JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Probability XII

A fair die is tossed repeatedly until a six is obtained. Let

X

denote the number of tosses required.

The probability that

X \geq 3

equals

(A) $\frac{125}{216}$
(B) $\frac{25}{36}$ ✓ Correct
(C) $\frac{5}{36}$
(D) $\frac{25}{216}$

Answer: (B)

Solution

The probability that

X \geq 3

is the complement of

X \leq 2

P (X \leq 2) = P (X = 1) + P (X = 2)

= \frac{1}{6} + \frac{5}{6} \times \frac{1}{6} = \frac{1}{6} + \frac{5}{36} = \frac{11}{36}

Therefore:

P (X \geq 3) = 1 - \frac{11}{36} = \frac{25}{36}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option

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