JEE Mains · Maths · STD 12 - 13. probability
If the sum and the product of mean and variance of a binomial distribution are \(24\) and \(128\) respectively, then the probability of one or two successes is.
- A \(\frac{33}{2^{32}}\)
- B \(\frac{33}{2^{29}}\)
- C \(\frac{33}{2^{28}}\)
- D \(\frac{33}{2^{27}}\)
Answer & Solution
Correct Answer
(C) \(\frac{33}{2^{28}}\)
Step-by-step Solution
Detailed explanation
\(np + npq =24\) \(np \cdot npq =128\) Solving \((1)\) and \((2)\): We get \(p =\frac{1}{2}, q =\frac{1}{2}, n =32\). Now, \(P ( X =1)+ P ( X =2)\) \(={ }^{32} C _{1} pq ^{31}+{ }^{32} C _{2} p ^{2} q ^{30}\) \(=\frac{33}{2^{28}}\)
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