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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

  1. A \(11\)
  2. B \(12\)
  3. C \(13\)
  4. D \(10\)
Verified Solution

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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