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: Understanding the Question:
We are given the sum of the mean and variance for a binomial distribution with $n=5$ trials. We need to find the probability of success, $p$.

Step 2: Key Formula or Approach:
For a binomial distribution:
Mean ($\mu$) $= np$
Variance ($\sigma^2$) $= npq = np(1-p)$

Step 3: Detailed Explanation:
According to the given condition:
$$np + np(1-p) = 1.8$$ Factor out $np$:
$$np(1 + 1 - p) = 1.8$$ $$np(2 - p) = 1.8$$ Substitute the number of trials $n = 5$:
$$5p(2 - p) = 1.8$$ Divide both sides by 5:
$$p(2 - p) = 0.36$$ $$2p - p^2 = 0.36$$ Rearrange into a standard quadratic equation:
$$p^2 - 2p + 0.36 = 0$$ Solve for $p$ using the quadratic formula:
$$p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(0.36)}}{2(1)}$$ $$p = \frac{2 \pm \sqrt{4 - 1.44}}{2}$$ $$p = \frac{2 \pm \sqrt{2.56}}{2}$$ $$p = \frac{2 \pm 1.6}{2}$$ This gives two possible values for $p$:
$$p = \frac{3.6}{2} = 1.8 \quad \text{or} \quad p = \frac{0.4}{2} = 0.2$$ Since the probability of an event cannot exceed 1, we reject $1.8$.

Step 4: Final Answer:
The probability of success is $0.2$, matching option (A).