Posted by **oscar** on Sunday, December 12, 2010 at 11:12am.

The red face cards and the black cards numbered 2-9 are put into a bag. Four cards are drawn at random without replacement. Find the following probabilities:

a) All 4 cards are red

b) 2 cards are red and two cards are black

c) At least one of the red cards is red.

d) All four cards are black

- statistics -
**PsyDAG**, Monday, December 13, 2010 at 3:12pm
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.

- statistics -
**Meli**, Thursday, February 2, 2012 at 6:16am
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))

- statistics -
**Bladimir**, Thursday, March 8, 2012 at 3:56am
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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