Question

Let $X$ follow Binomial distribution with parameters $12$ and $p$, let $q = 1 - p$. If \[ \sum_{x=0}^{12}(x - 12p)^2 \cdot {}^{12}C_x \cdot q^{12 - x} \cdot p^x = \frac{8}{3} \quad \text{and} \quad P(X>10) = \left(\frac{2}{3}\right)^K, \, (K>1), \] then $K =$

• A. \(\dfrac{14}{3}\)
• B. \(\dfrac{25}{9}\)
• C. \(\dfrac{16}{7}\)
• D. \(\dfrac{3}{2}\)

Hint:

Use binomial identities for variance and tail probabilities. Always check if given expressions match known moments or cumulative values.

Answer 1

The Correct Option is A

Given that \(X\) follows a binomial distribution with parameters \(12\) and \(p\), where \(q = 1 - p\). The expression \(\sum_{x=0}^{12}(x - 12p)^2 \cdot {}^{12}C_x \cdot q^{12-x} \cdot p^x = \frac{8}{3}\) is the variance of the binomial distribution, which is known to be \(12pq\). Set this equal to the given expression:
\[12pq = \frac{8}{3}\]
Thus, we solve for \(pq\):
\[pq = \frac{2}{9}\]
Given, \(P(X>10) = \left(\frac{2}{3}\right)^K\), which implies:
\[P(X>10) = P(X=11) + P(X=12)\]
Calculating individual probabilities using the binomial probability formula:
\[\begin{aligned} P(X=11) &= {}^{12}C_{11} \cdot p^{11} \cdot q^1 \\ &= 12p^{11}q \end{aligned}\]
\[\begin{aligned} P(X=12) &= {}^{12}C_{12} \cdot p^{12} \\ &= p^{12} \end{aligned}\]
Therefore, the sum \(P(X = 11) + P(X = 12)\) is:
\[P(X>10) = 12p^{11}q + p^{12} = p^{11}(12q + p)\]
Since \(q = 1 - p\), replace \(q\) to get:
\[P(X>10) = p^{11}(12(1-p) + p) = p^{11}\{12 - 11p\}\]
Equate it to the given value of \(P(X>10)\):
\[p^{11}(12-11p) = \left(\frac{2}{3}\right)^K\]
To find the value of \(K\):
Set \(pq = \frac{2}{9}\). This condition with \(p^{11}(12-11p)\) implies substitutions for \(p\) and \(q\) which best satisfy both relations. Test with potential fixed ratios \(p= \frac{1}{3}\), \(q= \frac{2}{3}\):
\[P(X>10) = \left(\frac{1}{3}\right)^{11}(12-\frac{11}{3}) = \frac{1}{3}\left(\frac{2}{3}\right)^{11}\left(\frac{25}{3}\right)\]
Substitute and simplify the expression:
\[(\frac{2}{3})^{K} = \frac{1}{3}\cdot\left(\frac{2}{3}\right)^{11}\cdot\frac{25}{3}\]
Equate powers of the same base \(\left(\frac{2}{3}\right)\):
\[K = \frac{14}{3}\]
Thus, the value of \(K\) is \(\dfrac{14}{3}\).