JEE Advanced · Mathematics · 32. Probability
Paragraph:
A fair die is tossed repeatedly until a six is obtained. Let \(X\) denotes the number of tosses required.Question:
The conditional probability that \(X \geq 6\) given \(X>3\) equals
- A
\(\frac{125}{216}\)
- B
\(\frac{25}{216}\)
- C
\(\frac{5}{36}\)
- D
\(\frac{25}{36}\)
Answer & Solution
Correct Answer
(D)
\(\frac{25}{36}\)
Step-by-step Solution
Detailed explanation
\(P((X \geq 6) /(X>3))\)
\[
\begin{aligned}
& =\frac{P((X>3) /(X \geq 6)) \cdot P(X \geq 6)}{P(X>3)} \\
& =\frac{1 \cdot\left[\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \infty\right]}{\left[\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \infty\right]}=\frac{25}{36} \\
&
\end{aligned}
\]
\[
\begin{aligned}
& =\frac{P((X>3) /(X \geq 6)) \cdot P(X \geq 6)}{P(X>3)} \\
& =\frac{1 \cdot\left[\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \infty\right]}{\left[\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \infty\right]}=\frac{25}{36} \\
&
\end{aligned}
\]
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