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

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)]

Hence, the probability of neither being blue given that neither is red is .

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