jeff bivin -- lzhs permutation, combinations, probability, oh my… jeff bivin lake zurich high...
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Jeff Bivin -- LZHS
Permutation, Combinations,
Probability, Oh My…
Jeff BivinLake Zurich High School
jeff.bivin@lz95.org
ICTM ConferenceOctober 21, 2011
Jeff Bivin -- LZHS
Jeff Bivin -- LZHS
Fundamental Counting PrincipalHow many different meals can be made if 2 main courses, 3 vegetables, and 2 desserts are available?
M1 M2
V1 V2 V3 V1 V2 V3
D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2
1 2 3 4 5 6 7 8 9 10 11 12
Let’s choose a
main course
Now choose a
vegetable
Finally choose
A dessert
Jeff Bivin -- LZHS
Linear Permutations
A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible?
president vice-president secretary treasurer
Jeff Bivin -- LZHS
A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible?
president vice-president secretary treasurer
30P4
Linear Permutations
Jeff Bivin -- LZHS
Permutation Formula
)!(!rnnPrn
!26!30
)!430(!30
430
P
27282930!26
!2627282930
Jeff Bivin -- LZHS
Linear Permutations
There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible?
seat 1 seat 2 seat 3 seat 4 seat 5
1.5511 x 1025
Jeff Bivin -- LZHS
Linear Permutations
There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible?
seat 1 seat 2 seat 3 seat 4 seat 5
25P25
1.5511 x 1025
Jeff Bivin -- LZHS
Permutation Formula
)!(!rnnPrn
!0!25
)!2525(!25
2525
P
!251!25
1.5511 x 1025
Jeff Bivin -- LZHS
Circular PermutationsThere are 5 people sitting at a round table, how many different seating arrangements are possible?
245120
5!5
straight line
Divide by 5
Jeff Bivin -- LZHS
Circular PermutationsThere are 5 people sitting at a round table, how many different seating arrangements are possible?
245120
5!5
straight line
Treat all permutations as if linear
Now consider the circular issue
When circular, divide by the number of items in the circle
Jeff Bivin -- LZHS
Circular PermutationsThere are 9 people sitting around a campfire, how many different seating arrangements are possible?
403209
3628809!9
straight line
Treat all permutations as if linear
Is it circular?
Yes, divide by 9
Jeff Bivin -- LZHS
There are 5 people sitting at a round table with a captain chair, how many different seating arrangements are possible?
More Permutations
120!5
straight line
NOTE:
Jeff Bivin -- LZHS
More PermutationsHow many ways can you arrange 3 keys on a key ring?
12
236
3!3
straight line
Treat all permutations as if linearIs it circular?
Now, try it. . .PROBLEM:Turning it over results in the same outcome.
Yes, divide by 3
So, we must divide by 2.
Jeff Bivin -- LZHS
More PermutationsHow many ways can you arrange the letters MATH ?
4 3 2 1 4! 24
How many ways can you arrange the letters ABCDEF ?
6 5 4 3 2 16! 720
Jeff Bivin -- LZHS
Permutations with RepetitionHow many ways can you arrange the letters AAAB?
4624
!324!4
Let’s look at the possibilities:
AAABAABAABAABAAA
Are there any others?What is the problem?
If a permutation has repeated items, we divide by the number of ways of arranging the repeated items (as if they were different).
Divide by 3!
Jeff Bivin -- LZHS
How many ways can you arrange 5 red, 7 blue and 8 white flags on the tack strip across the front of the classroom?
240,768,99!8!7!5!20
If all were different, how may ways could we arrange 20 items?
There are 5 repeated red flags Divide by 5!
There are 7 repeated blue flags Divide by 7!
There are 8 repeated white flags Divide by 8!
Jeff Bivin -- LZHS
How many ways can you arrange the letters in the non-word
A A B B C C C C D E F G G G G G G ?
800,940,145,5!6!4!2!2
!17
If all were different, how may ways could we arrange 17 items?
There are 2 repeated A’s Divide by 2!
There are 2 repeated B’s Divide by 2!
There are 4 repeated C’s Divide by 4!
There are 6 repeated G’s Divide by 6!
Jeff Bivin -- LZHS
Permutations ORDER
Multiply the possibilities
Divide by the numberof items in the circle
Divide by 2
Divide by the factorial of thenumber of each duplicated item
Assume the itemsare in a straight line! Use the nPr formula
(if no replacement)
or
Are the items in a circle??
