Question

If a random variable $X$ follows the Binomial distribution $\text{B}(33, \text{p})$ such that $3\text{P}(\text{X} = 0) = \text{P}(\text{X} = 1)$, then the variance of X is

• A. $\frac{11}{144}$
• B. $\frac{35}{48}$
• C. $\frac{121}{48}$
• D. $\frac{33}{144}$

Hint:

Use ratio of probabilities to find \(p\) quickly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A Binomial distribution $B(n, p)$ models the number of successes in $n$ independent trials, each with success probability $p$. We are given the number of trials $n$ and an equation relating the probabilities of 0 and 1 successes. We use this equation to solve for the success probability $p$. Once $p$ is known, we compute the variance using the standard formula. Step 2: Key Formula or Approach:
1. Probability Mass Function: $P(X = k) = \binom{n}{k} p^k q^{n-k}$, where $q = 1 - p$. 2. Variance of a Binomial distribution: $\text{Var}(X) = npq$. Step 3: Detailed Explanation:
The given distribution is $X \sim B(n, p)$ with $n = 33$. The condition given is $3P(X = 0) = P(X = 1)$. Using the Binomial PMF formula, let's write out the expressions for these probabilities: \[ P(X = 0) = \binom{33}{0} p^0 q^{33-0} = 1 \cdot 1 \cdot q^{33} = q^{33} \] \[ P(X = 1) = \binom{33}{1} p^1 q^{33-1} = 33 \cdot p \cdot q^{32} \] Substitute these explicitly into the given equation: \[ 3 \cdot q^{33} = 33 \cdot p \cdot q^{32} \] Assuming $q \neq 0$ (as that would mean $p=1$ and the probabilities would be trivially 0, not fitting the equation well), we divide both sides by $3q^{32}$: \[ q = 11p \] We also know the fundamental axiom that probabilities sum to 1, so $p + q = 1$. Substitute the expression for $q$ into this equation: \[ p + 11p = 1 \] \[ 12p = 1 \implies p = \frac{1}{12} \] Now find the corresponding value of $q$: \[ q = 1 - p = 1 - \frac{1}{12} = \frac{11}{12} \] We now possess all parameters ($n, p, q$) to compute the variance: \[ \text{Variance} = npq \] \[ \text{Variance} = 33 \cdot \left(\frac{1}{12}\right) \cdot \left(\frac{11}{12}\right) \] \[ \text{Variance} = \frac{33 \times 11}{144} = \frac{3 \times 11 \times 11}{12 \times 12} = \frac{11 \times 11}{4 \times 12} = \frac{121}{48} \] Step 4: Final Answer:
The variance of X is $\frac{121}{48}$.