Posted by oscar on Sunday, December 12, 2010 at 11:12am.
a) There are 16 red-faced cards and 16 black.
First Card red = 16/32
Second card red = 15/31
Third Card red = 14/30
Fourth card red = 13/29
The probability of all occurring is found by multiplying the individual probabilities.
b) 16/32, 15/31, 16/30, 15/29 Again multiply.
c) Firgure probability of one red card, two red cards, three red cards, or four red cards. Remember the probability of the remaning cards being black in each case, except the last. Since you are interested in "either-or", add those four probabilities.
d) Same as a.
There are 6 red cards not 16 ( only face cards: 2 kings 2 queens and 2 jacks) and 16 black. 22 in total.
a) (6/22)*(5/21)*(4/20)*(3/19)
b) (16/22)*(15/21)*(6/20)*(5/19)
c) 1-((16/22)*(15/21)*(14/20)*(13/19))
d)(16/22)*(15/21)*(14/20)*(13/19))
CARDS:red=6; black=16 ... total 22
a.(6 nCr 4)/(22 nCr 4) = .002
b.(6 nCr 2)(16 nCr 2)/(22 nCr 4)= .246
c.1-(.249) = .751 ... get d first
d.(16 nCr 22)/(22 nCr 4) = .249
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