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Suppose that 10balls are put into 5boxes, with each ball independently being put in box iwith probability pi,∑i=15pi=1

(a) Find the expected number of boxes that do not have any balls.

(b) Find the expected number of boxes that have exactly 1ball.

Short Answer

Expert verified

(a) The expected number of boxes that do not have any balls is∑i=151-pi10

(b) The expected number of boxes that have exactly 1 ball is∑i=1510pi1-pi9

Step by step solution

01

Given information(part a)

Given in the question that, Suppose that 10balls are put into 5boxes, with each ball independently being put in box iwith probability pi,∑i=15pi=1.

02

Step 2:Explanation (Part a)

Given:

10balls are put into 5boxes independently

Probability of a ball being placed in box iis pi.

∑i=15pi=1

The number of successes among a selected number of independent trials with a persistent probability of success observes a binomial distribution. Definition binomial probability:

P(X=k)=nk·pk·(1-p)n-k=n!k!(n-k)!·pk·(1-p)n-k

Let us evaluate the definition of binomial probability at n=10,p=piand k=0:

P(box iis empty )=P(X=0)

=100·pi0·1-pi10-0

=1·1·1-pi10

=1-pi10

The expected value (or mean) is the sum of the product of each possibility x(number of boxes) with its probability P(x).

E(empty boxes )=∑xP(x)

=∑i=151×1-pi10

=∑i=151-pi10

03

Final answer (Part a)

The expected number of boxes that do not have any balls is∑i=15 1−pi10

04

Given information(part b)

Suppose that10balls are put into 5 boxes, with each ball independently being put in boxi with probabilitypi,∑i=15pi=1

05

Step 5:  Explanation (Part b)

Now we define the indicator random variable

Ii=1 â¶Ä…â¶Ä…â¶Ä…;1thbox has one ball0 â¶Ä…â¶Ä…â¶Ä…;otherwise

Now the indicator random variable follows a binomial distribution

PIi=1=101pi11−pi9

Now, the number of boxes contains one ball is given by ∑i=15 Ii

Expected number of boxes that have exactly 1 ball =E∑i=15 Ii

=∑i=15 EIi

=∑i=15 101pi11−pi9

06

Step 6:Final answer

Expected number of boxes that have exactly 1 ball is∑i=15 101pi11−pi9

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