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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Mathematics
Chapter
Probability

JEE Advanced 2009 · Paper 1 · Mathematics · Probability

Paragraph: A fair die is tossed repeatedly until a six is obtained....

Paragraph: A fair die is tossed repeatedly until a six is obtained. Let XX denotes the number of tosses required.Question: The conditional probability that X≥6X \geq 6 given X>3X>3 equals

Options

  1. A125216\frac{125}{216}
  2. B25216\frac{25}{216}
  3. C536\frac{5}{36}
  4. D2536\frac{25}{36}

Correct answer

D. 2536\frac{25}{36}

Step-by-step solution

P((X \geq 6) /(X \gt 3)) =P((X>3)/(X≥6))⋅P(X≥6)P(X>3)=1⋅[(56)5⋅16+(56)6⋅16+…∞][(56)3⋅16+(56)4⋅16+…∞]=2536\begin{aligned}& =\frac{P((X>3) /(X \geq 6)) \cdot P(X \geq 6)}{P(X>3)}\\ & =\frac{1 \cdot\left[\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \infty\right]}{\left[\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \infty\right]}=\frac{25}{36}\\ & \end{aligned}
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