Question:
An unbiased die is tossed until 5 appears. If \( X \) denotes the number of tosses required, \( \frac{25}{P(X=5)} \) is equal to
An unbiased die is tossed until 5 appears. If \( X \) denotes the number of tosses required, \( \frac{25}{P(X=5)} \) is equal to
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For repeated trials until success, always use geometric distribution formula.
Updated On: Apr 21, 2026
- \( \frac{25}{36} \)
- \( \frac{125}{216} \)
- \( \frac{216}{125} \)
- \( \frac{36}{25} \)
- \( \frac{216}{25} \)
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The Correct Option is C
Solution and Explanation
Concept:
Geometric distribution:
\[
P(X=n) = \left(\frac{5}{6}\right)^{n-1}\left(\frac{1}{6}\right)
\]
Step 1: Find \( P(X=5) \).
\[ P(X=5) = \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} = \frac{625}{1296} \]
Step 2: Compute required value.
\[ \frac{25}{P(X=5)} = \frac{25}{625/1296} = 25 \cdot \frac{1296}{625} \] \[ = \frac{32400}{625} = \frac{216}{125} \]
Step 1: Find \( P(X=5) \).
\[ P(X=5) = \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} = \frac{625}{1296} \]
Step 2: Compute required value.
\[ \frac{25}{P(X=5)} = \frac{25}{625/1296} = 25 \cdot \frac{1296}{625} \] \[ = \frac{32400}{625} = \frac{216}{125} \]
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