Question

In a Binomial distribution, the sum of its mean and variance is $1.8$. If the event was conducted $5$ times, then the probability of two successes is:

• A. 0.3402
• B. 0.2048
• C. 0.1543
• D. 0.0564

Hint:

In binomial distributions, use the condition $\mu + \sigma^2$ carefully—it helps determine $p$ quickly.

Answer 1

The Correct Option is B

 

Step 1: Recall mean and variance of a binomial distribution. 
For $X \sim B(n,p)$, \[ \mu = np, \sigma^2 = np(1-p). \]

Step 2: Use given condition. 
\[ \mu + \sigma^2 = np + np(1-p) = np(2-p) = 1.8 \] Here $n=5$, so: \[ 5p(2-p) = 1.8. \]

Step 3: Solve for $p$. 
\[ 10p - 5p^2 = 1.8 \Rightarrow 5p^2 - 10p + 1.8 = 0. \] Divide by 5: \[ p^2 - 2p + 0.36 = 0. \] \[ p = \frac{2 \pm \sqrt{4 - 1.44}}{2} = \frac{2 \pm \sqrt{2.56}}{2} = \frac{2 \pm 1.6}{2}. \] \[ p = 0.2 \text{or} p = 1.8 \; (\text{not valid}). \] So, $p = 0.2$.

Step 4: Compute probability of 2 successes. 
\[ P(X=2) = \binom{5}{2} (0.2)^2 (0.8)^3 \] \[ = 10 \cdot 0.04 \cdot 0.512 = 0.2048. \]

Step 5: Conclusion. 
Thus, the probability of 2 successes is $0.2048$.