Answer to Question #185806 in Statistics and Probability for Betty

Question

1. There are 4 red balls, 6 blue balls and 2 white balls in a bag. Suppose on every single turn, you randomly select a ball, see the color of it and put it back in the bag. You keep doing this repeatedly.

a. What is the probability that you get the first red ball on the 5th turn?

b. How many turns are expected to get one non-white ball?

c. What is the variance of the number of turns required to get one blue ball?

2. There are 4 red balls, 6 blue balls and 2 white balls in a bag. Suppose on every single turn, you randomly select a ball, see the color of it and put it back in the bag. Let’s say, you do this 6 times.

a. What is the probability that you get exactly 3 blue balls after 6 turns?

b. What is the probability that you pick more than 4 blue balls after 6 turns?

c. What is the mean number of red balls picked after 48 turns?

d. What is the standard deviation of the number of white balls picked after 36 turns?

Answer 1

1. Since this is sampling with replacement, we use Binomial distribution.

$p_R = \frac{4}{12} \\ p_B = \frac{6}{12} \\ p_W = \frac{2}{12}$

a) the probability that you get the first red ball on the 5th turn

$= p_R(1 -p_R)^{5-1} \\ = \frac{4}{12}(1 - \frac{4}{12})^4 \\ = 0.0658$

b) turns are expected to get one non-white ball

The probability of getting one non-white ball:

$p_{NW} = \frac{4+6}{12} = \frac{10}{12} = \frac{5}{6}$

Let Y denotes the turns are expected to get one non-white ball. Then

Y follow geometric distribution with expectation (mean):

$E(Y) = \frac{1}{p} = \frac{1}{\frac{5}{6}} \\ = \frac{6}{5} \\ = 1.2$

c) the variance of the number of turns required to get one blue ball

$p_B = \frac{6}{12} = 0.5$

Let Y denotes the turns required to get one blue ball, then Y follows a geometric distribution with variance:

$V(Y) = \frac{1-p}{p^2} \\ = \frac{1-0.5}{0.5^2} \\ = 2$

2. Since this is sampling with replacement, we use Binomial distribution.

$p_R = \frac{4}{12} \\ p_B = \frac{6}{12} \\ p_W = \frac{2}{12}$

a) the probability that you get exactly 3 blue balls after 6 turns

$= \binom{6}{3}p_B^3(1 -p_B)^3 \\ = \binom{6}{3}(\frac{6}{12})^3(1 - \frac{6}{12})^3 \\ = 0.3125$

b) the probability that you pick more than 4 blue balls after 6 turns

$= \binom{6}{5}p_B^5(1 -p_B) + \binom{6}{6}p_B^6(1 -p_B)^0 \\ = \binom{6}{5}(\frac{6}{12})^5(1 - \frac{6}{12}) + \binom{6}{6}(\frac{6}{12})^6(1 -\frac{6}{12})^0 \\ = 0.109375$

c) the mean number of red balls picked after 48 turns

$np_R = 48(\frac{4}{12}) = 16$

d) the standard deviation of the number of white balls picked after 36 turns

$= \sqrt{np_W(1-p_W)} \\ = \sqrt{36 \times \frac{2}{12} \times \frac{10}{12}} \\ = 2.236$