Most states and Canadian provinces have government-sponsored lotteries. Here is

Question

Most states and Canadian provinces have government-sponsored lotteries. Here is a simple lottery wager, from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a three-digit number, 000 to 999. The state chooses a three-digit winning number at random and pays you $500 if your number is chosen.

a. Find the probability distribution of the random variable, X, defined to be the amount your ticket pays you.

b. Find the mean, variance, and standard deviation of X.

c. Find the probability distribution, mean, variance, and standard deviation of the random variable W defined to be your winnings if a ticket costs $1.

d. Suppose now that you buy a $1 ticket on each of two different days. The payoffs X and Y of the two tickets are independent because separate drawings are held each day. Find the probability distribution, mean and standard deviation of the total payoff, X + Y.

Explanation / Answer

Part (a)

Probability distribution of the random variable, X, defined to be the amount the ticket pays:

X = 500 if the ticket number tallies with the number picked by the lottery

   = 0 otherwise.

Since there are 1000 numbers involved, probability the ticket number tallies with the number picked by the lottery = 1/1000 = 0.001.

Thus, Probability distribution of the random variable, X, is:

500 with probability 0.001 and 0 with probability 0.999. ANSWER

Part (b)

Mean = (500 x 0.001) + (0 x 0.999) = 0.5 ANSWER 1

E(X2) = (5002 x 0.001) + (02 x 0.999) = 250

Variance of X = E(X2) – Mean2 = 249.75 ANSWER 2

Standard Deviation = sqrt(Variance) = 15.803 ANSWER 3

Part (c)

W = winning amount paid by the lottery – ticket cost = X – 1.

So,W = 499 if the ticket number tallies with the number picked by the lottery

         = - 1 otherwise.

Since there are 1000 numbers involved, probability the ticket number tallies with the number picked by the lottery = 1/1000 = 0.001.

Thus, Probability distribution of the random variable,W, is:

499 with probability 0.001 and - 1 with probability 0.999. ANSWER 1

Mean = (499 x 0.001) + (- 1 x 0.999) = - 0.5 ANSWER 2

E(W2) = (4992 x 0.001) + {(- 1)2 x 0.999) = 250

Variance of W = E(W2) – Mean2 = 249.75 ANSWER 3

Standard Deviation = sqrt(Variance) = 15.803 ANSWER 4

[Additional Inputs: since W = X – 1. E(W) = E(X) – 1 and SD(W) = SD(X)]

Part (d)

For convenience, let Z = X + Y.

Then,

Z = (500 + 500) if numbers on both tickets tally with the number picked by the lottery;

   = (500 + 0) if number of first ticket tallies with the number picked by the lottery, but

number of second ticket does not tally with the number picked by the lottery;

   = (0 + 500) if number of second ticket tallies with the number picked by the lottery, but

number of first ticket does not tally with the number picked by the lottery;

   = (0 + 0) if numbers on both tickets do not tally with the number picked by the lottery.

Thus, probability distribution of Z is:

P(Z = 1000) = 0.001 x 0.001 = 0.000001;

P(Z = 500) = 2 x (0.001 x 0.999) = 0.001998;

P(Z = 0) = 0.999 x 0.999 = 0.998001. ANSWER 1

Mean (Z) = (1000 x 0.000001) + (500 x 0.001998) + (0 x 998001) = 1 ANSWER 2

E(Z2) = (10002 x 0.000001) + (5002 x 0.001998) + (02 x 998001) = 500.5

V(Z) = 500.5 – 12 = 499.5 ANSWER 3

SD(Z) = sqrt(499.5) = 22.35 ANSWER 4

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