Lottery Probability question

douglasd3

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Apr 10, 2014
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HI - I have been using wiki to brush up on my math skills for a question I have been tinkering with.
Here it is and this is a real world scenario.

The state is issuing three licenses to qualified applicants. I have two applications in, there are a total of 9 viable applications (including my applications).
They are conducting a lottery to determine what applications are chosen to move forward.

What are the chances that one of mine will make it through the lottery (be chosen by the state) ?
What are the chances that both will make it in the lottery?

THANKS!
 
So there are nine "candidates", three of them are to be chosen, and you have two of them?

The probability one of yours will be chosen first is 2/9. Given that there are now 8 candidates and one of them is yours. The probability yours will NOT be chosen is 7/8. Given that yours is not chosen there are now 7 candidates and one of them is yours. The probability that your is not chosen is 6/7. So the probability that one of yours will be chosen on the first draw is (2/9)(7/8)(6/7)= 12/72= 1/6.

If you do the same to find the probability that exactly one yours will be chosen on the second or third draws you will get different fractions but the same numbers as numerators and denominators so the same product. Since there are three draws, the probability that exactly one of yours will be chosen on any draw is 3(1/6)= 1/2.

To find the probability that both of yours are chosen, do much the same thing. There are 9 "candidates" and you have two of them so the probability that one of yours will be chosen first is again 2/9. Now there are 8 candidates and on of them is yours so the probability yours will be chosen is 1/8. Given that that happens you cannot be chosen on the third draw. So the probability you will be chosen on both first and second draws s (2/9)(1/8)= 1/36. Again, you can show that the probability that both will be drawn on two other draws is the same, 1/36. There are again 3 such orders (writing "Y" for "yours" and "O" for "others" they are "YYO", "YOY", and "OYY") so the probability that both will be drawn is 3(1/36)= 1/12.

The probability that none of yours is chosen is (7/9)(6/8)(5/7)= 5/12 so the probability that "at least one of yours is chosen" is 1- 5/12= 7/12. Equivalently, 1/2+ 1/12= 6/12+ 1/12= 7/12.
 
So there are nine "candidates", three of them are to be chosen, and you have two of them?

The probability one of yours will be chosen first is 2/9. Given that there are now 8 candidates and one of them is yours. The probability yours will NOT be chosen is 7/8. Given that yours is not chosen there are now 7 candidates and one of them is yours. The probability that your is not chosen is 6/7. So the probability that one of yours will be chosen on the first draw is (2/9)(7/8)(6/7)= 12/72= 1/6.

If you do the same to find the probability that exactly one yours will be chosen on the second or third draws you will get different fractions but the same numbers as numerators and denominators so the same product. Since there are three draws, the probability that exactly one of yours will be chosen on any draw is 3(1/6)= 1/2.

To find the probability that both of yours are chosen, do much the same thing. There are 9 "candidates" and you have two of them so the probability that one of yours will be chosen first is again 2/9. Now there are 8 candidates and on of them is yours so the probability yours will be chosen is 1/8. Given that that happens you cannot be chosen on the third draw. So the probability you will be chosen on both first and second draws s (2/9)(1/8)= 1/36. Again, you can show that the probability that both will be drawn on two other draws is the same, 1/36. There are again 3 such orders (writing "Y" for "yours" and "O" for "others" they are "YYO", "YOY", and "OYY") so the probability that both will be drawn is 3(1/36)= 1/12.

The probability that none of yours is chosen is (7/9)(6/8)(5/7)= 5/12 so the probability that "at least one of yours is chosen" is 1- 5/12= 7/12. Equivalently, 1/2+ 1/12= 6/12+ 1/12= 7/12.


Thanks! That all makes sense. THE YYO YOY OYY was what I had forgotten from math class way back when.
 
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