Posted by Maria on Sunday, August 9, 2009 at 8:17pm.
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 club-dimaond OR dimaond-club.
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.