Question 12.11E1: Selecting Colored Balls with Replacement A basket contains 3......

a)  We will consider selecting a red ball a success and selecting a ball of any other color a failure. Since only 1 of the 3 balls is red, the probability of success on any single trial, p, is \frac{1}{3}. The probability of failure on any single trial, q, is 1  –  \frac{1}{3}  =  \frac{2}{3}. We are finding the probability of selecting 0 red balls, or 0 successes. Since x represents the number of successes, we let x = 0. There are 3 independent selections (or trials), so n = 3. In our calculations, we will need to evaluate (\frac{1}{3})^0. Note that any nonzero number raised to a power of 0 is 1. Thus, (\frac{1}{3})^0  =  1. We determine the probability of 0 successes, or P(0), as follows.

P(x)  =  (_nC_x)p^xq^{n-x}
P(0)  =  (_3C_0) \left( \frac{1}{3}\right)^0  \left( \frac{2}{3}\right)^{3-0}
=  (1)(1)\left( \frac{2}{3}\right)^3
= \left( \frac{2}{3}\right)^3  =  \frac{8}{27}

b)  We are finding the probability of obtaining exactly 1 red ball or exactly 1 success in 3 independent selections. Thus, x = 1 and n = 3. We find the probability of exactly 1 success, or P(1), as follows.

P(x)  =  (_nC_x)p^xq^{n-x}
P(1)  =  (_3C_1) \left( \frac{1}{3}\right)^1  \left( \frac{2}{3}\right)^{3-1}
=  3\left( \frac{1}{3}\right)\left( \frac{2}{3}\right)^2
=  3\left( \frac{1}{3}\right)\left( \frac{4}{9}\right)  =  \frac{4}{9}

c)  We are finding the probability of selecting exactly 2 red balls in 3 independent trials. Thus, x = 2 and n = 3. We find P(2) as follows.

P(x)  =  (_nC_x)p^xq^{n-x}
P(2)  =  _3C_2 \left( \frac{1}{3}\right)^2  \left( \frac{2}{3}\right)^{3-2}
=  3\left( \frac{1}{3}\right)^2\left( \frac{2}{3}\right)^1
=  3\left( \frac{1}{9}\right)\left( \frac{2}{3}\right)  =  \frac{2}{9}

d)  We are finding the probability of selecting exactly 3 red balls in 3 independent trials. Thus, x = 3 and n = 3. We find P(3) as follows.

P(x)  =  (_nC_x)p^xq^{n-x}
P(3)  =  (_3C_3) \left( \frac{1}{3}\right)^3  \left( \frac{2}{3}\right)^{3-3}
=  1\left( \frac{1}{3}\right)^3\left( \frac{2}{3}\right)^0
=  1\left( \frac{1}{27}\right)  (1)  =  \frac{1}{27}