Q.4.20 - A First Course in Probability

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 $(n-1)p\approx \lambda$,

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

Step-by-Step Solution

Answer

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

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 $\left[b\right]$