Which of the following statements are correct?
(A) The mean and variance of the Poisson distribution are equal.
(B) The mean and variance of a Binomial distribution are equal
(C) An unbiased die is thrown again and again until two sixes are obtained, then the probability of obtaining the second six in the 3rd throw is \(\displaystyle\frac{5}{108}\).
(D) If the variance of a Poisson distribution is 2, then \(P(X = 2) = 2e^{-2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
For a Poisson distribution with parameter \(\lambda\), mean = variance = \(\lambda\). For a binomial distribution, mean = \(np\) and variance = \(npq\), with \(q \lt 1\).
The second six comes on the 3rd throw when the first two throws give exactly one six and the 3rd throw is a six:
\( \displaystyle P = {}^{2}C_{1}\times\frac{1}{6}\times\frac{5}{6}\times\frac{1}{6} \)
\( \displaystyle = \frac{10}{216} = \frac{5}{108} \)
A Poisson variance of 2 gives \(\lambda = 2\):
\( \displaystyle P(X = 2) = \frac{e^{-2}\,2^2}{2!} = 2e^{-2} \)
So the Poisson mean equals its variance, the binomial variance \(npq\) is less than its mean \(np\), the probability in (C) is \(\frac{5}{108}\) and \(P(X = 2) = 2e^{-2}\).
Checking the statements
(A) The mean and variance of the Poisson distribution are equal. — Correct. Both are \(\lambda\).
(B) The mean and variance of a Binomial distribution are equal — Incorrect. \(npq \lt np\) because \(q \lt 1\).
(C) An unbiased die is thrown again and again until two sixes are obtained, then the probability of obtaining the second six in the 3rd throw is \(\frac{5}{108}\). — Correct. \(2 \times \frac{1}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{5}{108}\).
(D) If the variance of a Poisson distribution is 2, then \(P(X = 2) = 2e^{-2}\) — Correct. With \(\lambda = 2\), \(P(X = 2) = \frac{4e^{-2}}{2}\).