Question

In a Binomial distribution with $n=4$ if $2P(X=3)=3P(X=2)$, then the variance is

• A. $\frac{36}{169}$
• B. $\frac{144}{169}$
• C. $\frac{9}{169}$
• D. $\frac{16}{169}$

Hint:

Logic Tip: When solving binomial distribution equality equations, always try to factor out and cancel common terms like $p^k$ and $(1-p)^k$ as early as possible to avoid expanding into higher-degree polynomials.

Answer 1

The Correct Option is B

Concept:
For a Binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), the probability of $r$ successes is given by: $$P(X=r) = {}^nC_r p^r q^{n-r}$$ where $q = 1-p$. The variance of the binomial distribution is given by the formula $\text{Variance} = npq$.
Step 1: Express the given probabilities using the binomial formula.
Given $n = 4$, we can write expressions for $P(X=3)$ and $P(X=2)$: $$P(X=3) = {}^4C_3 p^3(1-p)^1 = 4p^3(1-p)$$ $$P(X=2) = {}^4C_2 p^2(1-p)^2 = 6p^2(1-p)^2$$
Step 2: Substitute these into the given condition to find p.
We are given the condition $2P(X=3) = 3P(X=2)$. Substituting our expressions: $$2[4p^3(1-p)] = 3[6p^2(1-p)^2]$$ $$8p^3(1-p) = 18p^2(1-p)^2$$ Since $p \neq 0$ and $p \neq 1$, we can divide both sides by $2p^2(1-p)$: $$4p = 9(1-p)$$ $$4p = 9 - 9p$$ $$13p = 9 \implies p = \frac{9}{13}$$
Step 3: Calculate the variance.
Now we find $q = 1 - p$: $$q = 1 - \frac{9}{13} = \frac{4}{13}$$ Using the variance formula $\text{Variance} = npq$: $$\text{Variance} = 4 \times \left(\frac{9}{13}\right) \times \left(\frac{4}{13}\right)$$ $$\text{Variance} = \frac{144}{169}$$