JEE Main · 2023 · 12 Apr 2023 · Shift 1 · 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

Answer: 10

Solution

Let

X

be the number of tosses. The probability of success (rolling less than

n

) on any toss is

\frac{n - 1}{n}

, and failure is

\frac{1}{n}

. This is a geometric distribution with

E [X] = \sum_{k = 1}^{\infty} k \cdot (\frac{1}{n})^{k - 1} \cdot \frac{n - 1}{n} = \frac{n - 1}{n} \cdot \frac{1}{( 1 - 1/ n ) ^{2}} = \frac{n}{n - 1}

. Setting this equal to

\frac{n}{9}

\frac{n}{n - 1} = \frac{n}{9} ⟹ n - 1 = 9 ⟹ n = 10.

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer