Probability 4 Question 8
8. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\frac{1}{3}$, then the correct option(s) with the possible values of $n _1$ and $n _2$ is/are
(a) $n _1=4$ and $n _2=6$
(b) $n _1=2$ and $n _2=3$
(c) $n _1=10$ and $n _2=20$
(d) $n _1=3$ and $n _2=6$
Passage II
Let $U _1$ and $U _2$ be two urns such that $U _1$ contains 3 white and 2 red balls and $U _2$ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from $U _1$ and put into $U _2$. However, if tail appears then 2 balls are drawn at random from $U _1$ and put into $U _2$. Now, 1 ball is drawn at random from $U _2$.
(2011)
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Answer:
Correct Answer: 8. (d)
Solution:
$\therefore P\left(\right.$ drawing red ball from $\left.B _1\right)=\frac{1}{3}$
$$ \begin{aligned} & \Rightarrow \frac{n _1-1}{n _1+n _2-1} \quad \frac{n _1}{n _1+n _2}+\frac{n _2}{n _1+n _2} \quad \frac{n _1}{n _1+n _2-1}=\frac{1}{3} \\ & \Rightarrow \quad \frac{n _1^{2}+n _1 n _2-n _1}{\left(n _1+n _2\right)\left(n _1+n _2-1\right)}=\frac{1}{3} \end{aligned} $$
Clearly, options (c) and (d) satisfy.
Passage II
$$ \left.\left(\begin{array}{c} 3 W \\ 2 R \end{array}\right) \quad\left(\begin{array}{c} 1 W \\ U _1 \end{array}\right) } \text { Initial } $$
Head appears
Tail appears
