Gordon-Poker

Exercises 4.8:

4.112 Poker. A poker hand consists of 5 cards dealt from an ordinary
deck of 52 playing cards.

a. How many poker hands are possible?

52C5 = 52!/5!(52-5)! = 525150494847!5!47! =
311,875,200/120 = 2,598,960 I feel pretty confident here.

How many different hands consisting of three kings and two queens are
possible? This where my mind goes bonk.

First, we know that 52C5 = 2,598,960, next we calculate 4C3
48C2for the possibility of selecting 3 of the four kings in the
deck. Then we divide that by 52C5. Then we multiply 4C2 48C3for
the possibility of selecting 2 queens from the deck and likewise divide
that by 52C5. Finally, we add? the two results to calculate the total
number of possible different hands consisting of three kings and two
queens.

4C3 4!/3!(4-3)! = 4!/3! = 24/6 = 4

48C2 48!/2!(48-2)! = 484746!/2!46! = 2,256/2 = 1,128

(4C3 48C2) = 4 1,1284,512

4C2 4!/2!(4-2)! = 4!/2! = 24/2 = 12

48C3 48!/3!(48-3)! = 48474645!/3!45! = 103,776/6 = 17,296

(4C2 48C3) = 12 17,296207,552

Adding the two together, we get a total of 212,064 possible full-house
combinations (4,512 + 207,552 = 212,064) in the deck. Since each suit is
comprised of 1/13th kings, and 1/13th queens, we must divide the total
by 13?. There are 16,312 different hands consisting of three kings and
two queens are possible (212,064/13 = 16,312).

The hand in part (b) is an example of a full house: 3 cards of one
denomination and 2 of another. How many different full houses are
possible?

212,064

d. Calculate the probability of being dealt a full house.

f/N = 212,064/2,598,960 = 0.08 , or 8%

4.116 Quality Assurance. There are 100 TV’s manufactured. We are to
assume that 6 of the 100 are defective. The Quality Assurance Engineer
is selecting a sample of 5 TV’s from the 100. Determine the
probability that the number of defective TVs obtained by the engineer
is:

a. Exactly one.

100C5 = 100!/5!(100-5)! = 100!/5!95! = 75,287,520

6C1 = 6!/1!(6-1)! = 6/1 = 6

939291/4321 = 73,188,024/24 = 3,049,501

6C1  94C4 = 63,049,501 = 18,297,006

f/N = (6C1  94C4) / 100C5 = 18,297,006 / 75,287,520 = 0.243

At most one. This is where I am lost.

94C4 = 94!/4!(94-4)! = 94939291/4321 = 73,188,024/24 =
3,049,501

f/N = 3,049,501/75,287,520 = 0.040



c. At least one. This is also where I am not very sure.

94C5 = 94!/5!(94-5)! = 9493929190/5! = 6,586,922,160/120 =
54,891,018

f/N = 54,891,018/75,287,520 = 0.729

Using the complementation rule: 1 – 0.729 = 0.271

There is a 27.1% probability that at least one TV will be defective.

Seems too high?