Can the itembe turned over??
Are there duplicateitems in your
arrangement??
Jeff Bivin -- LZHS
How many ways can you put 5 red and 7 brown beads on a necklace?
!7!5212!12
How may ways could we arrange 12 items in a straight line?
Is it circular? Yes divide by 12
Can it be turned over? Yes divide by 2
Are there repeated items? Yes divide by 5! and 7!
33
Jeff Bivin -- LZHS
How many ways can you arrange 5 red and 7 brown beads on a necklace that has a clasp?
!7!52!12
How may ways could we arrange 12 items in a straight line?
Is it circular? N0 the clasp makes it linear
Can it be turned over? Yes divide by 2
Are there repeated items? Yes divide by 5! and 7!
396
Jeff Bivin -- LZHS
How different license plates can have 2 letters followed by 3 digits (no repeats)?
A straight line?
Is it circular? No
Can it be turned over? No
Are there repeated items? No
468,000
26 ∙ 25 ∙ 10 ∙ 9 ∙ 8lette
rletter number number number
Jeff Bivin -- LZHS
How different license plates can have 2 letters followed by 3 digits with repeats?
A straight line?
Is it circular? No
Can it be turned over? No
Are there repeated items? Yes, but because we are using multiplication andnot factorials, we do not need to divide by anything.
676,000
26 ∙ 26 ∙ 10 ∙ 10 ∙ 10lette
rletter number number number
Jeff Bivin -- LZHS
Jeff Bivin -- LZHS
Combinations NO orderNO replacement
Use the
nCr formula
Typically
Jeff Bivin -- LZHS
Combinations An organization has 30 members and must select a committee of 4 people to plan an upcoming function. How many different committees are possible?
!)!(!rrn
nCrn
!4!26!30
!4)!430(!30
430
C
Jeff Bivin -- LZHS
!3!9!12
!3)!312(!12
312
C
Combinations
!)!(!rrn
nCrn
A plane contains 12 points, no three of which are co-linear. How many different triangles can be formed?
Jeff Bivin -- LZHS
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles?
5 35! 5!
(5 3)! 3! 3! 3!C
Combinations
!)!(!rrn
nCrn
5 red6 white9 blue
3 red
have want
Jeff Bivin -- LZHS
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 blue marbles?
!3!6!9
!3)!39(!9
39
C
Combinations
!)!(!rrn
nCrn
5 red6 white9 blue
3 blue
have want
Jeff Bivin -- LZHS
The OR factor.
10 84 94 5 red
6 white9 blue
3 redOR
3 blue
have want
5 3 9 35! 9!
(5 3)! 3! (9 3)! 3!C C
want
OR ADD
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles or 3 blue marbles?
Jeff Bivin -- LZHS
The OR factor.
5 70 75 5 red
8 blue
4 redOR
4 blue
have
5 4 8 45! 8!
(5 4)! 4! (8 4)! 4!C C
OR ADD
wantwant
An jar contains 13 marbles – 5 red and 8 blue. If four are selected at random, how many ways can you select 4 red marbles or 4 blue marbles?
Jeff Bivin -- LZHS
5 1 9 25! 9!
(5 1)! 1! (9 2)! 2!C C
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 1 red marble and 2 blue marbles?
The AND factor.
AND MULTIPLY
5 36 180 5 red
6 white9 blue
1 redand
2 blue
have want
Jeff Bivin -- LZHS
3B2NB or 4B1NB or 5B
At least
3 or 4 or 5 blue
591114921139 CCCCC 126111265584
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at least 3 blue marbles?
6132
Jeff Bivin -- LZHS
0R5Nr or 1R4NR
At most
0 or 1 red
4151551505 CCCC 1365530031
An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at most 1 red marbles?
