In a trial, the probability of success is twice the probability of failure. In six trials, the probability of at most two failure will be

Step-by-step solution

Let p be the probability of success and q be the probability of failure. Given p = 2q . Since p + q = 1 , we have 2q + q = 1 , which implies 3q = 1 , so q = 1 3 and p = 2 3 . The number of trials is n = 6 . We want the probability of at most two failures, which is P(X 2) where X is the number of failures. This is given by P(X = 0) + P(X = 1) + P(X = 2) . Using the binomial distribution formula P(X = k) = ^ n C_ k q^ k p^ n-k : P(X = 0) = ⁶C₀ ( 1 3 )⁰ ( 2 3 )⁶ = 1 1 64 729 = 64 729 P(X = 1) = ⁶C₁ ( 1 3 )¹ ( 2 3 )⁵ =