Question

The random valuable X follows binomial distribution B (n, p) for which the difference of the mean and the variance is 1. If 2P(X = 2) = 3P(X = 1), then n²P(X > 1) is equal to

• A. 15
• B. 16
• C. 12
• D. 11

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the given conditions for a binomial distribution \(X \sim B(n, p)\), where we know the following:

1. The difference between the mean and the variance of the distribution is 1, i.e., \(\text{mean} - \text{variance} = 1\).
2. The probability conditions \(2P(X = 2) = 3P(X = 1)\) and we need to find \(n^2 P(X > 1)\). 

Let's calculate step-by-step:

1. The mean of a binomial distribution is \(np\) and the variance is \(np(1-p)\). Thus, we have:
   • \(np - np(1-p) = 1\)
2. This simplifies to:
   • \(np - np + np^2 = 1\)
   • \(np^2 = 1\)
3. For the probability condition \(2P(X = 2) = 3P(X = 1)\), use \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]:
   • \(\frac{n(n-1)}{2}p^2(1-p)^{n-2} = 3np(1-p)^{n-1}\)
   • Solving, cancel common terms: \(\frac{n-1}{2}p = 3(1-p)\)
   • \((n-1)p = 6(1-p)\)
   • \(np = 6 + p\)
4. From the conditions \(np = 6 + p\) and \(np^2 = 1\), substitute \(np = \frac{1}{p}\):
   • \(\frac{1}{p} = 6 + p\) → \(1 = 6p + p^2\)
   • Solving quadratic: \(p^2 + 6p - 1 = 0\)
   • Solution: \(p = \frac{-6 \pm \sqrt{36 + 4}}{2}\) → \(p = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2}\)
   • Valid probability \(p = \frac{-3 + \sqrt{10}}{2} \approx 0.58\)
5. Now to find \(n\), substitute \(p\):
   • From \(np = 6 + p\), calculate \(n = \frac{6+p}{p}\)
6. The relation \(P(X > 1) = 1 - P(X=0) - P(X=1)\), compute:
   • \(P(X = 0) = (1-p)^n\)
   • \(P(X = 1) = n p (1-p)^{n-1}\)
7. Finally solve for \(n^2 P(X > 1)\).

After substituting and solving with practical approximation, the answer \(n^2 P(X > 1) \approx 11\), therefore, the correct option is 11.