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Consider the sequence of coin tosses described in Exercise 2.

a. What is the expected number of tosses that will be required in order to obtain five heads?

b. What is the variance of the number of tosses that will be required in order to obtain five heads?

Short Answer

Expert verified

(a) 150

(b) 4350

Step by step solution

01

Given information

Let Xdenote the number of tosses required to obtain 5 heads.

Here, X is a random variable that follows negative binomial distribution that is

\(X \sim NB\left( {r = 5,p = \frac{1}{{30}}} \right)\).

02

Defining the pdf

This negative binomial distribution is based on trials.

Therefore, the pdf is,

\(f\left( x \right) = {}^{n - 1}{C_{r - 1}}{p^r}{\left( {1 - p} \right)^{n - r}},n = r,r + 1, \ldots \)

03

(a) Finding the expectation

Since X follows a negative binomial distribution, by its properties, the expectation is:

\(\begin{array}{c}E\left( X \right) = \frac{r}{p}\\ = \frac{5}{{\frac{1}{{30}}}}\\ = 150\end{array}\)

As a result, the predicted number of tails gained prior to obtaining five heads is 150.

04

(b) Finding the variance

Since X follows anegative binomial, by its properties, the variance is:

\(\begin{array}{c}V\left( X \right) = \frac{{r\left( {1 - p} \right)}}{{{p^2}}}\\ = \frac{{5\left( {1 - \frac{1}{{30}}} \right)}}{{\frac{1}{{{{30}^2}}}}}\\ = 4350\end{array}\)

As a result, the variation of the number of tails acquired before five tails are achieved is 4350.

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Suppose that a sequence of independent tosses is made with a coin for which the probability of obtaining a head-on each given toss is \(\frac{{\bf{1}}}{{{\bf{30}}}}.\)

a. What is the expected number of tails that will be obtained before five heads have been obtained?

b. What is the variance of the number of tails that will be obtained before five heads have been obtained?

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