An urn contains 3 one-dollar bills, 1 five-dollar bill and 1 ten-dollar bill. A player draws bills one at a time without replacement from the urn until a ten-dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: (A) The probability of winning $12. (B) The probability of winning all bills in the urn. (C) The probability of the game stopping at the second draw.

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PART A

There are 3 $1 bills, 1 $5 bill, and 1 $10 bill. This gives us 5 total bills.

First, we want to find the probability of winning $12. Well, to win, you have to draw the $10 bill. You only have room for two dollars beforehand to equal $12 dollars after pulling out the ten. So, this is the probability of drawing two one dollar bills and the the ten. Let's calculate this below.

Where did I get these numbers from? Well 3 of the 5 bills are $1, so the first probability is 3/5. Then, if we draw one of the $1 bills, there are only 2 of those left and 4 total bills, so the probability is then one half. Finally, there would be only 3 left and you need to pick the $10 bill, which is a probability of 1/3.

The probability of winning $12 is 1/10 or 10%.

PART B

Now, we want to find the probability of picking every single bill before the ten. This means that we pick the three one dollar bills and the five dollar bill before the ten.

To pick the first $1 bill, our probability is 3/5, and then for the second it is 1/2. For the third, there are three total cards and 1 $1 bill, so the probability is 1/3. Then we have a 1/2 chance of picking the $5 bill over the $10 bill, giving us this solution.

The probability of winning all bills in the urn is 1/20 or 5%.

PART C

For this event, we want to get any bill that isn't the $10 and then we want the $10 on the second one.

Since there are 4 bills that aren't the $10, our first probability is 4/5. Then, we only have 4 left, with 1 being the $10, so our second probability is 1/4.

The probability of the game stopping at the second draw is 1/5 or 20%.

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