Probability 4 Question 9
9. 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) $\frac{11}{30}$
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Answer:
Correct Answer: 9. (b)
Solution:
- Now, probability of the drawn ball from $U _2$ being white is
$$ \begin{aligned} & \left.P \text { (white } / U _2\right)=P(H) \cdot \frac{{ }^{3} C _1}{{ }^{5} C _1} \times \frac{{ }^{2} C _1}{{ }^{2} C _1}+\frac{{ }^{2} C _1}{{ }^{5} C _1} \times \frac{{ }^{1} C _1}{{ }^{2} C _1} \\ & \quad+P(T) \frac{{ }^{3} C _2}{{ }^{5} C _2} \times \frac{C _2}{{ }^{3} C _2}+\frac{{ }^{2} C _2}{{ }^{5} C _2} \times \frac{{ }^{1} C _1}{{ }^{3} C _2}+\frac{{ }^{3} C _1 \cdot{ }^{2} C _1}{{ }^{5} C _2} \times \frac{{ }^{2} C _1}{{ }^{3} C _2} \end{aligned} $$
$$ \begin{aligned} =\frac{1}{2} & \frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2} \\ & +\frac{1}{2} \frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}=\frac{23}{30} \end{aligned} $$
