Mathematics, 31.05.2020 04:58 aleilyg2005
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
Answers: 1
![= 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}](/tpl/images/0672/5086/d633b.png)
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Rework problem 22 from section 3.2 of your text, involving the colored balls. For this problem, assu...
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