JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 5)

A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.

The probability that $$X\ge3$$ equals :

$${{125} \over {216}}$$

$${{25} \over {36}}$$

$${{5} \over {36}}$$

$${{25} \over {216}}$$

The probability $$P(X \ge 3)$$ is nothing but the probability of $$P(X \le 2)$$:

$${1 \over 6} + {5 \over 6} \times {1 \over 6} = {{11} \over {36}}$$

The required probability is

$$1 - {{11} \over {36}} = {{25} \over {36}}$$