Math: Help: probability
posted by bart .
Which answer, if any, is correct to:
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? My options are 13:204; 204:13, 13;191; or 191:13
I had two different people help me and here are the two different answers.
13/52*13/51= 169/2652= 13204 = 204:13
The other way was, but doesn't make sense is: P(club and diamond) = (13/52)(13/51)
P(not drawing a club and a diamond) = 1  ( /(52*51)) = (52*5113 )/(52*51)
odds against = P(against)/P(for) = (52*51 )/ = 2483:169= 191:13

prob(club and diamond) = 2(13/52)(13/51) = 13/102
prob(not club and diamond) = 89/102
odds against a club and diamond = 89 : 13 
Now I have four different answers.

It would be,
there are 13 diamonds and 13 clubs.
because 13/52 is 1/4 which is how many cards of diamonds, clubs, hearts, or spades each have.
SO, if a person draws one card (we're assuming this is a club) which could be 13/52 because there are 13 total clubs, then there would be 1 less card from the 52, so the next fraction would be 3/51 because one card has been tooken away.
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