A lottery ticket(cupon?) with a row of 3 numbers. Those three numbers can vary from 0-9. A ticket(cupon) might be 5,6,7.
A number is picked randomly among the numbers 0-9. The number that was picked is then put back into the bowl. The lottery ends when it has three unequal numbers, for instance, it can pull out 5 numbers like this: "1,3,3,1,5", which leaves the lottery result with 1, 3,5 as the winning numbers.
What is the probabilty of winning then you have delievered a cupon consisting of the numbers 2,3,4?
My shot at it:
First number is always 1/9 chance.
Second number is 1/9*1/9 chance.
Third number is 1/9*1/9*1/9 chance.
This probability would be right if the lottery could have 3 numbers, no matter if a number got picked twice, or "trice".
Need a bit help :)
April 26th 2010, 08:11 PM
I believe its 1/720. There are 10 possible numbers. You need to match all three. The odds of the first match is 1/10. However after that they will draw a new number if the first number comes up again. So for the second drawing there are basically only 9 possibilities. So the odds of hitting the second number is 1/9. Similarly for the third it is 1/8. So your chances of hitting all three are 1/10 * 1/9 * 1/8 = 1/720.
April 27th 2010, 06:52 AM
Haha, yes, actually, I did dream about this answer, that I should just multiply the latter(?) by reducing the 10 to 9, then to 8...
And yes, my bad, 1/10 possibilities on the first, 1/9 on the second one.....