1 , 2 , , ( 50 2 ) and for all i in this range define the indicator random variable:
I i = { 1 if pair  i  share a birthday 0 otherwise.
If X = #pairs sharing a birthday, then X = i I i and so,
E ( X ) = i = 1 ( 50 2 ) E ( I i ) by linearity of expectation = i = 1 ( 50 2 ) P ( pair  i  share a birthday ) = i = 1 ( 50 2 ) 1 365 = ( 50 2 ) 365 .
Step 2
(b) For i = 1 , 2 , , 365 , define the indicator random variable:
I i = { 1 if day  i  has at least  2  birthdays 0 otherwise.
If X = #days with at least  2  birthdays , then X = i I i and so,
E ( X ) = i = 1 365 E ( I i ) by linearity of expectation = i = 1 365 P ( day  i  has at least  2  birthdays ) = i = 1 365 ( 1 P ( day  i  has  0  birthdays ) P ( day  i  has  1  birthday ) ) = i = 1 365 ( 1 ( 50 0 ) ( 364 365 ) 50 ( 50 1 ) ( 364 365 ) 49 1 365 ) = 365 ( 1 ( 364 365 ) 50 50 ( 364 365 ) 49 1 365 ) .
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2020-11-01
Solve the following problems applying Polya’s Four-Step Problem-Solving strategy.
If six people greet each other at a meeting by shaking hands with one another, how many handshakes take place?
asked 2021-03-02
In how many ways can 7 graduate students be assigned to 1 triple and 2 double hotel rooms during a conference
asked 2022-09-26
How to show that these groups are isomorphic?
Show that group of all real matrices of form
[ x y y x ] , ( x , y ) ( 0 , 0 )
is isomorphic with/to C∖{0} under complex multiplication?
I know two ways to show isomorphism: 1) finding a homomorphic function 2) writing the multiplication table and comparing.
asked 2020-10-21
Understanding the Concepts and Skills Give correct answer for in using F-procedures to make inferences for two population standard deviations, why should the distributions (one for each population) of the variable under consideration be normally distributed or nearly so?
asked 2022-09-30
How can I show that PGL(2,9) is not isomorphic to S 6 ?
My primary idea is to compare the size of conjugacy classes of two well-chosen elements in these groups. Is there another simpler approach?
asked 2021-10-15
1). What is the difference between a t-test for independent samples and a t-test for dependent samples? 2). Provide one study that would be analyzed using a t-Test for dependent samples; then provide one study that would be analyzed using a t-Test for independent samples
asked 2022-09-12
First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: Q ( 2 3 , γ ), where γ 3 = 1 and γ 2 = 1 γ. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.
I know that my Galois group will consist of the following automorphisms:
σ i = identity
σ 1 : 2 3 2 3 γ and γ is fixed.
σ 2 : 1 3 2 3 γ 2 and γ is fixed.
σ 3 : γ γ 2 and 2 3 is fixed.
σ 4 : 2 3 2 3 γ and γ γ 2 .
σ 5 : 1 3 2 3 γ 2 and γ γ 2 .
If I didn't miss something or make some error.
My specific question is, when constructing my automorphisms, why exactly is it that I can't send 2 3 to something like 4 3 or γ or some other random thing? Why are these automorphisms so strictly defined?

New questions