Chapter 16, Problem 6E

(a)

To determine

To construct: the probability model for this carnival game.

Explanation of Solution

Suppose that the random variable which shows the number of darts.

Denote the cases:

H = Hit the target

H’= Fail to hit the target

The associating probability of hitting the target and fail to hit the target are.

P(H) = 0.10

P(H’) = 0.90

Calculating the probability model for the carnival game.

A person willing to throw a dart in 4 time for $20.

So, X takes the values from 1 to 4. That is, X = 1, 2, 3, 4.

The probability of X=1 is, hitting the target by dart.

That is, P (X = 1) = 0.10

Then the probability would be:

  $\begin{array}{c}P\left(X=2\right)=\left(0.90\right)\left(0.10\right)\\ =0.09\end{array}$

The probability of X=3 can be written as, missing the first two darts and hitting by the third dart.

Then the probability would be:

  $\begin{array}{c}P\left(X=3\right)=\left(0.90\right)\left(0.90\right)\left(0.10\right)\\ =0.081\end{array}$

similarly

  $\begin{array}{c}P\left(X=4\right)=\left(0.90\right)\left(0.90\right)\left(0.90\right)\left(0.90\right)+\left(0.90\right)\left(0.90\right)\left(0.90\right)\left(0.10\right)\\ =0.729\end{array}$

Then the probability model for the carnival game would be

X	1	2	3	4
P(X=x)	0.10	0.90	0.081	0.729

(b)

To determine

To find: the expected number of darts that will throw.

Answer to Problem 6E

3.439

Explanation of Solution

Formula used:

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

Calculation:

Computer the expected number of darts that can be thrown

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum x.P\left(x\right)}\\ =1\left(0.10\right)+2\left(0.09\right)+3\left(0.081\right)+4\left(0.729\right)\\ =0.1+0.18+0.243+2.916\\ =3.439\end{array}$

Hence, the expected number of darts that can be thrown is 3.439

(c)

To determine

To find: the expected winnings.

Answer to Problem 6E

17.197

Explanation of Solution

Formula used:

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

Calculation:

it costs to play $5 and if the person wins $100 they won’t return the $5.

Then the probability model becomes:

X	$95	$90	$85	$80	-$20
P(X=x)	0.10	0.90	0.081	0.729	0.656

The expected winning is,

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum x.P\left(x\right)}\\ =95\left(0.10\right)+90\left(0.09\right)+85\left(0.081\right)+80\left(0.729\right)-20\left(0.656\right)\\ =9.5+8.1+6.885+5.84-13.12\\ =17.197\end{array}$

Therefore, the expected winning is 17.197.