Question 197977: For the daily lottery game in Illinois, participations select three numbers between 0 and 9. A number cannot be selected more than once. So a winning ticket could be, say 307, but not 337. Purchasing one ticket allows you to select one set of numbers. The winning numbers are announced on tv each night.
a. How many different outcomes (Three different numbers) are possible?
b. If you purchased a ticket for the game tonight, what is the likelihood you will win?
c. Suppose you purchase three tickets for tonight’s drawing and select a different number for each ticket. What is the probability that you will not win with any of the tickets
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For the daily lottery game in Illinois, participations select three numbers between 0 and 9. A number cannot be selected more than once. So a winning ticket could be, say 307, but not 337. Purchasing one ticket allows you to select one set of numbers. The winning numbers are announced on tv each night.
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a. How many different outcomes (Three different numbers) are possible?
There are 10C3 = 120 "sets" of three numbers.
Each set can be arranged in 3! = 6 ways
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Total # of 3 digit numbers is 6*120 = 720
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b. If you purchased a ticket for the game tonight, what is the likelihood you will win?
Ans: 1/720
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c. Suppose you purchase three tickets for tonight’s drawing and select a different number for each ticket. What is the probability that you will not win with any of the tickets
Ans: 1 - (3/720) = 0.9958333.... or 99.58%
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Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The digits 0 through 9 represent 10 choices, so there are 10 ways to pick the first number, then 9 ways to pick the second number, and 8 ways to pick the third number, hence there are 10 times 9 times 8 or 720 different numbers.
One in 720
The probability that one ticket won't win is  . But once the first ticket doesn't win, there are only 719 possibilities left, only one of which wins, so the probability the second ticket doesn't win is  . Likewise, the probability the third ticket doesn't win is  , and the probability of all three events is the product of the three probabilities.
\left(\frac{718}{719}\right)\left(\frac{717}{718}\right))
But wait! There's more! You did win after all! You get to do your own arithmetic.
Super-Double-Plus Extra Credit. Without using pencil, paper, or calculator, how many tickets would you have to buy so that the chance of winning is equal to the chance of losing?
John
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