A fair die is thrown until 2 appears. Then the probability, that 2...

Let the probability of $2$ appearing in a throw be given by $P\left(2\right) and the probability of$2$not appearing be given by$P\left({2}^{'}\right)$.</p><p>$⇒ P\left(2\right) = \dfrac{1}{6}, P\left({2}^{'}\right) = \dfrac{5}{6}$</p><p>Now, we want$2$to appear in${2}^{nd}\text{throw or}{4}^{th}\text{throw or}{6}^{th}\text{throw} . . . \text{and so on}$</p><p>So, required probability is given by,</p><p>$P\left(E\right) = \dfrac{5}{6} \times \dfrac{1}{6} + {\left(\dfrac{5}{6}\right)}^{3} \times \dfrac{1}{6} + {\left(\dfrac{5}{6}\right)}^{5} \times \dfrac{1}{6} + . . .$</p><p>$⇒ P\left(E\right) = \dfrac{1}{6} \times \dfrac{\dfrac{5}{6}}{1 - \dfrac{25}{36}}$</p><p>$⇒ P\left(E\right) = \dfrac{1}{6} \times \dfrac{\dfrac{5}{6}}{\dfrac{11}{36}}$</p><p>$⇒ P\left(E\right) = \dfrac{5}{6} \times \dfrac{36}{11} \times \dfrac{1}{6}$</p><p>$⇒ P\left(E\right) = \dfrac{5}{11}$