| Money For Lunch? |
In my town there are four restaurants. The price for lunch at the four restaurants is $5.00, $10.00, $15.00, and $20.00.
In my wallet I have four bills. They might be ones, fives, tens, or any combination of those (with each type of bill being equally likely) but I don't know what they are.
If I randomly select a restaurant, what is the probability I'll have enough money for lunch? |
Problem Moderated by: Douglas |
| Problem Solution |
First, we need to determine how many possibilities there are for the bills in my wallet. There are four bills, and 3 possibilities for each, so the total combinations is 34 = 81.
Consider each case separately
Five Dollar Restaurant
The only way I won't have enough money is if all four bills are ones. This can happen in exactly one way, so the probability is 1/81.
Ten Dollar Restaurant
In addition to the possibility described above, the only other way I won't have enough money is if three bills are ones, and one is a five. There are four ways this to happen, so the probability is 1/81 + 4/81 = 5/81
Fifteen Dollar Restaurant
In addition to the possibilities above, I won't have enough money if I have a ten and four ones, or 2 ones and 2 fives. The probability then is:
1/243 + 4/81 + 4/81 + 6/81 = 15/81
Twenty Dollar Restaurant
In addition to the possibilities above, I won't have enough money if I have 3 fives and a one, or a ten, a five, and 2 ones. So the probability is:
1/81 + 4/81 + 4/81 + 6/81 + 4/81 + 12/81= 31/81
Since each restaurant is equally likely, we divide each probability by four and then add them. This is the probability that I won't have enough money: 13/81.
So the probability I will have enough money is: 68/81 |
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