IIT-JEE 2009 · Paper 1 · Mathematics Q38

MathematicsSingle correct+4 / -1Question 38 of 60

Passage

Paragraph for Question Nos. 36 to 38

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

The conditional probability that $X\ge6$ given $X>3$ equals

\frac{125}{216}

\frac{25}{216}

\frac5{36}

\frac{25}{36}

Answer and worked solution

Official answer: D

Memorylessness. $P(X>k)=\left(\frac56\right)^k$ , so
$P(X\ge6\mid X>3)=\frac{P(X>5)}{P(X>3)}=\left(\frac56\right)^2=\frac{25}{36}.$

(D)

Topics:Conditional probability Independent events

IIT-JEE 2009 · Paper 1, Mathematics question 38