Under which of the following conditions the Poisson distribution is the limiting case of, binomial distribution:

(A) The number of trials is indefinitely large.
(B) The probability of success for each trial is indefinitely small.
(C) The product of the number of trials and the probability of success for each trial is finite.
(D) The probability of success for each trial is indefinitely large.

Choose the correct answer from the options given below:

(A), (B) and (C) only

The correct answer is Option (1) → (A), (B) and (C) only

$\text{Binomial probability: }P_{n}(X=k)=\frac{n!}{k!(n-k)!}\,p^{k}(1-p)^{\,n-k}.$

$\text{Assume }n\to\infty,\ p\to 0,\ np=\lambda\ (\text{finite}).\ \text{Put }p=\frac{\lambda}{n}.$

$\frac{n!}{k!(n-k)!}\,p^{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac{\lambda}{n}\right)^{k}=\frac{\lambda^{k}}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).$

$\text{As }n\to\infty\ \text{the product }\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)\to 1\ \Rightarrow\ \frac{n!}{k!(n-k)!}\,p^{k}\to\frac{\lambda^{k}}{k!}.$

$(1-p)^{\,n-k}=\left(1-\frac{\lambda}{n}\right)^{n}\left(1-\frac{\lambda}{n}\right)^{-k}\to e^{-\lambda}\cdot 1=e^{-\lambda}\ \text{as }n\to\infty.$

$\ \lim_{n\to\infty}P_{n}(X=k)=e^{-\lambda}\frac{\lambda^{k}}{k!},\ k=0,1,2,\dots\ $

$\text{Hence the Poisson distribution arises from the binomial distribution under the conditions:}$

$\text{(A) }n\to\infty\ (\text{True}),\quad\text{(B) }p\to 0\ (\text{True}),\quad\text{(C) }np=\lambda\ \text{finite}\ (\text{True}),\quad\text{(D) }p\ \text{large (False).}$