Chapter 9, Problem 16PS

a.

To determine

Find the expected value of the ticket.

Answer to Problem 16PS

  $-\$0.71$

Explanation of Solution

Calculation:

Here, we will find the value of a ticket.

As,

Five numbers can be chosen from $47$ numbers in $C{}_5$ ways.

One number can be chosen from $27$ numbers in $C{}_1$ ways.

As, $x_1,x_2,\mathrm{.......},x_n$ are the values of the outcomes, and $p_1,p_2,\mathrm{.......},p_n$ are the probabilities.

Now, one value of the outcome is the winning amount of $\$12,000,000$ and another value of the outcome is the outgo of $\$1$ for the ticket amount. So, in this probability we will consider the probability of the complement, which is not a probability of winning.

Here, we know that the expected value $V$ is given by:

  $\begin{array}{l}V=p_1{x}_1+p_2{x}_2\mathrm{......}+p_n{x}_n\\ V=\left(\frac{1}{C{}_{5}C{}_1}\right)\left(120000000\right)+\left(1-\frac{1}{C{}_{5}C{}_1}\right)\left(-1\right)\\ V=\left(\frac{1}{C{}_5\times 27}\right)\left(120000000\right)-\left(1-\frac{1}{C{}_5\times 27}\right)\\ V=-0.71\end{array}$

Hence, the expected value of the ticket is approximately $-\$0.71$ .

b.

To determine

Find the expected value of each turn.

Answer to Problem 16PS

  $V=2.53,24turns$

Explanation of Solution

Calculation:

Here, we will find the expected value of each turn.

Now, the values for each dice are: $1,2,3,4,5,6$

So, the probability of each turn of two dice thrown together is:

  $p=\frac{1}{6}\times \frac{1}{6}$

  $p=\frac{1}{36}$

Now, we will consider the values where number of each turn on each dice is the same. So, the numbers of possible values are:

  $\left\{1,1\right\},\left\{2,2\right\},\left\{3,3\right\},\left\{4,4\right\},\left\{5,5\right\},\left\{6,6\right\}$

Here, your score is the product of the dice when numbers are the same. Therefore, the values are:

  $\begin{array}{l}{x}_1=1\\ x_2=4\\ x_3=9\\ x_4=16\\ x_5=25\\ x_6=36\end{array}$

Thus, the probability of complement that is turns of the dice do not produce the same number

  $\begin{array}{l}p\text{'}=1-\frac{1}{36}\\ p\text{'}=\frac{30}{36}\end{array}$

Here, we know that the expected value $V$ is given by:

  $\begin{array}{l}V=p_1{x}_1+p_2{x}_2+\mathrm{....}+p_n{x}_n\\ V=\frac{1}{36}\left(1\right)+\frac{1}{36}\left(4\right)+\frac{1}{36}\left(9\right)+\frac{1}{36}\left(16\right)+\frac{1}{36}\left(25\right)+\frac{1}{36}\left(36\right)+\frac{1}{36}\left(0\right)\\ V=2.53\end{array}$

Now, the number of turns on an average to score $60$ is:

  $\begin{array}{l}=\frac{60}{2.53}\\ =24turns\end{array}$

Hence, the expected value of each turn is approximately $V=2.53$ and number of turns on an average to score $60$ is $24turns$ .