Question

Rework problem 22 from section 3.2 of your text, involving the colored balls. For this problem, assume the box contains 6 blue balls, 6 red balls, and 4 white balls, and that we choose two balls at random from the box. What is the probability of neither being blue given that neither is red

Answer 1

Answer: $\dfrac{7}{20}$

Step-by-step explanation:

Given , there are 6 blue balls, 6 red balls, and 4 white balls, and that we choose two balls at random from the box.

P(neither blue nor red) = 1-P(blue or red)

= 1- [P(blue)+P(red)-P(blue and red)]

$= 1-\left [\dfrac{6}{16}+\dfrac{6}{16}-\dfrac{^{6}C_1\times\ ^{6}C_1}{^{16}C_2}\right]\\\\=1-\left[\dfrac{3}{4}-\dfrac{6\times6}{\dfrac{16\times15\times14!}{2!14!}}\right]\\\\=1-\left[\dfrac{3}{4}-\dfrac{12}{8\times15}\right]\\\\=1-\left[\dfrac{3}{4}-\dfrac{1}{2\times5}\right]\\\\=1-\dfrac{13}{20}\\\\=\dfrac{7}{20}$

Hence, the probability of neither being blue given that neither is red is $\dfrac{7}{20}$ .