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:
Given that the drawn ball from \(U_2\) is white, the probability that head appeared on the coin is
- A
\(\frac{17}{23}\)
- B
\(\frac{11}{23}\)
- C
\(\frac{15}{23}\)
- D
\(\frac{12}{23}\)
Answer & Solution
Correct Answer
(D)
\(\frac{12}{23}\)
Step-by-step Solution
Detailed explanation
\(P\) (head appeared/white from \(V_2\) )
\[
\begin{aligned}
& =P(H) \cdot \frac{\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\}}{\frac{23}{30}} \\
& =\frac{1}{2} \frac{\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\}}{\frac{23}{30}}=\frac{12}{23} \\
&
\end{aligned}
\]
\[
\begin{aligned}
& =P(H) \cdot \frac{\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\}}{\frac{23}{30}} \\
& =\frac{1}{2} \frac{\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\}}{\frac{23}{30}}=\frac{12}{23} \\
&
\end{aligned}
\]
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