Chapter 5.2, Problem 2CYU

a.

To determine

Compute the probability that exactly four of them have heard of “Modern Warfare 2” using binomial probability distribution.

Answer to Problem 2CYU

The probability that exactly four of them have heard of “Modern Warfare 2”is 0.2153.

Explanation of Solution

Calculation:

It was found that 20% of the gamers have heard of the game “Modern Warfare 2” in the recent year. A random sample of 18 gamers is considered.

Define the random variable X as the numbers of gamers have heard of the game “Modern Warfare 2”.

Here, a random sample (n) of 18 gamers is taken. Each gamer is independent of the other. Also, there are two possible outcomes, the gamer have heard of the game “Modern Warfare 2”or not (success or failure). It was found that 20% of the gamers have heard of the game. Thus, the probability of success (p) is 0.20. Thus, X follows binomial distribution.

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$P\left(x\right)={}_n{C}_x{p}^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n$

Where, p is the probability of success.

Substitute $n=18\text{ and }p=0.20$.

$P\left(x\right)={}_{18}{C}_x{\left(0.20\right)}^x{\left(1-0.20\right)}^{18-x}$

The probability that exactly four of them have heard of “Modern Warfare 2”is the probability at the point $x=4$. That is,

$\begin{array}{c}P\left(x=4\right)={}_{18}{C}_4{\left(0.20\right)}^4{\left(1-0.20\right)}^{18-4}\\ =\left[\frac{18!}{4!\left(18-4\right)!}{\left(0.20\right)}^4{\left(0.80\right)}^{14}\right]\\ =0.2153\end{array}$

Thus, the probability that exactly four of them have heard of “Modern Warfare 2”is 0.2153.

b.

To determine

Compute the probability that fewer than three of them have heard of “Modern Warfare 2” using Table A.1.

Answer to Problem 2CYU

The probability that fewer than three of them have heard of “Modern Warfare 2” is 0.271.

Explanation of Solution

Calculation:

The probability that fewer than three of them have heard of “Modern Warfare 2” is,

$P(x<3)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)$

The probability values are obtained from the binomial table.

Procedure:

• From Table A.1 of binomial probabilities, select the section for n = 18.
• Take the value corresponding to $x=0,1,2$ and $p=0.20$.

Thus, from the table, $P\left(x=0\right)=0.018,P\left(x=1\right)=0.081,P\left(x=2\right)=0.172$

Substituting the values in the equation,

$\begin{array}{c}P\left(x<3\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)\\ =0.018+0.081+0.172\\ =0.271\end{array}$

Hence, the probability that fewer than three of them have heard of “Modern Warfare 2” is 0.271.

c.

To determine

Find the probability that more than 3of the gamers have heard of “Modern Warfare 2” using any valid method.

Answer to Problem 2CYU

The probability that more than 3of the gamers have heard of “Modern Warfare 2” is 0.4990.

Explanation of Solution

Calculation:

The probability that more than 3of the gamers have heard of “Modern Warfare 2”is obtained as shown below:

$P\left(x>3\right)=1-P\left(x\le 3\right)$

Software procedure:

Step by step procedure to obtain the probability $P\left(x\le 3\right)$ using MINITAB software is given below:

• Choose Calc > Probability Distributions>Binomial Distribution.
• Choose Cumulative Probability.
• Enter Number of trials as 18 and Event probability as 0.20.
• In Input constant, enter 3.
• Click OK.

The output using the Minitab software is given below:

Substitute the value $P\left(x\le 3\right)=0.5010$, the probability is,

$\begin{array}{c}P\left(x>3\right)=1-P\left(x\le 3\right)\\ =1-0.5010\\ =0.4990\end{array}$

Thus, the probability that more than 3of the gamers have heard of “Modern Warfare 2” is 0.4990.

d.

To determine

Compute the probability that the number who have heard of “Modern Warfare 2” is between 2 and 6 inclusive using any valid method.

Answer to Problem 2CYU

The probability that the number who have heard of “Modern Warfare 2” is between 2 and 6 inclusive is 0.8496.

Explanation of Solution

Calculation:

The probability that the number who have heard of “Modern Warfare 2” is between 2 and 6 inclusive is obtained as shown below:

$\begin{array}{c}P\left(2<x\le 6\right)=P\left(x\le 6\right)-P\left(x<2\right)\\ =P\left(x\le 6\right)-P\left(x\le 1\right)\end{array}$

Software procedure:

Step by step procedure to obtain the probability $P\left(x\le 6\right)$ using MINITAB software is given below:

• Choose Calc>Probability Distributions>Binomial Distribution.
• Choose Cumulative Probability.
• Enter Number of trials as 18 and Event probability as 0.20.
• In Input constant, enter 6.
• Click OK.

The output using the Minitab software is given below:

Software procedure:

Step by step procedure to obtain the probability $P\left(x\le 1\right)$ using MINITAB software is given below:

• Choose Calc>Probability Distributions>Binomial Distribution.
• Choose Cumulative Probability.
• Enter Number of trials as 18 and Event probability as 0.20.
• In Input constant, enter 1.
• Click OK.

The output using the Minitab software is given below:

Substitute the values $P\left(x\le 6\right)=0.94873$ and $P\left(x\le 1\right)=0.09908$, the probability is,

$\begin{array}{c}P\left(2\le x\le 6\right)=P\left(x\le 6\right)-P\left(x\le 1\right)\\ =0.94873-0.09908\\ =0.8496\end{array}$

Thus, the probability that the number who have heard of “Modern Warfare 2” is between 2 and 6 inclusive is 0.8496.