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?

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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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please help
i am stuck:0
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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I think it is true his response
thank you
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.
and because it is in between 1-60
thank you sooooooo much I'm so bad at probability :D
It's okay. Tag me if you need more help, as I've done a little probability and like it.
Don't forget to medal, and then close this question so it's off the list, as now you no longer need help :)
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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