Joint Distribution Problem

  • Thread starter asdf71
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  • #1
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Hello,

I have a problem I need help with:

Homework Statement



There are 10 different types of coupons, each with prob 1/10 of being chosen. A total of N>=1 coupons are collected.
Xi= 1 if type-i coupon is among the N coupons, i =1, ..., 10
0 otherwise
Let S=X1+X2+X3 be the number of different types out of the subset {1,2,3} contained in the collection.
a) Find P(S=k), k=0,1,2,3


Homework Equations



Joint pmf: p(x1,x2,x3)=p(x1)*p(x2|x1)*p(x3|x1,x2)


The Attempt at a Solution



I know that:

P(Xi=0)=(9/10)^N
P(Xi=1)=1-(9/10)^N

Therefore

P(X1=0,X2=0,X3=0)= p(0)*p(0|0)*p(0|0,0)= (9/10)^N * (8/9)^N * (7/8)^N = (7/10)^N
p(0|0)=(8/9)^N because if we know that there are no type-1 coupon among the N, we know that there are only 9 types.

P(X1=0,X2=0,X3=1)= p(0)*p(0|0)*p(1|0,0)= (9/10)^N * (8/9)^N * (1-7/8)^N = (8/10)^N - (7/10)^N

Next probability is more difficult:
P(X1=0,X2=1,X3=1)= p(0)*p(1|0)*p(1|0,1)=

How do I find p(1|0,1)?

This is probability that there is a type-3 coupon among the N, given that there is no type-1 coupon, but there is type-2 coupon.
There is no type-1 coupon, so there is only 9 types left.
There is a type-2 coupon, so there is at least one type-2 coupon among the N, but I don't know how many... so I can only assume that there is one type-2 coupon? This would be p(1|0,1)=(1-(8/9)^(N-1)) ?

After that P(X1=1,X2=1,X3=1) is 1 minus all the other probabilities, and my problem is complete.

The only probability I can't find is p(1|0,1) and if anyone could help me it would be very appreciated! Thanks!!
 

Answers and Replies

  • #2
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I think it goes something like this:
if you have an infinite pile of coupons from which you select n, or if you put the coupon back after you look at it then:
Notice that P(x_1)=P(x_2)=P(x_3) because they are independent. Then you are performing three bernouli trials. What kind of probability is that?
 

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