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1. ## lottery ticket problem

ok, Lottery tickets have 4 digits and they go from 0000 to 9999. There are therefore 10,000 different tickets. Suppose you buy a ticket. what is the probability you match only the last two digits of the winning ticket?

2. I believe this is how its done:

P(matching only the last two) = P(match the last two) - P(match the last three) - P(match all four)

1/100 - 1/1000 - 1/10000 = (100-10-1)/10000= 89/10000

3. Isn't it just P(not matching first) x P(not matching second) x P(matching third) x P(matching fourth) = 9/10 x 9/10 x 1/10 x 1/10 = 81/10000?

4. Thanks guys...... The actex book says its 90/10000........ it says that the thousand place can be any of 10 numbers and that the 100 place can be any 9 making 90 total....... I thought it was 81 too.....9 wrong for the first and 9 wrong for the second digit.

5. I dont get how the 1000 place can be any of 10 numbers because then 1 number would match the winning??

6. Any answers? Is the book wrong?

first divide this set into two parts, the first two digits and last two digits.

1. the probability of matching the last two digits is 1/100.

2. either matching the first or second digit fails to fulfill your goal to match the last two digits only. Therefore, this part is 10/10 * 9/10

then 9/10 * 1/100 = 90/10000

Can someone be able to explain it?

8. Wouldn't that imply that there is a 100% chance that you fail to match the first digit? I sure don't want to play that lottery.

9. I mean either first or second digit matched is considered as a failure in this question. (once one of the first two digits is matched, you don't need to match another one to be failed.)

10. Is it possible for you to write out word for word the acetex solution? I remember this problem and I know I figured out what they meant, but I've forgotten how.

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