In a binomial distribution, the sum of its mean and variance is 1.8. If the event was conducted 5 times, then the probability of 3 successes is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.0512
For a binomial distribution with n trials, mean = np and variance = npq, where q = 1 − p. Here n = 5:
\( \displaystyle np+npq=1.8 \)
\( \displaystyle 5p+5p(1-p)=1.8 \)
\( \displaystyle 10p-5p^2=1.8 \)
\( \displaystyle p^2-2p+0.36=0 \)
\( \displaystyle p=\frac{2\pm\sqrt{4-1.44}}{2} \)
\( \displaystyle p=\frac{2\pm 1.6}{2} \)
p = 1.8 is impossible for a probability, so p = 0.2 and q = 0.8.
\( \displaystyle P(X=3)={}^{5}C_{3}\,p^3q^2 \)
\( \displaystyle =10\times(0.2)^3\times(0.8)^2 \)
\( \displaystyle =10\times 0.008\times 0.64=0.0512 \)
Why the other options are incorrect
Option (2) 0.2048 — Incorrect. 0.2048 is \( P(X=2)={}^{5}C_{2}(0.2)^2(0.8)^3 \), the probability of 3 failures instead of 3 successes.
Option (3) 0.3056 — Incorrect. The probability of 3 successes is 0.0512, not 0.3056.
Option (4) 0.1054 — Incorrect. The probability of 3 successes is 0.0512, not 0.1054.