Basic Math
posted by Maria .
I'm really stuck in this question can someone please help me out in this one.
Two cards are drawn without replacement from an ordinary deck of 52 playing cards. What are the odds against drawing a club and a diamond?

first find the prob of getting a club and a diamond.
DC > (13/52)*(13/51) = 169/2652 = 13/204
CD > (13/52)*(13/51) = 13/204
so prob of club and diamond = 13/102
OR prob of club and diamond = C(13,1)*C(13/1)/C(52/2) = 13/102
so prob of NOT gettting a club and diamond = 89
so the odds AGAINST a club and diamond
= 89:13 
Reiny this are my mult. choice that I need to choose from I'm studying for my exam for next week.
A) 204 : 13
B) 191 : 13
C) 13 : 204
D) 13 : 191 
i have worked it I got the answer like you I guess I just need to go with what you and I got right?

odds in favour of some event
= (prob of that event happening):(prob of that event not happening)
since the prob of a club and a diamond is less than the prob of NOT a club and a diamond, we can automatically eliminate choices C and D
choice B would be correct if the wording had been ...
"What are the odds against drawing a club and a diamond in that order ?"
but from the given wording I concluded it could be clubdimaond OR dimaondclub.
The wording should have been more explicit.
BTW, my line of :
so prob of NOT gettting a club and diamond = 89
should have been "
so prob of NOT gettting a club and diamond = 89/102
but I think you figured that out already.