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Probability

Question
CBSEENMA12036282
Wired Faculty App

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

  • 37/256

  • 219/ 256

  • 128/256

  • 28/256

Solution

D.

28/256

Given that mean = 4
np = 4 and variance = 2
npq = 2 ⇒ 4q = 2
⇒ q = 1/2
∴ p = 1 - q = 1 - 1/2 = 1/2
also n = 8 Probability of 2 successes
straight P left parenthesis straight X space equals 2 right parenthesis space equals space straight C presuperscript 8 subscript 2 straight p squared straight q to the power of 6 space equals space fraction numerator 8 factorial over denominator 2 factorial space straight x space 6 factorial end fraction space straight x space left parenthesis 1 divided by 2 right parenthesis squared space straight x space left parenthesis 1 divided by 2 right parenthesis to the power of 6 space equals space 28 space straight x 1 over 2 to the power of 8 space equals space 28 over 256

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