# 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?

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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 : $\dfrac{25}{216}$

Concept:

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)

Calculation:

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 = $25\over216$