JEE Mains · Maths · STD 12 - 13. probability
Let \(X\) have a binomial distribution \(B ( n , p )\) such that the sum and the product of the mean and variance of \(X\) are \(24\) and \(128\) respectively. If \(P ( X > n -3)=\frac{ k }{2^{ n }}\), then \(k\) is equal to.
- A \(528\)
- B \(529\)
- C \(629\)
- D \(630\)
Answer & Solution
Correct Answer
(B) \(529\)
Step-by-step Solution
Detailed explanation
Let \(\alpha=\) Mean \(\& \quad \beta=\) Variance \((\alpha>\beta)\) So, \(\alpha+\beta=24, \quad \alpha \beta=128\) \(\alpha=16 \quad \& \quad \beta=8\) \(np =16 \quad npq =8 \Rightarrow q =\frac{1}{2}\) \(\therefore p =\frac{1}{2}, n =32\)…
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