A box B_1 contains 3 blue balls and 6 red bal | vedclass

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(A) Let $E_1$ be the event of selecting box $B_1$ and $E_2$ be the event of selecting box $B_2$. Since the box is selected at random,$P(E_1) = P(E_2)

Vedclass · Accepted Answer

$\frac{1}{7} \leq p < \frac{3}{5}$

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(A) Let $E_1$ be the event of selecting box $B_1$ and $E_2$ be the event of selecting box $B_2$. Since the box is selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.Let $R$ be the event that the selected ball is red.The probability of drawing a red ball from $B_1$ is $P(R|E_1) = \frac{6}{3+6} = \frac{6}{9} = \frac{2}{3}$.The probability of drawing a red ball from $B_2$ is $P(R|E_2) = \frac{n}{n+8}$.Using Bayes' theorem,the probability $p$ that the red ball is from $B_2$ is:$p = P(E_2|R) = \frac{P(E_2)P(R|E_2)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)}$$p = \frac{\frac{1}{2} 	imes \frac{n}{n+8}}{\frac{1}{2} 	imes \frac{2}{3} + \frac{1}{2} 	imes \frac{n}{n+8}} = \frac{\frac{n}{n+8}}{\frac{2}{3} + \frac{n}{n+8}} = \frac{3n}{2(n+8) + 3n} = \frac{3n}{5n+16}$.Since $n \in N$,the minimum value of $n$ is $1$. For $n=1$,$p = \frac{3(1)}{5(1)+16} = \frac{3}{21} = \frac{1}{7}$.As $n 	o \infty$,$p = \lim_{n 	o \infty} \frac{3n}{5n+16} = \frac{3}{5}$.Since $p$ is an increasing function of $n$,the range of $p$ is $\frac{1}{7} \leq p < \frac{3}{5}$.

$A$ 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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