If the mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly 5 successes in the distribution is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( ^{16}C_{5}\left(\displaystyle\frac{1}{4}\right)^5\left(\displaystyle\frac{3}{4}\right)^{11} \)
For a binomial distribution, mean = np and variance = npq.
\( \displaystyle q=\frac{npq}{np}=\frac{3}{4} \)
\( \displaystyle p=1-q=\frac{1}{4} \)
\( \displaystyle n=\frac{np}{p}=\frac{4}{1/4}=16 \)
The probability of exactly r successes is
\( \displaystyle P(X=r)={}^{n}C_{r}\,p^{r}q^{n-r} \)
\( \displaystyle P(X=5)={}^{16}C_{5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^{11} \)
Why the other options are incorrect
Option (1) \( ^{16}C_{5}\left(\displaystyle\frac{1}{4}\right)^8\left(\displaystyle\frac{3}{4}\right)^8 \) — Incorrect. p must carry the power r = 5 and q the power 16 − 5 = 11; here both powers are 8.
Option (2) \( ^{16}C_{5}\left(\displaystyle\frac{1}{4}\right)^{11}\left(\displaystyle\frac{3}{4}\right)^5 \) — Incorrect. The powers of p and q are interchanged; p = 1/4 must carry the power 5, the number of successes.
Option (4) \( ^{16}C_{5}\left(\displaystyle\frac{1}{4}\right)^7\left(\displaystyle\frac{3}{4}\right)^9 \) — Incorrect. The powers 7 and 9 add up to 16, but p must carry the power 5, not 7.