Question
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 k?
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 k?
C. 1 4
The probability mass function of a binomial distribution is given by P(X = r) = ^ n C_ r p^r q^ n-r , where q = 1 - p. Given n = 6, p = k, and q = 1 - k. We have 9P(X = 4) = P(X = 2). Substituting the probabilities: 9 (^ 6 C_ 4 ) k^4 (1 - k)^2 = (^ 6 C_ 2 ) k^2 (1 - k)^4 Since ^ 6 C_ 4 = ^ 6 C_ 2 = 15, we can cancel it from both sides: 9 k^4 (1 - k)^2 = k^2 (1 - k)^4 Assuming k 0 and k 1, dividing both sides by k^2 (1 - k)^2 gives: 9 k^2 = (1 - k)^2 Taking the square root of both sides: 3k = 1 - k 4k = 1 k = 1 4 The other root 3k = -(1 - k) 2k = -1 k = - 1 2 is rejected since probability cannot be negative. Answer: 1 4
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