# A fair die is tossed until a number greater than 4 appears. The probability that an even number of tosses shall be required is

$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{3}{5} \\ (C)\;\large\frac{1}{5} & \quad (D)\;\large\frac{2}{3} \end {array}$

## 1 Answer

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• Sum of infinite G.P.$=\large\frac{a}{1-r}$ where $a$ is $1^{st}$ term and $r=$common ratio.
Probability of getting a number greater than $4=\large\frac{2}{6}$
Probability of getting a number not greater than $4=1-\large\frac{2}{6}=\frac{4}{6}$
Let a number greater than 4 appears after exactly 'n' tosses.
Then the probability of this event is $\bigg( \large\frac{4}{6} \bigg)^n \bigg( \large\frac{2}{6} \bigg)= \bigg( \large\frac{2}{3} \bigg)^n \bigg( \large\frac{1}{3} \bigg)$
Given that $n$ takes even values.
$\Rightarrow \:n$ can take values 0, 2, 4, 6....
Hene the required probability is $\large\frac{1}{3} \bigg[ 1+\large\frac{4}{9}+\large\frac{16}{81}+.... \bigg]$
$1+\large\frac{4}{9}+\frac{16}{81}+........$ is infinite G.P. with $r=\large\frac{4}{9}$
We know that sum of infinite G.P.$=\large\frac{a}{1-r}$ where $a$ is $1^{st}$ term.
$\Rightarrow\:$Required probability $= \large\frac{1}{3} \times \large\frac{1}{1-\large\frac{4}{9}}$
$= \large\frac{3}{5}$
Ans : (B)
answered Jan 8, 2014
edited Mar 26, 2014

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