Consider the following for the two (02) items that follow: Let X be a random variable following binomial distribution with parameters n = 6 and p = k . Further, 9P(X = 4) = P(X = 2) . What is the value of P(X = 3) ?

Step-by-step solution

For a binomial distribution, the probability of r successes is given by P(X = r) = ^ n C_ r p^r q^ n-r , where q = 1 - p . Given n = 6 and p = k , we have q = 1 - k . The given condition is 9P(X = 4) = P(X = 2) . Substituting the probabilities, we get: 9 ⁶C₄ k^4 (1-k)^2 = ⁶C₂ k^2 (1-k)^4 Since ⁶C₄ = ⁶C₂ = 15 , we can cancel 15 k^2 (1-k)^2 from both sides (as k 0 and k 1 ): 9k^2 = (1-k)^2 Taking the square root on both sides (since k > 0 and 1-k > 0 ): 3k = 1 - k 4k = 1 k = 1 4 Now, we need to find P(X = 3) : P(X =