JEE Mains · Maths · STD 12 - 13. probability
A fair \(n(n > 1)\) faces die is rolled repeatedly until a number less than \(n\) appears. If the mean of the number of tosses required is \(\frac{ n }{9}\), then \(n\) is equal to
- A \(11\)
- B \(12\)
- C \(13\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
Mean \(=1 \cdot \frac{ n -1}{ n }+2 \frac{1}{ n }\left(\frac{ n -1}{ n }\right)+3\left(\frac{1}{ n }\right)^2\left(\frac{ n -1}{ n }\right)\) \(\frac{ n }{9}=\left(\frac{ n -1}{ n }\right)\left(1+2\left(\frac{1}{ n }\right)+3\left(\frac{1}{ n }\right)^2 \ldots \ldots\right)\)…
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