Question 1191648: Two cards are selected without replacement from a usual deck of 52 cards.(a) How many points (simple events) are in the sample space?(b) Find the following probabilities: (i) both cards are clubs; (ii) both cards are kings; (iii) one ofthe cards is the king of clubs; (iv) at least one of the cards is a king or a club; (v) exactly oneof the cards is a face card (J, Q or K).(c) What is the probability that the second card selected has a lower rank than the first cardselected? [Consider an ace as the highest rank only.](d) If neither of the two cards selected were spades, what is the probability both were diamonds?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! a.-52C2=26*51=1326 possible outcomes
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b1- 13C2/52C2=78/1326=0.0588. OR 13/52 chances first is a club*12/51 for the second=156/2652=78/1326.
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b2-4C2/52C2=6/1326 or 1/221.
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b3 (52C1)(1C1)/2C2=52/1326 OR (51/52)(1/51)
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b4- probability no king or no club includes 16 cards, 13 clubs, 4 kings minus 1 KC.
this is probability 36/52*35/51=1260/2652 and may be reduced.
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b5-probability none is would be 43/52*42/51.
probability both are is 9/52*8/51.
That sum is (1806+72)/2652=1888/2652. The complement is 764/2652 and may be reduced.
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Lower rank. 6 cards per suit are lower, 6 higher, and one is the same. So the probability a card is lower rank is 6/13.
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That excludes 13 cards so the sample space is 39 cards. Thirteen are diamonds, so the probalbity is (13/39)(12/38)=12/114 or 2/19.
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