A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is:
$A. 1/2$
$B. 3/5$
$C. 1/5$
$D. 2/3$
What I did: The probability should be $(2/3)*(1/3)+(2/3)*(2/3)*(2/3)*(1/3)+(2/3)*(2/3)(2/3)*(2/3)*(2/3)*(1/3)+...$, which should be $2/5$.
But this answer doesn't match with the options. Where did I go wrong?

You didn't go wrong. The probability of "success" is 1/3, so the probability of succeding in $2k$ tosses  is
$$\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}$$
And
$$\sum_{k=1}^\infty\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}=\frac{1}{3} \frac{3}{2}\sum_{k=1}^\infty\left(\frac{4}{9}\right)^k =\frac{1}{2} \frac{4/9}{1-4/9}=\frac{2}{5}$$
