Question

The number of trials conducted in a binomial distribution is 6. If the difference between the mean and variance of this variate is \(\frac{27}{8}\), then the probability of getting at most 2 successes is:

• A. \(\frac{106}{4^6}\)
• B. \(\frac{144}{4^6}\)
• C. \(\frac{126}{4^6}\)
• D. \(\frac{154}{4^6}\)

Hint:

Use mean and variance relation to find \(p\), then calculate cumulative binomial probability.

Answer 1

The Correct Option is D

Given: Number of trials \(n=6\), Mean = \(np\), Variance = \(np(1-p)\), and \[ \text{Mean} - \text{Variance} = \frac{27}{8}. \] Calculate: \[ np - np(1-p) = np^2 = \frac{27}{8}. \] Let \(np^2 = \frac{27}{8}\). Since \(n=6\), \[ 6p^2 = \frac{27}{8} \implies p^2 = \frac{27}{48} = \frac{9}{16} \implies p = \frac{3}{4}. \] Calculate probability of at most 2 successes: \[ P(X \leq 2) = \sum_{k=0}^2 \binom{6}{k} p^k (1-p)^{6-k}. \] Here, \(p = \frac{3}{4}\), \(q = \frac{1}{4}\). Calculate each term: \(k=0\): \[ \binom{6}{0} \left(\frac{3}{4}\right)^0 \left(\frac{1}{4}\right)^6 = 1 \times 1 \times \frac{1}{4^6} = \frac{1}{4^6}. \] \(k=1\): \[ \binom{6}{1} \left(\frac{3}{4}\right)^1 \left(\frac{1}{4}\right)^5 = 6 \times \frac{3}{4} \times \frac{1}{4^5} = \frac{18}{4^6}. \] \(k=2\): \[ \binom{6}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^4 = 15 \times \frac{9}{16} \times \frac{1}{4^4} = \frac{135}{4^6}. \] Sum: \[ \frac{1 + 18 + 135}{4^6} = \frac{154}{4^6}. \]