In a binomial distribution, the mean is \(\frac{2}{3}\) and the variance is \(\frac{5}{9}\). What is the probability that X = 2?

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = ⁿC_k p^k (1 - p)^{(n - k)}

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

Properties:

• Mean of the distribution (μ_X) = n × p
• The variance (σ²_x) = n × p × (1 - p)
• Standard deviation (σ_x) = √{np(1 - p)}

Calculation:

Given:

Mean = n × p = 2/3    …. (1)

Variance = n × p × (1 - p) = 5/9    …. (2)

Dividing equation (1) by equation (2)

\(\frac{{{\rm{np}}\left( {1 - {\rm{p}}} \right)}}{{{\rm{np}}}} = \frac{{\frac{5}{9}}}{{\frac{2}{3}}}\)

⇒ (1 - p) = 5/6

⇒ p = 1/6

Put p = 1/6 in equation (1)

n × (1/6) = 2/3

⇒ n = 4

Now, P(X = 2) = ⁴C₂ × (1/6)² × (1 - 1/6)^{(4 - 2)}

P(X = 2) = 6 × (1/6)² × (5/6)²

\(\Rightarrow {\rm{P}}\left( {{\rm{x}} = 2} \right) = \frac{{25}}{{{6^4}}}\)

\(\Rightarrow {\rm{P}}\left( {{\rm{X}} = 2} \right) = \frac{{25}}{{216}}\)