Question

Consider $n$coins, each of which independently comes up heads with probability $p$. Suppose that $n$is large and $p$is small, and let $\lambda =np$. Suppose that all $n$coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all coins, and so on. That is, we stop the first time that at least one of the $n$coins come up heads. Let $X$denote the total number of heads that appear. Which of the following reasonings concerned with approximating $P\{X=1\}$is correct (in all cases, $Y$is a Poisson random variable with parameter $\lambda )$?

(a) Because the total number of heads that occur when all $n$coins are rolled is approximately a Poisson random variable with parameter $\lambda$,

$P\{X=1\}\approx P\{Y=1\}=\lambda e^{-\lambda }$

(b) Because the total number of heads that occur when all $n$coins are rolled is approximately a Poisson random variable with parameter $\lambda$, and because we stop only when this number is positive,

$P\{X=1\}\approx P\{Y=1\mid Y>0\}=\frac{\lambda e^{-\lambda }}{1-e^{-\lambda }}$

(c) Because at least one coin comes up heads, $X$will equal 1 if none of the other $n-1$coins come up heads. Because the number of heads resulting from these $n-1$coins is approximately Poisson with mean $(n-1)p\approx \lambda$,

$P\{X=1\}\approx P\{Y=0\}=e^{-\lambda }$

Short answer

The correct option is$\left[b\right]$

Step-by-step solution

Step 1:Given information

Given in the question that,consider $n$coins, each of which independently comes up heads with probability $p$. Suppose that n is large and p is small, and let $\lambda =np$. Suppose that all n coins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all$n$coins, and so on. That is, we stop the first time that at least one of the n coins come up heads. Let $X$denote the total number of heads that appear

Step 2:Explanation

Here '$n$'coins are tossed. Also mentioned that ' $n$' is large and the probability that it turns up head is ' $p$' which is small.

Let $\lambda =np$.

Let ' $X$'denotes the total number of heads that appear.

Let $Y$denote the number of heads that occur when all $n$coins are tossed.

Let ' $Y$' is a Poisson random variable with parameter $\lambda$

This implies, the probability density function of '' $Y$' is $P\left(Y\right)=\frac{e^{-\lambda }{\lambda }^y}{y!}$

Therefore,

$P(Y=1)=\frac{e^{-\lambda }{\lambda }^1}{1!}=e^{-\lambda }\lambda$

$P(Y>0)=1-P(Y=0)$

$=1-\frac{e^{-\lambda }{\lambda }^0}{0!}($since$0!=1)$

$=1-e^{-\lambda }$

Since$X$is distributed as the conditional distribution of $Y$given that $Y>0$,

$P\{X=1\}=P\{Y=1\mid Y>0\}$

$=\frac{P\{\mathrm{Y}=1\}}{P\{Y>0\}}$

$\approx \frac{\lambda e^{-\lambda }}{1-e^{-\lambda }}$

The experiment is repeated until at least one head appears.

So, the correct option is [b].

Step 3:Final answer

The correct option is$\left[b\right]$