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January 27, 2015

January 27, 2015

Posted by **Anonymous** on Friday, August 24, 2012 at 2:58pm.

P(five of clubs) = 1

P(five) = 2

P(club) = 3

P(jack) = 4

P(spade) = 5

P(jack of spades) = 6

P(five and a jack) = 7

P(five or a jack) = 8

P(heart and a jack) = 9

P(heart or a jack) = 10

0

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- math -
**David Q/R**, Saturday, August 25, 2012 at 5:53amBecause there's no sample space diagram here to clarify the point, my first question is "Does an ordinary deck of cards contain a joker?", because that affects the answer to every subsequent question. Assuming that it doesn't:

P(five of clubs): there's only one five of clubs in a deck, so the answer is 1/52.

P(five): there are four fives in a deck, so the answer is 4/52 = 1/13.

P(club): there are 13 clubs in a deck, so the answer is 13/52 = 1/4.

And so on. Note on question 7 that a single card can't be both a five and a jack.

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