IIT-JEE 2009 · Paper 1 · Mathematics Q36

MathematicsSingle correct+4 / -1Question 36 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 $X=3$ equals

\frac{25}{216}

\frac{25}{36}

\frac5{36}

\frac{125}{216}

Answer and worked solution

Official answer: A

Geometric distribution. The first two tosses are not six, the third is:
$P(X=3)=\left(\frac56\right)^2\frac16=\frac{25}{216}.$

(A)

Topics:Independent events

IIT-JEE 2009 · Paper 1, Mathematics question 36