The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P(X = 1) is

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  1. \( \frac{1}{4} \)
  2. \(\frac{1}{32}\)
  3. \(\frac{1}{16} \)
  4. \(\frac{1}{8}\)

Answer (Detailed Solution Below)

Option 2 : \(\frac{1}{32}\)
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Detailed Solution

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Calculation 

Given mean = 4

Variance = 2

∴ np = 4

and npq = 2

∴ 4q = 2, q = \(\frac{2}{4}\) = \(\frac{1}{2}\)

p = 1 - q = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)

∴ n(\(\frac{1}{2}\)) = 4; n = 8

Thus

P(X = 1) = \({^8C_1 (\frac{1}{2}) (\frac{1}{2})^7}\)

= 8(\(\frac{1}{2}\))8 = \(\frac{8}{256}\) = \(\frac{1}{32}\)

Hence option 2 is correct

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