Transcript Powerpoint for combination symbols

Combination Symbols
A supplement to Greenleaf’s QR Text
Compiled by Samuel Marateck ©2009
1
How many 4-card hands consisting of
1 king and 3 queens can be chosen
from a deck?
2
How many 4-card hands consisting of
1 king and 3 queens can be chosen
from a deck?
Since order does not matter and there are
four kings and four queens in the deck,
the answer is:
( 4 1) ( 4 3)
3
What is the meaning of ( 4 1)?
It’s the number of ways we can choose one
thing from four, independent of the order.
It is pronounced “four choose one”.
4
Similarly ( 4 3) is the number of ways we can
choose three things from four independent
of the order. It is pronounced “four choose
three”.
5
In ( 4 1) ( 4 3), why do we multiply the
two?
6
For each king there are three queen pairings.
These are the pairings for the king of spades:
k♠ Q♠ Q♣ Q♥
k♠ Q♠ Q♣ Q♦
k♠ Q♣ Q♥ Q♦
k♠ Q♥ Q♦ Q ♠
7
But there are also k♥, k♣ and k♦. So there
are 16 different combinations, four for each
King.
8
What is the probability of choosing
4-card hands consisting of 1 king and 3
queens from a deck?
9
What is the probability of choosing
4-card hands consisting of 1 king and 3
queens from a deck?
( 4 1) ( 4 3) / ( 52 4)
10
We divide by ( 52 4) since this is the number
of ways we can choose four cards at
random from a deck.
11
Let’s evaluate ( 4 1) ( 4 3) / ( 52 4)
12
( 4 1) ( 4 3) / ( 52 4) is:
16/(52*51*50*49/(4*3*2*1))
=0.00006 or .006%
13
Out of how many hands would you expect
to get this hand?
14
Out of how many hands would you expect
to get this hand?
0.00006 is 6x 10-5 , so in 105 hands you
would expect to get 6 such hands or
in one out of 16,666 hands you would get
this hand.
15
How many 5-card hands can you get that
have three aces?
16
How many 5-card hands can you get that
have three aces?
The number of ways we can choose three
aces is ( 4 3) . How many cards are left in
the deck?
17
How many non-aces are in the deck?
There are 48 non-aces left in the deck and
there are two more cards to choose for our
hand.
18
So there are ( 4 3) ( 48 2) ways we can get
three aces:
4*48*47/2 = 4*47*24 = 4512 ways.
19
What is the probability of getting three
aces in a 5-card hand?
20
What is the probability of getting three
aces in a 5-card hand?
( 4 3) ( 48 2) / ( 52 5) =
4512/((52*51*50*49*48)/(5*4*3*2*1)) =
4512/2598960 = .00174
21
What is the probability of winning the
lottery?
22
What is the probability of winning the
lottery?
There are 54 numbers that you can choose
from; the numbers 1 to 54. You must choose
the five correct numbers independent of
their order. The answer is:
23
P(winning) = 1/( 54 5)
( 54 5) = 54*53*52*51*50/120
1/( 54 5) = 3.16 x 10-7
24
If there are 6 pegs distributed in a circle and
a line is drawn from each peg to each other
peg, how many lines are there?
25
For each peg 5 lines are drawn; but there
are 6 pegs. Since, however, each line
connects two pegs, we are overcounting
by 2, so we must divide by 2.
What is the answer?
26
# of lines is 5*6/2 or 15.
27
Another way of looking at this is:
From the first peg, 5 lines are drawn. From
the second peg, 4 lines are drawn since it
is already connected to the first peg. From
the third peg, 3 lines are drawn, since it is
connected to the first two, and so on,
28
For the six pegs, 5+4+3+2+1 or 15 lines are
drawn. For n pegs n-1 + n-2 + n-3 +..+ 1
lines are drawn. We know what the sum
from 1 to m is.
29
The sum is: m(m+1)/2.
Substituting n-1 for m, the sum from 1 to
n-1 is (n-1)(n-1 +1)/2 =?
30
(n-1)(n-1 +1)/2 = n(n-1)/2 which is the
answer we got before.
Can we do this with combination symbols?
31
If there are 6 pegs distributed in a circle and
a line is drawn from each peg to each other
peg, how many lines are there?
32
There are 6 slots:
. . . . . ..
1 2 3 4 5 6
How many ways can we place two item
in these slots?
33
How many ways can we place two item
in these slots?
The answer is ( 6 2).
For n pegs it’s ?
34
For n pegs it’s ( n 2).
35
How many ways can we choose a 5-card
hand so that no two cards have the same
face values?
36
How many ways can we choose a 5-card
hand so that no two cards have the same
face values?
For the first card we have ( 52 1) ways we
can choose the first card. How many
choices do we have for the second card?
37
How many choices do we have for the
second card?
48, since one face value has been
eliminated. So the number of ways we can
choose the second card is:
38
( 48 1).
The third card is?
39
( 44 1).
So the final answer is:
( 52 1) ( 48 1)( 44 1) ( 40 1) ( 36 1).
What is the probability?
40
P(each card has a different face value) =
( 52 1) ( 48 1)( 44 1) ( 40 1) ( 36 1)
( 52 5)
41
In a class of 25, what is the probability that
two or more people have the same
birthdate?
42
In a class of 25, what is the probability that
two or more people have the same
birthdate?
We will first calculate the probability that no
one has the same birthdate.
43
Given the first person, the probability that
the second one has a different birth date is
364/365. That the first, second and third
ones have different birth dates is:
1* 364/365*363/365.
For all 25 people?
44
For all 25 people?
P(different birth dates) =
364*363*362*361…341/36524 = 0.47
45
P(2 or more have same birth dates) = .53
46
There are 25 people to be chosen for a
Committee or 5. What is my probability of
being chosen?
47
What is the probability of my being chosen?
( 1 1) ( 24 4)/ ( 25 5).
48
An urn contains 10 red balls and 40 black
ones. What is the probability you will draw
2 red balls.
49
( 10 2) ( 40 0)/ ( 50 2) = 10*9/2 /(50*49/2)
= 45/1225
50
An urn contains 17 red balls and 33 black
ones. What is the probability you will draw
7 red balls if you choose 10 randomly?
51
( 17 7) ( 33 3)/ ( 50 10)
52
A jury pool contains 98 men and 75 women.
12 jurors are chosen at random. What is
the probability that 6 will be women
53
( 98 6) ( 75 6)/ ( 173
12)
54