The sum and the product of the mean and variance of binomial distribution are 1.8 and 0.8 respectively. The probability of atleast one success is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1-\left(\displaystyle\frac{4}{5}\right)^5 \)
Let the mean be np = m and the variance be npq = v.
\( \displaystyle m+v=1.8 \)
\( \displaystyle mv=0.8 \)
So m and v are the roots of
\( \displaystyle t^2-1.8t+0.8=0 \)
\( \displaystyle (t-1)(t-0.8)=0 \)
\( \displaystyle t=1 \text{ or } 0.8 \)
In a binomial distribution the variance is less than the mean (q < 1), so np = 1 and npq = 0.8.
\( \displaystyle q=\frac{npq}{np}=0.8=\frac{4}{5} \)
\( \displaystyle p=\frac{1}{5} \)
\( \displaystyle n=\frac{1}{p}=5 \)
\( \displaystyle P(X\ge 1)=1-P(X=0) \)
\( \displaystyle =1-q^5=1-\left(\frac{4}{5}\right)^5 \)
Why the other options are incorrect
Option (1) \( \left(\displaystyle\frac{4}{5}\right)^5 \) — Incorrect. \( \left(\frac{4}{5}\right)^5 \) is P(X = 0), the probability of no success.
Option (2) \( \left(\displaystyle\frac{4}{5}\right)^4 \) — Incorrect. This uses n = 4 and gives P(X = 0), not P(X ≥ 1); here n = 5.
Option (4) \( 1-\left(\displaystyle\frac{4}{5}\right)^4 \) — Incorrect. The complement is right, but n = 5, not 4.