The mean and standard deviation of a Binomal distribution are 4 and $$\sqrt{\frac83}$$ respectively. With usual notations, values of n and p are​

Correct option is B

​Solution:

$$\textbf{Given:} \\\text{Mean} = 4 \quad \text{and} \quad \text{Standard deviation} = \sqrt{\frac{8}{3}} \\[8pt]\textbf{Let } n = \text{number of trials, } p = \text{probability of success, } q = 1 - p \\[8pt]\text{Mean of binomial distribution: } \mu = n \cdot p = 4 \quad \text{(1)} \\[8pt]\text{Standard deviation: } \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{\frac{8}{3}} \\[8pt]\Rightarrow n \cdot p \cdot q = \frac{8}{3} \quad \text{(2)} \\[8pt]\text{Divide (2) by (1):} \\\frac{n \cdot p \cdot q}{n \cdot p} = \frac{8/3}{4} \\[8pt]\Rightarrow q = \frac{2}{3} \Rightarrow p = 1 - \frac{2}{3} = \frac{1}{3} \\[8pt]\text{Substitute } p = \frac{1}{3} \text{ into (1):} \\n \cdot \frac{1}{3} = 4 \Rightarrow n = 12 \\[8pt]\boxed{n = 12, \quad p = \frac{1}{3}}$$

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