AP EAMCET · Maths · Probability
The number of trials conducted in a binomial distribution is 6. If the difference between the mean and variance of this variate is \(\frac{27}{8}\), then the probability of getting atmost 2 successes is
- A \(\frac{106}{4^6}\)
- B \(\frac{144}{4^6}\)
- C \(\frac{126}{4^6}\)
- D \(\frac{154}{4^6}\)
Answer & Solution
Correct Answer
(D) \(\frac{154}{4^6}\)
Step-by-step Solution
Detailed explanation
\(n = 6\) \(\mu - \sigma^2 = np - npq = np(1-q) = np^2\) \(np^2 = \frac{27}{8}\) \(6p^2 = \frac{27}{8}\) \(p^2 = \frac{27}{48} = \frac{9}{16}\) \(p = \frac{3}{4}\) \(q = 1 - p = 1 - \frac{3}{4} = \frac{1}{4}\) \(P(X \le 2) = P(X=0) + P(X=1) + P(X=2)\)…
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