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Consider ncoins, each of which independently comes up heads with probability p. Suppose that nis large and pis small, and let =np. Suppose that all ncoins 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 ncoins come up heads. Let Xdenote the total number of heads that appear. Which of the following reasonings concerned with approximating P{X=1}is correct (in all cases, Yis a Poisson random variable with parameter )?

(a) Because the total number of heads that occur when all ncoins are rolled is approximately a Poisson random variable with parameter ,

P{X=1}P{Y=1}=e-

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

P{X=1}P{Y=1Y>0}=e-1-e-

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

P{X=1}P{Y=0}=e-

Short Answer

Expert verified

The correct option is[b]

Step by step solution

01

Step 1:Given information

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

02

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 =np.

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

Let Ydenote the number of heads that occur when all ncoins are tossed.

Let ' Y' is a Poisson random variable with parameter

This implies, the probability density function of '' Y' is P(Y)=e-yy!

Therefore,

P(Y=1)=e-11!=e-

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

=1-e-00!(since0!=1)

=1-e-

SinceXis distributed as the conditional distribution of Ygiven that Y>0,

P{X=1}=P{Y=1Y>0}

=P{Y=1}P{Y>0}

e-1-e-

The experiment is repeated until at least one head appears.

So, the correct option is [b].

03

Step 3:Final answer

The correct option is[b]

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