South Carolina Pick 3 In South Carolina’s Pick 3 lottery game, you can pay \(1 to select a sequence of three digits, such as 227. If you buy only one ticket and win, your prize is \)500 and your net gain is $499.

a. If you buy one ticket, what is the probability of winning?

b. If you play this game once every day, find the mean number of wins in years with exactly 365 days.

c. If you play this game once every day, find the probability of winning exactly once in 365 days.

d. Find the expected value for the purchase of one ticket.

The number of moons of all the 8 planets is listed.

a.

The total number of possible sequences is computed below:

\(10 \times 10 \times 10 = 1000\)

The number of winning sequences = 1

The probability of winning is computed below:

\(\begin{array}{c}P\left( {{\rm{winning}}} \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{winning}}\;{\rm{sequences}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{sequences}}}}\\ = \frac{1}{{1000}}\\ = 0.001\end{array}\)

Therefore, the probability of winning is equal to 0.001.

b.

The number of days in a year is equal to n=365.

The probability of winning a game is equal to p=0.001.

The mean number of wins in a year if one game is played every day is computed below:

\(\begin{array}{c}\mu = np\\ = 365\left( {0.001} \right)\\ = 0.365\end{array}\)

Thus, the mean number of wins in a year if one game is played every day is equal to 0.365.

c.

Let X denote the number of wins in a year.

So, X follows a Poisson distribution with mean equal to 0.365.

The probability of winning once in a year is computed below:

\[\begin{array}{c}P\left( {X = x} \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( {X = 1} \right) = \frac{{{{\left( {0.365} \right)}^1}{{\left( {2.71828} \right)}^{ - 0.365}}}}{{1!}}\\ = 0.253\end{array}\]\(\left( {{x_{{\rm{loss}}}}} \right)\)

Thus, the probability of winning once in a year is equal to 0.253.

d.

The amount received on winning\(\left( {{x_{{\rm{win}}}}} \right)\)is equal to $499.

The probability of winning\(\left( {{p_{win}}} \right)\)is equal to 0.001.

The amount lost on losing is equal to -1dollar.

The probability of losing is equal to:

\(\begin{array}{c}{p_{{\rm{loss}}}} = 1 - {p_{win}}\\ = 1 - 0.001\\ = 0.999\end{array}\)

The expected value lost/gained on the purchase of one ticket is computed below:

\(\begin{array}{c}E = {x_{win}}{p_{win}} + {x_{loss}}{p_{loss}}\\ = \left( {499} \right)\left( {0.001} \right) + \left( { - 1} \right)\left( {0.999} \right)\\ = - 0.50\end{array}\)

Thus, the expected value for the purchase of one ticket is equal to – 0.50 dollars or – 50 cents.