Question

Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box i with probability pi, 5 i=1 pi = 1. (a) find the expected number of boxes that do not have any balls. (b) find the expected number of boxes that have exactly 1 ball.

Answer 1

a) 5 balls
b) 5 boxs.

Answer 2

Hey.
Here is your answer.

We first describe informally the probability model. We grab the first ball, and choose at random a box to put this ball into, with all choices equally likely. We then independently go through the same procedure with the second ball, the third ball, and so on.

Expected Number of Empty Boxes: For i=1,2,…,5i=1,2,…,5, define the random variable XiXi by Xi=1Xi=1 if Box ii ends up with zero balls, and by Xi=0Xi=0 otherwise. Let

X=X1+X2+X3+X4+X5.X=X1+X2+X3+X4+X5.

Then XX is the total number of boxes that end up with zero balls in them. Note that

E(X)=E(X1+X2+⋯+X5)=E(X1)+E(X2)+⋯+E(X5).E(X)=E(X1+X2+⋯+X5)=E(X1)+E(X2)+⋯+E(X5).

Next we calculate E(Xi)E(Xi). For any ii, Xi=1Xi=1 if 1010 times in a row we chose one of the other boxes. Thus P(Xi=1)=(4/5)10P(Xi=1)=(4/5)10. It follows that

E(Xi)=(45)10.E(Xi)=(45)10.

Now the calculation of E(X)E(X) is easy:

E(X)=5(45)10.

Expected Number with 11 Ball: The same idea works. Let random variable YiYi have value 11 if Box ii ends up with 11 ball, and value 00 otherwise. Let Y=Y1+Y2+⋯+Y5Y=Y1+Y2+⋯+Y5. Then YY is the number of boxes with precisely 11 ball. We want E(Y)E(Y).

The probability that Box ii has precisely one ball is given by

P(Yi=1)=(101)(15)1(45)9.P(Yi=1)=(101)(15)1(45)9.

Then E(Yi)=P(Yi=1)E(Yi)=P(Yi=1), and E(Y)=5E(Yi)E(Y)=5E(Yi).

Thanks.