# Probability Question that involves alot of counting

1. Jul 23, 2004

### relinquished™

I can't seem to put the Fundamentals of Counting to good use... I have such a hard time answering probability questions with the rules of counting. Here's one question that blew my mind:

There are 4 persons. The sample space E consists of the events E={E1, E2, E3, E4, E5 }. Let E1 be the event that all four person have the same birthmonth, E2 be the event that exactly 3 of them have the same birthmonth, E3 be the event that exactly 2 of them have the same birthmonth while the remaining 2 have different birthmonths, E4 be the event that two have the same birthmonths and the other two have the same birthmonths, and E5 be the event that all four person have different birthmonths. Assign probability measures to each and find the probability that at least 2 persons have the same birthmonth. Note that the sample space is E1 U E2 U E3 U E4 U E5 (meaning that they are disjoint).

I know that you need to use Baye’s Theorem and the Rule of Elimination to find the probabilities of the difficult event… the only problem is I find it difficult to find the probabilities of events E1 to E5 ‘cause my knowledge of the fundamentals of counting suck (I hate to admit it). Can anyone help me out and give pointers on when to use the fundamentals of counting when given questions such as the one above?

2. Jul 23, 2004

### theFuture

For E1={all four people have the same birthmonth}. To count this, I'd first reason that there is only 1 way they can all have the same birthmonth. But you have to chose a 1 month from 12 so #E1=12

For E2={all four have different birthmonths}. To count this I chose the first person's birthmonth from any of the 12 month (12C1}, the second person's from the remaining 11 (11C1), third from the remaning 10 (10C1), and fourth from the remaining 9 (9C1). This leaves me with #E2=(12C1)(11C1)(10C1)(9C1)

3. Jul 24, 2004

### relinquished™

I think you mean E5 is 12C1 * 11C1 * 10C1 * 9C1

but i still dont know how to get the probabilities of each event ^_^;;; or should i say, the sample space of E and the cardinality of E1 to E5

4. Aug 26, 2004

### supundika

The answers for the questions with explanations

Here I have explained in detail how I got to the probability and how I calculated each one. But we can use (12C1 etc.) and solve it as well.

If we take the four people as -> a,b,c,d

E1=[abcd] = 12 { If all four people have the same birth month then
because of the 12 months the number of chances it can happen is 12 }

E2=[abc][d],[adc]...... = 12*11*4 = 528 { If 3 of them have the same birth month they could be born in any of the 12 months and the remainning person could be in any of the 11 months. So I have multiplied 12*11. And there could be 4 different ways where 3 and only 3 people could be born in the same month like [abc][d],[adc]...... SO I have multiplied it by 4. So it should be 12*11*4 = 528 }

E3=[ab][c][d],[ac][d]....... = 6*10*10*12 = 7200
{ If two of them are born in the same month and one of the others in a
particular month then the next month could be in any of the other 10
months. And the other person could be in 10 different months as well.
So I have multiplied 10*10. And the 2people who are born in the same
month could have been born in any of the 12 months. So I multiplied it
by 12. And there are 6 different ways where 2 and only 2 people could
be born in the same month like [ab][c][d],[ac][d]....... SO I have
multiplied it by 6. So it should be 6*10*10*12 = 7200}

E4=[ab][cd],[ac][bd]........ = 12*11*6 = 792 { If two of them are born in the same month and one of the others in a particular month then the next month should be in the same month as well.So there are 11 different months where the other person could be in. And the 2people who are born in the same month could have been born in any of the 12 months. So I multiplied it by 12. And there are 6 different ways where 2 and only 2 people could be born in the same month SO I have multiplied it by 6. So it should be [ab][cd],[ac][bd]........ = 12*11*6 = 792 }

E5=[a][c][d]....... = 12*11*10*9 = 11880 { If all four of them are born in different months then one could be born in any of the 12 months and another could be in any of the other 11 and the other could be ain any of the other 10 and the other could be in any of the other 9 months. So I have mulitiplied 12*11*10*9 = 11880 }

E=E1+E2+E3+E4+E5= 20412
{Here the total number of things that could happen is 20412} And the probabilities are given below:

P[E1]=12/20412

P[E2]=528/20412

P[E3] = 7200/20412

P[E4] = 792/20412

P[E5] = 11880/20412

At least 2 people have the same birth month =
P[E1+E2+E3+E4] = 9324/20412