If the difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 5, then the probability of success is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{3}\)
For a binomial distribution, mean = \(np\) and variance = \(npq\):
\( np - npq = 1 \)
\( (np)^2 - (npq)^2 = 5 \)
Using the difference of squares:
\( (np + npq)(np - npq) = 5 \)
\( np + npq = 5 \)
Adding and subtracting this and \(np - npq = 1\):
\( np = 3,\quad npq = 2 \)
\( \displaystyle q = \frac{npq}{np} = \frac{2}{3} \)
\( \displaystyle p = 1 - \frac{2}{3} = \frac{1}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{3}\) — Incorrect. This is q, the probability of failure, not of success.
Option (3) \(\displaystyle\frac{2}{5}\) — Incorrect. It would need \(q = \frac{3}{5}\), but \(q = \frac{2}{3}\).
Option (4) \(\displaystyle\frac{3}{5}\) — Incorrect. It would need \(q = \frac{2}{5}\), but \(q = \frac{2}{3}\).