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

Step-by-step solution

Let the probability of 2 appearing in a throw be given by P 2 and the probability of 2 not appearing be given by P 2 ' . ⇒ P 2 = 1 6 , P 2 ' = 5 6 Now, we want 2 to appear in 2 n d throw or 4 t h throw or 6 t h throw . . . and so on So, required probability is given by, P E = 5 6 × 1 6 + 5 6 3 × 1 6 + 5 6 5 × 1 6 + . . . ⇒ P E = 1 6 × 5 6 1 - 25 36 ⇒ P E = 1 6 × 5 6 11 36 ⇒ P E = 5 6 × 36 11 × 1 6 ⇒ P E = 5 11