A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

The correct option is 5/11.

Let us solve the problem step-by-step.
We are looking for the probability that the number 2 appears for the first time in an even number of throws of a fair die.

Let $p$ be the probability of getting a 2 in a single throw of a fair die. Since a fair die has 6 faces and only one of them is 2, we have:

$p=\frac{1}{6}$

Let $q$ be the probability of not getting a 2 in a single throw:

$q=1-p=1-\frac{1}{6}=\frac{5}{6}$

We want the number 2 to appear for the first time in an even number of throws. This means the first 2 can appear on the 2nd throw, the 4th throw, the 6th throw, and so on.
Let us write down the probabilities for each of these mutually exclusive cases:

1. The first 2 appears on the 2nd throw: This means we fail on the 1st throw and succeed on the 2nd throw. The probability is:
$qp$

2. The first 2 appears on the 4th throw: This means we fail on the first 3 throws and succeed on the 4th throw. The probability is:
$q^{3}p$

3. The first 2 appears on the 6th throw: The probability is:
$q^{5}p$

In general, the probability that the first 2 appears on the $2k$-th throw (where $k\ge 1$) is $q^{2k-1}p$.

The total probability $P$ is the sum of these probabilities:

$P=qp+q^{3}p+q^{5}p+\cdots$

We can factor out $qp$ from the infinite series:

$P=qp\left(1+q^2+q^4+\cdots \right)$

The terms inside the parentheses form an infinite geometric progression (GP) with the first term $a=1$ and common ratio $r=q^2$.
Since $\left|r\right|<1$, the sum of this infinite GP is:

$S=\frac{1}{1-q^2}$

Therefore, the probability $P$ is:

$P=\frac{qp}{1-q^2}$

Now, substitute the values of $p=\frac{1}{6}$ and $q=\frac{5}{6}$ into the formula:

$P=\frac{\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)}{1-{\left(\frac{5}{6}\right)}^2}$

Simplify the numerator and denominator:

$P=\frac{\frac{5}{36}}{1-\frac{25}{36}}$

$P=\frac{\frac{5}{36}}{\frac{36-25}{36}}$

$P=\frac{\frac{5}{36}}{\frac{11}{36}}$

$P=\frac{5}{11}$

Thus, the probability that 2 appears in an even number of throws is 5/11.