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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
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Verified Answer
The correct answer is:
\(\frac{1}{7} \leq p < \frac{3}{5}\)
According to given information, the required
\(\text {probability }=\frac{\frac{1}{2} \times \frac{n}{n+8}}{\left(\frac{1}{2} \times \frac{n}{n+8}\right)+\left(\frac{1}{2} \times \frac{6}{3+6}\right)}\)
(According to Baye's theorem)
\(=\frac{\frac{n}{n+8}}{\frac{n}{n+8}+\frac{2}{3}}=\frac{3 n}{3 n+2 n+16}=\frac{3 n}{5 n+16}=p \text { (given) }\)
For minimum value of \(p, n\) must be \(\mathrm{l}\) (as \(n \in N\) )
So, \(\quad p \geq \frac{3}{21} \Rightarrow p \geq \frac{1}{7}\)
For maximum value of \(p, n \rightarrow \infty\)
So, \(\quad p < \frac{3}{5}\)
\(\Rightarrow \quad \frac{1}{7} \leq p < \frac{3}{5}\)
Hence, option (a) is correct.
\(\text {probability }=\frac{\frac{1}{2} \times \frac{n}{n+8}}{\left(\frac{1}{2} \times \frac{n}{n+8}\right)+\left(\frac{1}{2} \times \frac{6}{3+6}\right)}\)
(According to Baye's theorem)
\(=\frac{\frac{n}{n+8}}{\frac{n}{n+8}+\frac{2}{3}}=\frac{3 n}{3 n+2 n+16}=\frac{3 n}{5 n+16}=p \text { (given) }\)
For minimum value of \(p, n\) must be \(\mathrm{l}\) (as \(n \in N\) )
So, \(\quad p \geq \frac{3}{21} \Rightarrow p \geq \frac{1}{7}\)
For maximum value of \(p, n \rightarrow \infty\)
So, \(\quad p < \frac{3}{5}\)
\(\Rightarrow \quad \frac{1}{7} \leq p < \frac{3}{5}\)
Hence, option (a) is correct.
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