A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is k 3 11 , then k is equal to

Step-by-step solution

Given that, A pair of dice is thrown 5 times. Required Sample space is 1 , 1 , 1 , 2 , 1 , 3 . . . . . . . . . . . 6 , 6 Total number of observations = 36 Now, a sum of 5 is observed in 1 , 4 , 4 , 1 , 2 , 3 , 3 , 2 No of favourable outcomes = 4 P (success) p = 4 36 = 1 9 P (failure) q = 1 - 1 9 = 8 9 We know that for a binomial distribution, B n , p P X = k = C k n p k q n - k where k ≥ 0 and p + q = 1 ∴ Required probability ⇒ P X ≥ 4 = P X = 4 + P X = 5 = C 4 5 1 9 4 8 9 + C 5 5 1 9 5 = 5