Urn 1contains 5white and 6black balls, while urn 2contains 8white and 10black balls. Two balls are randomly selected from urn 1and are put into urn 2. If 3balls are then randomly selected from urn 2, compute the expected number of white balls in the trio.

Hint: LetXi = 1if the i th white ball initially in urn 1is one of the three selected, and let Xi = 0otherwise. Similarly, let Yi = 1if the i the white ball from urn 2is one of the three selected, and let Yi = 0otherwise. The number of white balls in the trio can now be written as15Xi+18Yi

Short Answer

Expert verified

The total number of chosen white balls is,

N=i=15Xi+i=18Yi
E(N) =147110

Step by step solution

01

Given Information

No.of balls in urn 1are White balls and 6Black ball. No. of balls in urn 2are 8White balls and 10Black balls. LetXi = 1if the i th white ball initially in urn 1is one of the three selected, and let Xi = 0otherwise. Similarly,letYi =1if the i th white ball from urn 2is one of the three selected, and let Yi = 0otherwise. The number of white balls in the trio can now be written as15Xi+18Yi

02

Explanation

Define indicator random variables Xi that are equal to 1if and only if i th white ball from the urn 1 has been chosen in the trio, i=1, ..., 5. Similarly, define indicator random variables Yi that are equal to 1 if and only if i th white ball from the urn 2has been chosen in the trio, i=1, ..., 8. The total number of chosen white balls is

N=i=15Xi+i=18Yi

and using the linearity of the expectation, we have that

E(N)=i=15E(Xi)+i=18E(Yi)

03

Explanation

The probability that a certain White ball from urn 1is contained in the trio is equal to

E(Xi)=P(Xi=1)=211320

The probability that a certain White ball from urn 2is contained in the trio is equal to

E(Yi)=P(Yi=1)=320

Therefore

E(N)=5211320+8320=147110

04

Final Answer

The total number of chosen white balls is,

E(N)=i=15E(Xi)+i=18E(Yi)

E(N) =147110

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