check please

[OptimusPrime]

There are 100 dogs participating in a competition (two from each state)

a. If two dogs are chosen at random, what is the probability they are from the same state?

100/100 x 1/99 = 1/99

b. If a collection of 10 dogs is chosen at random, what is the probability no two are from the same state?

100/100 x 98/99 x 96/98 x 94/97 x 92/96 x... x 82/91

Am I right?

Thanks

[UrbanXrisis]

nope

a. 100/100 * 49/99

you just chose one from a state, and now you want to choose one from that state again...there are now 99 dogs left, and 49 of the dogs are from that state to choose from

try b by yourself and apply the same logic

[HallsofIvy]

nope

a. 100/100 * 49/99

you just chose one from a state, and now you want to choose one from that state again...there are now 99 dogs left, and 49 of the dogs are from that state to choose from

try b by yourself and apply the same logic

?? What?? There are only 2 dogs from each state. If you have already chosen one, how can there be 49 dogs left from THAT state?

Using UrbanXrisis's method, though, the first dog chosen is from SOME state. There are 99 dogs left that could be chosen second and exactly one of those is from the same state as the first: the probability that that dog will be chose is exactly 1/99 which is what Optimus Prime said originally?

In problem b, I think Optimus Prime is saying: The first dog can be from any state: 100/100. There are now 99 dogs left, 98 from states other than the first so the probability that the second dog is not from the same state is 98/99. There are now 97 dogs left, 96 of them not from the state of either of the first two: probability the third dog is not from the same state, 96/97, etc. The product you have looks good.

I will make a comment on notation: I am sure you recognize that 100(99)(98)...(91) is 100!/90!. You can also write 100(98)(96)...(82) as 50(2)(49)(2)(48)(2)...(41)(2)= 210(100!/40!).

The product you have is (100!/90!)(40!/100!)(2-10)= 2-10(40!/90!).