Let a die rolled till 2 is obtained. The probability that 2 obtained on even numbered toss is equal to

To find the probability that a 2 is obtained on an even-numbered toss of a die, we can break down the problem step-by-step.

First, let us find the probability of success (getting a 2) and failure (getting any other number than 2) on a single roll of a fair six-sided die:
The probability of getting a 2 is:
$p=\frac{1}{6}$
The probability of not getting a 2 is:
$q=1-p=\frac{5}{6}$

The die is rolled until a 2 is obtained. We want the probability that the 2 is obtained on an even-numbered toss (i.e., on the 2nd, 4th, 6th, 8th, ... toss).
Let us write down the probability for each of these cases:
- The 2 is obtained on the 2nd toss (1st toss is a failure, 2nd is a success):
$P_2=qp$
- The 2 is obtained on the 4th toss (first 3 tosses are failures, 4th is a success):
$P_4=q^{3}p$
- The 2 is obtained on the 6th toss (first 5 tosses are failures, 6th is a success):
$P_6=q^{5}p$
and so on.

The total probability $P$ of obtaining 2 on an even-numbered toss is the sum of these probabilities:
$P=qp+q^{3}p+q^{5}p+...$

This is an infinite geometric progression (GP) where:
- The first term is $a=qp$
- The common ratio is $r=q^2$

The sum of an infinite geometric progression is given by the formula:
$S=\frac{a}{1-r}$

Substituting the values of $a$ and $r$ into the formula:
$P=\frac{qp}{1-q^2}$

Now, substitute the values of $p=\frac{1}{6}$ and $q=\frac{5}{6}$:
$P=\frac{\left(\frac{5}{6}\right)\times \left(\frac{1}{6}\right)}{1-{\left(\frac{5}{6}\right)}^2}$

Simplify the expression:
$P=\frac{\frac{5}{36}}{1-\frac{25}{36}}$
$P=\frac{\frac{5}{36}}{\frac{11}{36}}$
$P=\frac{5}{11}$

Therefore, the probability that 2 is obtained on an even-numbered toss is 5/11.