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

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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)

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