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February 1, 2015

February 1, 2015

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

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 8:28pmfirst 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 8:30pmReiny 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 8:34pmi 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 8:54pmodds 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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