Chapter 16, Problem 4E

(a)

To determine

To construct: a probability model for the amount that win.

Explanation of Solution

Suppose a fair die is rolled.

Let X be the amount winning for the game.

The amount winning for the 6 occurrences in first time is $100 and 6 occurrences in second time is $50.

the probability that getting 6 is 1/6

The probability that not getting 6 is,

  $P\left(\text{Not}\text{getting}6\right)=\frac{5}{6}$

The probability that getting 6 in first time is 1/6.

The probability the getting 6 in second time is,

  $\begin{array}{c}P\left(\text{6}\text{getin}\text{second}\text{time}\right)=\frac{5}{6}\times \frac{1}{6}\\ =\frac{5}{36}\end{array}$

The probability that not getting 6 in two throws is.

  $\begin{array}{c}P\left(\text{Not}\text{gettin}\text{6}\text{in}\text{two}\text{times}\right)=\frac{5}{6}\times \frac{5}{6}\\ =\frac{25}{36}\end{array}$

The probability model is

Roll	1	2	3
Win	$100	$50	$0
probability	1/6	5/36	25/36

(b)

To determine

To find: the expected amount for the win.

Answer to Problem 4E

$23.611

Explanation of Solution

Formula used:

  $E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}$

Calculation:

Use the following formula to compute the expected amount that will win:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ =100\left(\frac{1}{6}\right)+50\left(\frac{5}{36}\right)+0\left(\frac{25}{36}\right)\\ =\frac{600+250}{36}\\ =23.611\end{array}$

Hence, the expected amount that would win is $23.611.

(c)

To determine

To find: that would be willing to pay to play this game.

Explanation of Solution

when a die is rolled and getting 6 than amount that would be willing to pay the game is $23.611 or less than $23.611.