An urn contains 3 red and 2 white balls. A ball is selected, replaced, and one of the opposite
color is added. This is repeated three times. Find the range distribution, mean and variance of the
number X of red balls drawn.
Solution
The simplest way is to determine the probability of all 8 selections, then collect probabilities.
RRR
RRW
RWR
RWW
WRR
WRW
WWR
WWW
RRR
We start with 3 red balls and 2 white balls.
First probability of red is 3/5
Selecting a red, we add a white, so we have 3 reds and 3 whites
Probability of red is 3/6
Selecting a red, we add a white, so we have 3 reds and 4 whites
Probability of red is 3/7
We then select a red so we add another white, so we have 3 red balls.
RRW
We start with 3 red balls and 2 white balls.
First probability of red is 3/5

Selecting a red, we add a white, so we have 3 reds and 3 whites
Probability of red is 3/6
Selecting a red, we add a white, so we have 3 reds and 4 whites
Probability of white is 4/7
We then select a white, so we add another red, so we have 4 red balls.
RWR
We start with 3 red balls and 2 white balls.
First probability of red is 3/5
Selecting a red, we add a white, so we have 3 reds and 3 whites
Probability of white is 3/6
Selecting a white, we add a red, so we have 4 reds and 3 whites
Probability of red is 4/7
We then select a red, so we add another white, so we have 4 red balls.
RWW
We start with 3 red balls and 2 white balls.
First probability of red is 3/5
Selecting a red, we add a white, so we have 3 reds and 3 whites
Probability of white is 3/6
Selecting a white, we add a red, so we have 4 reds and 3 whites
Probability of white is 3/7
We then select a white, so we add another red, so we have 5 red balls.
WRR
We start with 3 red balls and 2 white balls.
First probability of white is 2/5
Selecting a white, we add a red, so we have 4 reds and 2 whites
Probability of red is 4/6
Selecting a red, we add a white, so we have 4 reds and 3 whites
Probability of red is 4/7
We then select a red so we add another white, so we have 4 red balls.

WRW
We start with 3 red balls and 2 white balls.
First probability of white is 2/5
Selecting a white, we add a red, so we have 4 reds and 2 whites
Probability of red is 4/6
Selecting a red, we add a white, so we have 4 reds and 3 whites
Probability of white is 3/7
We then select a white, so we add another red, so we have 5 red balls.
WWR
We start with 3 red balls and 2 white balls.
First probability of white is 2/5
Selecting a white, we add a red, so we have 4 reds and 2 whites
Probability of white is 2/6
Selecting a white, we add a red, so we have 5 reds and 2 whites
Probability of red is 5/7
We then select a red, so we add another white, so we have 5 red balls.
WWW
We start with 3 red balls and 2 white balls.
First probability of white is 2/5
Selecting a white, we add a red, so we have 4 reds and 2 whites
Probability of white is 2/6
Selecting a white, we add a red, so we have 5 reds and 2 whites
Probability of white is 2/7
We then select a white, so we add another red, so we have 6 red balls.
P(3) = P(RRR) = 3/5 * 3/6 * 3/7 = 9/70
P(2) = P(RRW) + P(RWR) + P(WRR) =
3/5*3/6*4/7 + 3/5*3/6*4/7+ 2/5*4/6*4/7 = 52/105
P(1) = P(RWW) + P(WRW) + P(WWR) =
3/5*3/6*3/7 + 2/5*4/6*3/7 + 2/5*2/6*5/7 = 71/210

P(0) = 2/5*2/6*2/7 = 4/105
We can select between 0 and 3 reds, so the range is 3 - 0 = 3
Mean =9/70*3 + 52/105 * 2 + 71/210 * 1 + 4/105 * 0 = 12/7, or 1.7142857
E(x^2) =9/70*3^2 + 52/105 * 2^2 + 71/210 * 1^2 + 4/105 * 0^2 = 73/21, or 3.47619047619048
Then, the variance isE(x^2) - (E(x))^2 =73/21 - (12/7)^2 = 79/147 = 0.537414965986393