anonymous
  • anonymous
In a state lottery, a player selects ten different numbers between 1 and 60 inclusive. If those ten numbers are drawn in the lottery, the player wins millions of dollars. Nguyen is considering buying a ticket, and his friend Brad suggests, "Why don't you pick 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10?" Nguyen replies, "Are you kidding? Those will never come up!" Brad comes back, "They have just as good a chance as anything else!" Is Brad's claim true or false?
Probability
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
please help
anonymous
  • anonymous
i am stuck:0
horsegirl27
  • horsegirl27
Well, it seems to me that he might be right. There is as good a chance a number like 20 will be picked as there is for a number like 3.

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anonymous
  • anonymous
I think it is true his response
anonymous
  • anonymous
thank you
horsegirl27
  • horsegirl27
But, picking them in that order would be near impossible. But there is a chance those numbers could be picked at random, just like there is a chance 11,12,13,14,15,16,17,18,19 and 20 could be picked.
anonymous
  • anonymous
and because it is in between 1-60
anonymous
  • anonymous
thank you sooooooo much I'm so bad at probability :D
horsegirl27
  • horsegirl27
It's okay. Tag me if you need more help, as I've done a little probability and like it.
horsegirl27
  • horsegirl27
Don't forget to medal, and then close this question so it's off the list, as now you no longer need help :)
ybarrap
  • ybarrap
Think about the problem like this. Think about ALL probability problems like this. Always, if you can, make them simpler so you can get some intuition. Rather than having numbers 1-60, let's just have numbers 1-10. Now if your friend says pick the number "1" out of the 10. Is this any less likely than say a "5" ? Hopefully you'll see that every possibility is equally likely and one choice isn't more likely than another. Now choose two numbers, say 1 and 2. Is this more likely than say 3 and 7 ? The system is the same so we don't expect one pair of numbers to be more likely than another because of the symmetry of the problem. When I say "symmetry" I mean, think of all combinations, for example how many pairs can we make with only 10 numbers? 12 21 14 38 39 .... and so on There are 10*9=90 permutations, but half of these are duplicates, like 12 and 21, so really there are 90/2 = 45 combinations There is a formula for this, if you are not aware: $$ {{10}\choose{2}}=\cfrac{10!}{2!(10-2)!}=\cfrac{10!}{2!8!}=45 $$ http://www.wolframalpha.com/input/?i=10+choose+2 So the chance of say 14 being the winning lottery ticket is 1/45, the same as if you were to choose 12. So by "symmetry" I mean that each choice is, from a probability perspective, the same. For your problem, the number of combinations for this lottery is $$ {{60}\choose{10}}=75,394,027,566 $$ and the chance of winning is 1/75,394,027,566 or about \(1.3× 10^{-11}\) or 0.000000000013. So combinations like 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 and 11,12,13,14,15,16,17,18,19 and 20 are each equally likely, by symmetry. Hope this helps.

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