Selecting Colored Balls with Replacement
A basket contains 3 balls: 1 red, 1 blue, and 1 yellow. Three balls are going to be selected with replacement from the basket. Find the probability that
a) no red balls are selected.
b) exactly 1 red ball is selected.
c) exactly 2 red balls are selected.
d) exactly 3 red balls are selected.
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}