A simple "Pick-3" lottery consists of players picking three digits from 000 to 999 with repeats allowed. It costs \$1$1dollar sign, 1 to play, and a player wins \$500$500dollar sign, 500 if they match all three numbers drawn. Jaime decides to buy 101010 tickets to improve their chances of winning. The table below displays the probability distribution of XXX, the amount of money Jaime profits from this strategy.

Answer:

Step-by-step explanation:

To determine the probability distribution of Jaime's profit, we first need to calculate the probability of winning and the probability of losing.

There are 1,000 possible outcomes in the lottery, from 000 to 999. Since repeats are allowed, there are 1,000 x 1,000 x 1,000 = 1,000,000,000 possible combinations of numbers that can be drawn.

To win, Jaime's three-digit number must match the three digits drawn in the lottery, in the correct order. There is only one winning combination out of 1,000,000,000 possible combinations, so the probability of winning is:

P(win) = 1/1,000,000,000

To lose, Jaime's number must not match the winning number. There are 999,999,999 possible losing combinations, so the probability of losing is:

P(loss) = 999,999,999/1,000,000,000 = 0.999999999

Now we can calculate the expected profit for each ticket:

E(profit per ticket) = P(win) x (500 - 1) + P(loss) x (-1) = (1/1,000,000,000) x 499 - (999,999,999/1,000,000,000) x 1 = -0.499999501

This means that Jaime can expect to lose about 50 cents per ticket on average.

Since Jaime is buying 10,000 tickets, the expected profit for this strategy is:

E(total profit) = E(profit per ticket) x number of tickets = -0.499999501 x 10,000 = -4,999.99501

So Jaime can expect to lose about $4,999.99 on average by buying 10,000 tickets. The probability distribution of XXX, the amount of money Jaime profits from this strategy, would have a single outcome of -4,999.99 with probability 1, since this is the expected profit for this strategy.