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A lottery involves selecting four digits from 0 to 9. The price of the bet is equal to $1. If the lottery is won, $5000 is collected.

a.

The total number of digits available is equal to 10.

The number of digits to be selected is equal to 4.

As the order of the selections is important, the formula of permutation is utilized as follows.

$$\begin{gathered} {}^{10}{P}_4=\frac{10!}{\left(10-4\right)!} \\ =\frac{10!}{6!} \\ =5040 \end{gathered}$$

Therefore, the number of selections possible is equal to 5040.

b.

The total number of selections is equal to 5040.

The number of combinations of digits that will result in a win is equal to 1.

The probability of winning is computed below.

$$\begin{gathered} P\left(\text{winning}\right)=\frac{1}{5040} \\ =0.000198 \end{gathered}$$

Therefore, the probability of winning is equal to 0.000198.

c.

The net profit on a $1 bet is computed below.

$$\begin{gathered} \text{Net}\text{profit}=\text{Amount}\text{won}-\text{Bet}\text{amount} \\ =5000-1 \\ =4999\text{dollars} \end{gathered}$$

Thus, the net profit is equal to 4999 dollars.

d.

The expected value of the bet is equal to the expected amount that can be won or lost.

It is computed as follows.

$$\begin{gathered} \text{Expected}\text{value}=\text{Net}\text{profit}\left(\text{Probability}\text{of}\text{winning}\right)-\text{Net}\text{loss}\left(\text{Probability}\text{of}\text{losing}\right) \\ =\left(5000-1\right)\left(0.000198\right)-\left(1\right)\left(1-0.000198\right) \\ =-0.00794\text{dollars} \end{gathered}$$

Thus, the expected value of a $1 bet on the Ohio lottery is equal to -0.00794 dollars.

e.

The game that has a higher expected value is considered beneficial.

The expected value of betting on the pass line in craps is equal to -1.4 cents or -0.0140 dollars.

As the expected value of betting on the pass line in craps (-0.0140 dollars) is greater than the expected value of betting on the Ohio Pick lottery (-0.00794 dollars), betting on the pass line in craps is better and more suitable.