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 conditional probability that

X \geq 6

given

X > 3

equals

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

Answer: (D)

Solution

We need

P (X \geq 6∣ X > 3) = \frac{P ( X \geq 6 )}{P ( X > 3 )}

.

For a geometric distribution with success probability

p = \frac{1}{6}

P (X \geq k) = (1 - p)^{k - 1} = (\frac{5}{6})^{k - 1}

Thus:

P (X \geq 6) = (\frac{5}{6})^{5}

P (X > 3) = P (X \geq 4) = (\frac{5}{6})^{3}

Therefore:

P (X \geq 6∣ X > 3) = \frac{( 5/6 ) ^{5}}{( 5/6 ) ^{3}} = (\frac{5}{6})^{2} = \frac{25}{36}

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

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