If the sum and difference of squares of mean and variance of a Binomial distribution is \(\displaystyle\frac{225}{256}\) and \(\displaystyle\frac{63}{256}\) respectively, then \(P(X \ge 2)\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5}{32}\)
For a binomial distribution, the mean is \(M = np\) and the variance is \(V = npq\). Given
\( \displaystyle M^2 + V^2 = \frac{225}{256}, \quad M^2 - V^2 = \frac{63}{256} \)
Adding and subtracting:
\( \displaystyle 2M^2 = \frac{288}{256}, \quad M^2 = \frac{144}{256}, \quad M = \frac{3}{4} \)
\( \displaystyle 2V^2 = \frac{162}{256}, \quad V^2 = \frac{81}{256}, \quad V = \frac{9}{16} \)
Dividing the variance by the mean:
\( \displaystyle q = \frac{npq}{np} = \frac{9/16}{3/4} = \frac{3}{4} \)
\( \displaystyle p = 1 - q = \frac{1}{4} \)
\( \displaystyle n = \frac{M}{p} = \frac{3/4}{1/4} = 3 \)
So X is binomial with \(n = 3\), \(p = \frac{1}{4}\):
\(P(X \ge 2) = P(X = 2) + P(X = 3)\)
\( \displaystyle = {}^{3}C_{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right) + \left(\frac{1}{4}\right)^3 \)
\( \displaystyle = \frac{9}{64} + \frac{1}{64} = \frac{10}{64} = \frac{5}{32} \)