Probability Ques 119
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)
The probability of the drawn ball from $U _2$ being white, is
(a) $\frac{13}{30}$
(b) $\frac{23}{30}$
(c) $\frac{19}{30}$
(d) $\frac{11}{30}$
Show Answer
Answer:
Correct Answer: 119.(b)
Solution:
- Now, probability of the drawn ball from $U _2$ being white is
$ \begin{aligned} & \left.P \text { (white } / U _2\right)=P(H) \cdot[ \frac{{ }^{3} C _1}{{ }^{5} C _1} \times \frac{{ }^{2} C _1}{{ }^{2} C _1}+\frac{{ }^{2} C _1}{{ }^{5} C _1} \times \frac{{ }^{1} C _1}{{ }^{2} C _1} ]\\ & \quad+P(T) [\frac{{ }^{3} C _2}{{ }^{5} C _2} \times \frac{C _2}{{ }^{3} C _2}+\frac{{ }^{2} C _2}{{ }^{5} C _2} \times \frac{{ }^{1} C _1}{{ }^{3} C _2}+\frac{{ }^{3} C _1 \cdot{ }^{2} C _1}{{ }^{5} C _2} \times \frac{{ }^{2} C _1}{{ }^{3} C _2}] \end{aligned} $
$ \begin{aligned} =\frac{1}{2} & [\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}] \\ & +\frac{1}{2} [\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}]=\frac{23}{30} \end{aligned} $
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