JEE Advanced · Mathematics · 32. Probability
Paragraph:
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 \(U_2\).Question:
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
\(\frac{11}{30}\)
Answer & Solution
Correct Answer
(B)
\(\frac{23}{30}\)
Step-by-step Solution
Detailed explanation
Now, probability of the drawn ball from \(\mathrm{V}_2\) being white is
\[
\Rightarrow P \text { (white/ } V_2 \text { ) }
\]
\[
\begin{aligned}
& \left.\Rightarrow P \text { (white/ } V_2\right) \\
& =P(H) \cdot\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\} \\
& +P(T)\left\{\frac{{ }^3 C_2}{{ }^5 C_2} \times \frac{{ }^3 C_2}{{ }^3 C_2}+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{{ }^1 C_1}{{ }^3 C_2}\right. \\
& \left.+\frac{{ }^3 C_1 \cdot{ }^2 C_1}{{ }^5 C_2} \times \frac{{ }^2 C_1}{{ }^3 C_2}\right\} \\
& \text { Now, } P\left(\text { white } / V_2\right)=\frac{1}{2}\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\} \\
& +\frac{1}{2}\left\{\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}\right\}=\frac{23}{30} \\
&
\end{aligned}
\]
\[
\Rightarrow P \text { (white/ } V_2 \text { ) }
\]
\[
\begin{aligned}
& \left.\Rightarrow P \text { (white/ } V_2\right) \\
& =P(H) \cdot\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\} \\
& +P(T)\left\{\frac{{ }^3 C_2}{{ }^5 C_2} \times \frac{{ }^3 C_2}{{ }^3 C_2}+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{{ }^1 C_1}{{ }^3 C_2}\right. \\
& \left.+\frac{{ }^3 C_1 \cdot{ }^2 C_1}{{ }^5 C_2} \times \frac{{ }^2 C_1}{{ }^3 C_2}\right\} \\
& \text { Now, } P\left(\text { white } / V_2\right)=\frac{1}{2}\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\} \\
& +\frac{1}{2}\left\{\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}\right\}=\frac{23}{30} \\
&
\end{aligned}
\]
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