Let X be a random variable following binomial distribution with parameters n=5 and p=k. Further, P(X=1)=0.4096 and P(X=2)=0.2048. What is the value of k?

Explanation

For a binomial distribution, P(X=x) = \binom{n}{x}p^xq^{n-x}. Dividing P(X=1) by P(X=2) yields \frac{5pq^4}{10p^2q^3} = \frac{q}{2p} = \frac{0.4096}{0.2048} = 2. Thus, q = 4p. Since p+q=1, we get p + 4p = 1 \implies 5p = 1 \implies p = 0.2. Therefore, k = 0.2.

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