Question

An urn contains 4 red balls and 3 green balls. Two balls are sampled randomly.

Let W denote the number of green balls in the sample when the draws are done with replacement. Give the possible values and the PMF of W.

Answer 1

Let W denote the number of green balls

W takes values 0,1,2,3.

We need to find the probability that there are 0, 1, 2, and 3 green balls.

Total balls in the urn = 4 red balls + 3 green balls = 7 Balls.

Method 1:

The probability that there are no green balls drawn is .

The probability that exactly one green ball is drawn is where the first term gives the probability of drawing the green as the first ball, the second term as the second ball, and third term as the third ball.

Similarly, the probability that exactly two green balls are drawn is

And the probability that exactly three green balls are drawn is

Method 2:

Alternatively, the total combinations of three balls drawn without replacement is . To get 0 green we have combinations.

P(W=0)= P(getting no green ball) =

P(W=1)= P(getting 1 green ball) =

P(W=2)= P(getting 2 green balls) =

P(W=3)= P(getting 3 green balls) =

the PMF of W:

w	P(W=w)
0	4/35
1	18/35
2	12/35
3	1/35

If we counted correctly, these numbers should add up to 1. From the final solution of Method 1, we have 64+144+108+27= 343. Good.

From Method 2 we have 4+ 18+ 12+ 1 =35