Suppose that a bag \(A\) contains \(n\) red and 2 black balls and another bag \(B\) contains 2 red and \(n\)…
Suppose that a bag \(A\) contains \(n\) red and 2 black balls and another bag \(B\) contains 2 red and \(n\) black balls. One of the two bags is selected at random and two balls are drawn from it at a time. When it is known that the two balls drawn are red, if the probability that those two balls drawn are from bag \(A\) is \(\frac{6}{7}\), then \(n=\)
6
4
8
7
Solution
Let \(E_1\) be the event that the ball is drawn from bag \(A. E_2\) be the event that it is drawn from bag \((B)\) and \(E\) that ball is red.
Given,
\(P\left(\frac{E_1}{E}\right)=\frac{6}{7}\)
By Baye's, theorem,
\(\begin{gathered}
P\left(\frac{E_1}{A}\right) \\
\frac{\frac{1}{2}\left(\frac{{ }^n C_2}{{ }^{n+2} C_2}\right)}{\frac{1}{2} \frac{{ }^n C_2}{{ }^{n+2} C_2}+\frac{1}{2} \frac{{ }^2 C_2}{{ }^{n+2} C_2}}=\frac{6}{7} \\
\Rightarrow \quad \frac{\frac{1}{2}\left(\frac{{ }^n C_2}{{ }^{n+2} C_2}\right)}{\frac{1}{2}\left[\frac{{ }^n C_2}{{ }^{n+2} C_2}+\frac{{ }^2 C_2}{n+2}\right]}=\frac{6}{7} \\
\Rightarrow \quad \frac{\frac{n(n-1)}{2}}{\frac{2+n(n-1)}{2}}=\frac{6}{7}
\end{gathered}\)
\(\begin{aligned}
& \Rightarrow \quad \frac{n(n-1)}{n^2-n+2}=\frac{6}{7} \\
& \Rightarrow \quad 7\left(n^2-n\right)=6 n^2-6 n+12 \\
& \Rightarrow \quad 7 n^2-7 n-6 n^2+6 n=12 \\
& \Rightarrow \quad n^2-n=12 \\
& \Rightarrow \quad n^2-n-12=0 \\
& \Rightarrow \quad n^2-4 n+3 n-12=0 \\
& \Rightarrow \quad n(n-4)+3(n-4)=0 \\
& \Rightarrow \quad(n-4)(n+3)=0 \\
& \Rightarrow \quad n=4 \quad[\because n > 0]
\end{aligned}\)