Probability with dates?

Question

On September 11, 2002, the 1st anniversary of the terrorist attack on the World Trade Center, the New York State lottery's daily number came up 9-1-1.

1) What is the probability that the winning numbers match the date on any given day? (Note that for this problem and all other problems, months with only one digit should not place a zero before the digit. In other words, if the date is January 1, then the lotter numbers would be 1-0-1, NOT 0-1-1. In October through December for all dates in which the day has two digits, there is no possible way in which those dates can be picked.)

2) What is the probability of an entire year passing without this happening?

3) What is the probability that the date and winning lottery number match at least once during any year?

4) If every one of the 50 states has a 3 digit lottery, what is the probability that at least one of them will come up 9-1-1 on Sept 11?


Thank you for any help. Explanations are appreciated.

Answer 1

The winning number is always a three digit number, of which there are 1000, not 100.

There are 65 days when the winning number cannot possibly match the date. Those days are

October 10 through October 31
November 10 through November 30
December 10 through December 31

On the remaining 300 days, the probability that the winning number will match the date is 1/1000. So I would say that on "any given day" the probability of a match is

(300/365) * (1/1000)

Gary H, below, forgot that in October, November, and December, the first nine days of each month are good, so he is short by 27 days. There are 300 good days, not 273.

For part 2), the probability that we don't get a match on any of the 300 days when a match is possible is 999/1000. So the probability of that happening on all 300 such days is

(999/1000)^300

For part 3), I agree with Gary H that it is 1 minus the answer to part 2).

In part 4), Gary H's reasoning cannot be correct. Let's say that there were 1001 states instead of 50. Then his reasoning would lead to a probability of greater than 1, clearly nonsense.

What you must do instead is ask, what is the probability of 9-1-1 NOT coming up on September 11 in Maine? It's 999/1000. What is the probability of it NOT coming up in Vermont? 999/1000 again. What is the probability of it NOT coming up in ANY of the 50 states?

(999/1000)^50

What is the probability that it comes up in at least one state?

1 - (999/1000)^50

This reasoning always leads to a probability less than 1, for no matter how many states you have you can never be sure that 9-1-1 will come up on September 11.

Answer 2

Unfortunately, you can't use some of the methods the person above used to answer this question (although the exact way he phrased it seems ok).
For #1, the person is absolutely correct... when the date is possible to be picked. I think you need to use a weighted average with this probability and 0, using the number of days when this is possible, and the number of days when it isn't possible. However, if you don't think the question wants you to be specific, then ignore this part, and just use 1/100 like the person above.
For #2, you can never just multiply the attempts by the probability to get the answer, although it seems like you should be able to. This is because the other way assumes that you are only winning once, while this one lets you win more than once, which is what you need. For example, if you had 13 cans of coke, and each one had a 1/12 chance of winning a prize, that doesn't mean you won for sure. Instead, take (1 - your answer to #1)^365 for the correct answer to this problem, since this is the chance of you NEVER winning.
For #3: The probability of it occuring ever in a year is 1 - (1 - your answer to #1)^365, of course. This is because it is the total probability (100%) minus the probability of #2... since the events are "mutually exclusive."
4.) On September 11, the date is possible. Therefore, you can use the 1/100 probability. Then, you know that you have fifty chances. So, you NEED to use the method above, NOT just 50/100. That means, do this: 1 - (1 - 1/100)^50.

Answer 3

It's a little ambiguous, but it sounds like the problem allows for a two-digit month, 1-digit day to be a match.

The other tricky thing about this problem is that NY State has TWO three-digit number lotteries per day (midday and evening).

Thus, for question #1, the probability of a match on any given day will be almost one in 500 (1=0.999^2=0.1999% to be exact) if the "given day" has a legal match.

For question #2, if the year is not a leap year, the probability is 0.999^(2*300) (300 is the number of days that can have a match)=about 54.8646907%

For question #3, the probability is the reverse of the previous question, or about 45.1353093%.

For question #4, if all 50 states had only a single 3-digit lottery, then the probability would be 1-.999^50, or about 4.8794372%. If they all had two 3-digit lotteries, it would be 1-.999^100, about 9.5207853%

Answer 4

Since you seemed to have asked this for the second time, I guess you really need the answer, huh? :^)

Ok, first figure all possible matching dates of a year -

1-0-1 thru 1-3-1 , total of 31
2-0-1 thru 2-2-8 , total of 28 plus 2-2-9 for a leap year
3-0-1 thru 3-3-1 , total of 31
4-0-1 thru 4-3-0 , total of 30
5-0-1 thru 5-3-1 , total of 31
6-0-1 thru 6-3-0 , total of 30
7-0-1 thru 7-3-1 , total of 31
8-0-1 thru 8-3-1 , total of 31
9-0-1 thru 9-3-1 , total of 30

for a total of 273 possible matches/yr, 274 in leapyears

1) 1/1000 for 273 of the days, 0 for the other 92, so

total probability of matching any given day would =

(1/1000)*(273/365), or (1/1000)*(273/366) in LY,
or [3(1/1000)*(273/365) + (1/1000)*(273/366)]/4 for total

2) (273/1000)*(273/365), or (273/1000)*(273/366) in LY
or [3(273/1000)*(273/365) + (273/1000)*(273/366)]/4 for total

3) 1 minus the answer to #2

4) (1/1000) * 50, aka 1 in 20

I think I've got these right.... good luck

Answer 5

1) Lets assume the winning number is always a 3 digit number.
There are 100 possible numbers.
Each day, the probability of any given number is 1/100

2) There are 365 day/year. 1/100 * 365 = 365/100 = 3.65
On average there will be 3.65 days/ year when the day matches the number.

Answer 6

You have severely too much time on your hands. Although I do rank as "model".... heh. 6'3? Your a model! 6'4? OMG YOU FREAKING WNBA LESBIAN AHHHHHHHHHhh