IIT-JEE 2009 · Paper 1 · Mathematics Q37

MathematicsSingle correct+4 / -1Question 37 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 probability that $X\ge3$ equals

\frac{125}{216}

\frac{25}{36}

\frac5{36}

\frac{25}{216}

Answer and worked solution

Official answer: B

$X\ge3$ means the first two tosses are not six:
$P(X\ge3)=\left(\frac56\right)^2=\frac{25}{36}.$

(B)

Topics:Independent events

IIT-JEE 2009 · Paper 1, Mathematics question 37