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  4. A Box B1 contains 3 blue balls and 6 red balls Another Box B2 contains 8 blue balls and 'n' red balls (n ∈ N) . A ball selected at random from a box is found to be red. If p is the probability that this red ball drawn is from box B2, then

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Q. A Box $B_{1}$ contains $3$ blue balls and $6$ red balls Another Box $B_{2}$ contains $8$ blue balls and 'n' red balls $(n \in N) .$ A ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_{2}$, then

AP EAMCETAP EAMCET 2019

Solution:

According to given information, the required probability
$=\frac{\frac{1}{2} \times \frac{n}{n+8}}{\left(\frac{1}{2} \times \frac{n}{n+8}\right)+\left(\frac{1}{2} \times \frac{6}{3+6}\right)}$
(According to Baye's theorem)
$=\frac{\overline{n+8}}{\frac{n}{n+8}+\frac{2}{3}}=\frac{3 n}{3 n+2 n+16}=\frac{3 n}{5 n+16}=p$(given)
For minimum value of $p, n$ must be 1 (as $n \in N)$
So, $ p \geq\, \frac{3}{21} $
$\Rightarrow \,p \geq \frac{1}{7}$
For maximum value of $p, n \to \infty$
So,$ p<\,\frac{3}{5} $
$\Rightarrow \, \frac{1}{7} \leq\, p<\,\frac{3}{5}$