Posted by **Maria** on Sunday, August 9, 2009 at 8:17pm.

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?

- Basic Math -
**Reiny**, Sunday, August 9, 2009 at 8:28pm
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

- Basic Math -
**George**, Sunday, August 9, 2009 at 8:30pm
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

- Basic Math -
**George**, Sunday, August 9, 2009 at 8:34pm
i have worked it I got the answer like you I guess I just need to go with what you and I got right?

- Basic Math -
**Reiny**, Sunday, August 9, 2009 at 8:54pm
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.

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