Chapter 7, Problem 121CR

Kyle and Lygia are going to play a series of Trivial Pursuit games. The first person to win four games will be declared the winner. Suppose that outcomes of successive games are independent and that the probability of Lygia winning any particular game is 0.6. Define a random variable x as the number of games played in the series.

1. a. What is p(4)? (Hint: Either Kyle or Lygia could win four straight games.)
2. b. What is p(5)? (Hint: For Lygia to win in exactly five games, what has to happen in the first four games and in Game 5?)
3. c. Determine the probability distribution of x.
4. d. On average, how many games will the series last?

To determine

Calculate p(4).

Answer to Problem 121CR

The value of p(4) is 0.1552.

Explanation of Solution

Calculation:

It is given that, K and L play a series of T games. The first person who wins the first four games is declared the winner. The outcomes of the successive games are independent to each other. Also, the probability of L winning any game is 0.6.

Thus, the probability of K winning any game is 0.4.

Define the random variable x as the number of games played in the series.

The probability p(4) represents that either K or L win the first four games.

$\begin{array}{c}p\left(4\right)=P\left(\text{L to win the series}\right)+P\left(\text{K to win the series}\right)\\ =\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.6\right)+\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.4\right)\\ =\text{0}\text{.1296}+\text{0}\text{.0256}\\ =\text{0}\text{.1552}\end{array}$

Thus, the value of p(4) is 0.1552.

To determine

Calculate p(5).

Answer to Problem 121CR

The value of p(5) is 0.2688.

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(5\right)=P\left(\text{L to win the series}\right)+P\left(\text{K to win the series}\right)\\ =\underset{{\text{4C}}_1\text{ ways}}{\underbrace{\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.4\right)}}\left(0.6\right)+\underset{{\text{4C}}_1\text{ ways}}{\underbrace{\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.6\right)}}\left(0.4\right)\\ =4\times {\left(0.6\right)}^3\left(0.4\right)\left(0.6\right)+4\times {\left(0.4\right)}^3\left(0.6\right)\left(0.4\right)\\ =\text{0}\text{.20736}+\text{0}\text{.06144}\\ =\text{0}\text{.2688}\end{array}$

Thus, the value of p(5) is 0.2688.

To determine

Obtain the probability distribution of x.

Answer to Problem 121CR

The probability distribution of x is given below:

x	4	5	6	7
p(x)	0.1552	0.2688	0.29952	0.27648

Explanation of Solution

Calculation:

Here, the series is completed when any one wins four games. Hence, the random variable x takes values 4, 5, 6, and 7.

The value of p(6) is obtained as given below:

$\begin{array}{c}p\left(6\right)=P\left(\text{L to win the series}\right)+P\left(\text{K to win the series}\right)\\ =\underset{{\text{5C}}_3\text{ ways}}{\underbrace{\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.4\right)\left(0.4\right)}}\left(0.6\right)+\underset{{\text{5C}}_3\text{ ways}}{\underbrace{\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.6\right)\left(0.6\right)}}\left(0.4\right)\\ =10\times {\left(0.6\right)}^3{\left(0.4\right)}^2\left(0.6\right)+10\times {\left(0.4\right)}^3{\left(0.6\right)}^2\left(0.4\right)\\ =\text{0}\text{.20736}+\text{0}\text{.09216}\\ =\text{0}\text{.29952}\end{array}$

The value of p(7) is obtained as given below:

$\begin{array}{c}p\left(7\right)=P\left(\text{L to win the series}\right)+P\left(\text{K to win the series}\right)\\ =\underset{{\text{6C}}_3\text{ ways}}{\underbrace{\left(0.6\right)\left(0.6\right)\left(0.6\right)\left(0.4\right)\left(0.4\right)\left(0.4\right)}}\left(0.6\right)+\underset{{\text{6C}}_3\text{ ways}}{\underbrace{\left(0.4\right)\left(0.4\right)\left(0.4\right)\left(0.6\right)\left(0.6\right)\left(0.6\right)}}\left(0.4\right)\\ =20\times {\left(0.6\right)}^3{\left(0.4\right)}^3\left(0.6\right)+20\times {\left(0.4\right)}^3{\left(0.6\right)}^3\left(0.4\right)\\ =\text{0}\text{.165888}+\text{0}\text{.110592}\\ =\text{0}\text{.27648}\end{array}$

Thus, the probability distribution of x is given below:

x	4	5	6	7
p(x)	0.1552	0.2688	0.29952	0.27648

To determine

Find the average number of games for the series to last.

Answer to Problem 121CR

The average number of games for the series to last is approximately 6.

Explanation of Solution

Calculation:

Mean of the random variable x is obtained as given below:

x	p(x)	$x\cdot p\left(x\right)$
4	0.1552	0.6208
5	0.2688	1.344
6	0.29952	1.79712
7	0.27648	1.93536
		${\displaystyle \sum _{x=4,\mathrm{...},7}x\cdot p\left(x\right)}=5.6973$

From the above table, the mean is,

$\begin{array}{c}{\mu }_x={\displaystyle \sum _{x=4,\mathrm{...},7}x\cdot p\left(x\right)}\\ =5.6973\end{array}$

Thus, the average number of games for the series to last is approximately 6.