Chapter 8, Problem 8.76RE

(a)

To determine

To show that this is a binomial setting.

Explanation of Solution

Given:

  $\begin{array}{l}n=25\\ p=0.25\end{array}$

We know that the Binomial theorem is as:

  $P(X=k)=C{}_k\times p^k\times {(1-p)}^{n-k}$

The four conditions for a binomial setting are: binary (success/failure), independent trials, fixed number of trials and the probability of success is the same for each trial.

Binary: Satisfied because success is “Win game” and failure is “Lose game”.

Independent trials: Satisfied, assuming that the different games are independent.

Fixed number of trials: Satisfied, because the fixed number of trials are $25$ games.

Probability of Success: Satisfied, because the probability of success is $p=0.25$ .

Thus the setting is binomial.

(b)

To determine

To calculate and interpret $P(Y=5)$ .

Answer to Problem 8.76RE

The probability is $16.45\%$ .

Explanation of Solution

Given:

  $\begin{array}{l}n=25\\ p=0.25\end{array}$

We know that the Binomial theorem is as:

  $P(X=k)=C{}_k\times p^k\times {(1-p)}^{n-k}$

Evaluate at $k=5$ :

  $\begin{array}{l}$ P(X=5)=C{}_5\times {(0.25)}^5\times {(1-0.25)}^{25-5}$=53130\times {0.25}^5\times {0.75}^{20}\\ \approx 0.1645\\ =16.45\%\end{array}$

  $16.45\%$ of all samples of $25$ games, will result in exactly five games of the $25$ games being won.

(c)

To determine

To calculate and interpret $P(Y>5)$ .

Answer to Problem 8.76RE

Explanation of Solution

Given:

  $\begin{array}{l}n=25\\ p=0.25\end{array}$

We know that the Binomial theorem is as:

  $P(X=k)=C{}_k\times p^k\times {(1-p)}^{n-k}$

Evaluate at $k=0,1,2,3,4,5$ :

  $\begin{array}{}$ P(X=0)=C{}_0\times {(0.25)}^0\times {(1-0.25)}^{25-0}\approx 0.0008$$ P(X=1)=C{}_1\times {(0.25)}^1\times {(1-0.25)}^{25-1}\approx 0.0063$$ P(X=2)=C{}_2\times {(0.25)}^2\times {(1-0.25)}^{25-2}\approx 0.0251$$ P(X=3)=C{}_3\times {(0.25)}^3\times {(1-0.25)}^{25-3}\approx 0.0641$$ P(X=4)=C{}_4\times {(0.25)}^4\times {(1-0.25)}^{25-4}\approx 0.1175$$ P(X=5)=C{}_5\times {(0.25)}^5\times {(1-0.25)}^{25-5}\approx 0.1645$\end{array}$

Use the addition rule:

  $\begin{array}{l}P(x\le 5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\ =0.0008+0.0063+0.0251+0.0641+0.1175+0.1645\\ =0.3783\end{array}$

Use the compliment rule:

  $\begin{array}{l}P(X>5)=1-P(x\le 5)\\ =1-0.3783\\ =0.6217\\ =62.17\%\end{array}$

  $62.17\%$ of all samples of $25$ games, will result in more than five games of the $25$ games being won.

(d)

To determine

To explain how much money should the player expect to win or lose in $25$ games.

Answer to Problem 8.76RE

Lose $\$0$ in $25$ games.

Explanation of Solution

In the previous exercise, we found that you were expected to lose $\$0$ per game.

Thus, you are then expected to lose:

  $25\times \$0=\$0$

You have to lose $\$0$ in $25$ games.