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In a binomial distribution, the mean is \(\dfrac{2}{3}\) and variance is \(\dfrac{5}{9}\). What is the probability that random variable D = 2?
1. \(\dfrac{5}{36}\)
2. \(\dfrac{25}{36}\)
3. \(\dfrac{25}{54}\)
4. \(\dfrac{25}{216}\)

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Correct Answer - Option 4 : \(\dfrac{25}{216}\)


For the binomial distribution of probability

  • Mean = np
  • Variance = npq

Where n is the total number of cases, p is probability of favorable cases and q is probability of unfavorable cases(1 - p)


Given mean = np = \(2\over3\) and variance = npq = \(5\over9\)

\(\rm variance \over mean\) = \({5\over9}\over{2\over3}\)

q = \(5\over6\)

p = 1 - q = \(1\over6\)

np = \(2\over3\)

n = 4

∴ The probability that random variable D = 2

P = \(\rm {^nC_2}p^2q^{n-2}\)

P = \(\rm {^4C_2}\times\left({1\over6}\right)^2\times\left({5\over6}\right)^2\)

P = \(\rm 6\times{1\over36}\times{25\over36}\)

P = \(25\over216\)

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