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January 31, 2015

January 31, 2015

Posted by **bart** on Sunday, May 16, 2010 at 5:43pm.

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*51-13 )/(52*51)

odds against = P(against)/P(for) = (52*51- )/ = 2483:169= 191:13

- Math: Help: probability -
**Reiny**, Sunday, May 16, 2010 at 6:10pmprob(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

- Math: 4 different answers? -
**bwb**, Sunday, May 16, 2010 at 7:54pmNow I have four different answers.

- Math: Help: probability -
**Jennifer**, Tuesday, October 12, 2010 at 7:26pmIt 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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