Answer: [tex]\dfrac{7}{20}[/tex]
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)]
[tex]= 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}[/tex]
Hence, the probability of neither being blue given that neither is red is [tex]\dfrac{7}{20}[/tex] .
