Question

If the sum of mean and variance of a binomial distribution for 5 trials is 1.8, then probability of a success is

• A. $0.2$
• B. $0.6$
• C. $0.4$
• D. $0.8$

Hint:

For binomial distributions, variance is always less than the mean because $\sigma^2 = \mu \times (1-p)$. Since $(1-p)$ is a fraction, multiplying the mean by it reduces the value.

Answer 1

The Correct Option is A

Step 1: Recall binomial mean and variance.
For $n$ trials with success chance $p$ and failure chance $q = 1 - p$: mean $= np$ and variance $= npq$. Here $n = 5$.

Step 2: Write the given sum.
\[ np + npq = 1.8 \]
Step 3: Take $np$ common.
\[ np(1 + q) = 1.8 \quad\Rightarrow\quad 5p(1 + q) = 1.8 \]
Step 4: Replace $q$ with $1 - p$.
\[ 5p(1 + 1 - p) = 1.8 \quad\Rightarrow\quad 5p(2 - p) = 1.8 \]
Step 5: Form the quadratic.
\[ 10p - 5p^2 = 1.8 \quad\Rightarrow\quad 5p^2 - 10p + 1.8 = 0 \] Dividing by 5: $p^2 - 2p + 0.36 = 0$.

Step 6: Solve and pick the valid root.
This factors to $(p - 0.2)(p - 1.8) = 0$. Since $p$ must be at most 1, we take $p = 0.2$. \[ \boxed{p = 0.2 \text{ (Option 1)}} \]