Weird Probability question

Styrofoam02

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There is a raffle, and we know the total number of tickets: 315. There are also going to be 13 unique winning tickets, each getting the same prize.

If I bought 11 tickets, what is the probability that I will win at least once?

For what it is worth, I can calculate the probability that I will win exactly once

1-(11/315) * (11/315) ... 13 times.


What would I need to do find out all the chances to win?
 
There is a raffle, and we know the total number of tickets: 315. There are also going to be 13 unique winning tickets, each getting the same prize.

If I bought 11 tickets, what is the probability that I will win at least once?

For what it is worth, I can calculate the probability that I will win exactly once

1-(11/315) * (11/315) ... 13 times.


What would I need to do find out all the chances to win?

What is the probability that you will not win even once?

That is usually the key to "at least one" problems.

I don't think your calculation for "exactly once" is correct; can you explain your thinking? I may not be interpreting it as you intend.
 
There is a raffle, and we know the total number of tickets: 315. There are also going to be 13 unique winning tickets, each getting the same prize. If I bought 11 tickets, what is the probability that I will win at least once?

For what it is worth, I can calculate the probability that I will win exactly once
1-(11/315) * (11/315) ... 13 times.
What would I need to do find out all the chances to win?

Your chance of winning exactly one is \(\displaystyle \dfrac{\binom{13}{1}\cdot\binom{302}{10}}{\binom{315}{11}}\) SEE HERE
Is that what you got?

Did you read Prof. Peterson suggestion? What do you think the following means?
\(\displaystyle 1-\dfrac{\binom{302}{11}}{\binom{315}{11}}\) SEE HERE
 
What is the probability that you will not win even once?

That is usually the key to "at least one" problems.

I don't think your calculation for "exactly once" is correct; can you explain your thinking? I may not be interpreting it as you intend.


here's how i was thinking:

The odds of winning with only 1 ticket drawn, is 11/315, So the odds of not winning, are 304/315, or aprox .965

I based my calculations on the 20 eating at a restaurant 20 times in a row with a 5% of getting sick, the odds of getting sick. Which was (.95)(.95)(.95) ... 20 times.

The winning tickets are all drawn at the same time, so I am not sure if that changes anything, but lets say the tickets are numbered 1-315, and there is a randomized that randomly puts names of entrants into the list, and the first 13 are winners. Does this change the math?
 
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