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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2011 · Paper 1 · Mathematics · Probability

Paragraph: Let U_1 and U_2 be two urns such that U_1 contains 3...

Paragraph: Let U1U_1 and U2U_2 be two urns such that U1U_1 contains 3 white and 2 red balls and U2U_2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1U_1 and put into U2U_2. However, if tail appears then 2 balls are drawn at random from U1U_1 and put into U2U_2. Now, 1 ball is drawn at random from U2U_2.Question: Given that the drawn ball from U2U_2 is white, the probability that head appeared on the coin is

Options

  1. A1723\frac{17}{23}
  2. B1123\frac{11}{23}
  3. C1523\frac{15}{23}
  4. D1223\frac{12}{23}

Correct answer

D. 1223\frac{12}{23}

Step-by-step solution

P(headappeared/whitefrom(head appeared/white fromV_2))=P(H)⋅{3C15C1×2C12C1+2C15C1×1C12C1}2330=12{35×1+25×12}2330=1223\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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