9828
Jeff Bivin -- LZHS
And More
Jeff Bivin -- LZHS
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
Let’s look at (x + y)p
(x + y)7 = _x7 + _x6y + _x5y2 + _x4y3 + _x3y4 + _x2y5 + _xy6 + _y7
Jeff Bivin -- LZHS
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
Let’s look at (x + y)p
Jeff Bivin -- LZHS
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
Let’s look at (x + y)p
Jeff Bivin -- LZHS
(x + y)7 = 1 7 21 35 35 21 7 1
(x + y)0 = 1
(x + y)1 = 1 1
(x + y)2 = 1 2 1
(x + y)3 = 1 3 3 1
(x + y)4 = 1 4 6 4 1
(x + y)5 = 1 5 10 10 5 1
(x + y)6 = 1 6 15 20 15 6 1
(x + y)8 = 1 8 28 56 70 56 28 8 1
Let’s look at (x + y)p
Jeff Bivin -- LZHS
(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + 1y7
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
Let’s look at (x + y)p
Jeff Bivin -- LZHS
12! 479001600 19,958,4002! 3! 2! 24
In how many ways can you arrange the letters in the word M A T H E M A T I C A L ?
Jeff Bivin -- LZHS
52 yx
351445040
!3!4!7
212405040
!5!2!7
34 yx
In how many ways can you arrange the letters in the non-word xxxxyyy?
In how many ways can you arrange the letters in the non-word xxyyyyy?
(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7
Jeff Bivin -- LZHS
Let’s look closer(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7
071!0!7!7 yx 167
!1!6!7 yx 2521
!2!5!7 yx
3435!3!4!7 yx 4335
!4!3!7 yx 5221
!5!2!7 yx
617!6!1!7 yx 701
!7!0!7 yx
Jeff Bivin -- LZHS
An alternate look(x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7
77!0!7!7 C 67!1!6
!7 C 57!2!5!7 C
47!3!4!7 C 37!4!3
!7 C 27!5!2!7 C
17!6!1!7 C 07!7!0
!7 C
Jeff Bivin -- LZHS
(2x - y)4 = 16x4 - 32x3y + 24x2y2 - 8xy3 + y4
40444 161)()2(1
!0!4!4 xyxC
)(84)()2(4!1!3!4 313
34 yxyxC
)(46)()2(6!2!2!4 2222
24 yxyxC
)(24)()2(4!3!1!4 331
14 yxyxC
)(1)()2(1!4!0!4 440
04 yyxC
Jeff Bivin -- LZHS
(3x + 2y)5 = 243x5 + 810x4y + 1080x3y2 + 720x2y3 + 240xy4 + 32y5
50555 2431)2()3(1
!0!5!5 xyxC
yxyxC 2815)2()3(5!1!4!5 414
45
222335 42710)2()3(10
!2!3!5 yxyxC
323225 8910)2()3(10
!3!2!5 yxyxC
44115 1635)2()3(5
!4!1!5 yxyxC
55005 3211)2()3(1
!5!0!5 yyxC
Jeff Bivin -- LZHS
In how many ways can you arrange the
letters in the non-word
x x x x x y y y y y y y y y y ?
Given: (x + y)15
105515 3003
!10!5!15 yxC
1053003 yx
What is the coefficient of the term ____ x5y10 ?
Jeff Bivin -- LZHS
Given: (4x - 3y)10
In how many ways can you arrange the
letters in the non-word
x x x x x x x y y y ?37
710 )3()4(120!3!7!10 yxC
37 2716384120 yx
What is the coefficient of the term ____ x7y3 ?
37160,084,53 yx
Jeff Bivin -- LZHS
Expand: (x + y + z)2
(x + y + z) (x + y + z)
x2 + xy + xz + yx + y2 + yz + zx + zy + z2
x2 + 2xy + 2xz + y2 + 2yz + z2
Jeff Bivin -- LZHS
x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3
(x + y + z)2 (x + y + z)
x3 + x2y + x2z + 2x2y + 2xy2 + 2xyz + 2x2z + 2xzy + 2xz2 + y2x + y3 + y2z
+2yzx + 2y2z + 2yz2 + z2x + z2y + z3
Expand: (x + y + z)3
(x2 + 2xy + 2xz + y2 + 2yz + z2)(x + y + z)
We did this in the last example
Jeff Bivin -- LZHS
zyx 6!1!1!1!3
x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3
Given: (x + y + z)3
In how many ways can you arrange the letters in the non-word xyz ?
What is the coefficient of the term ____ xyz ?
In how many ways can you arrange the letters in the non-word xxz ?
zx 23!1!0!2
!3
What is the coefficient of the term ____ x2z ?
RE
ME
MB
ER
Jeff Bivin -- LZHS
Given: (x + y + z)15
In how many ways can you arrange the
letters in the non-word
xxyyyyyyyzzzzzz ?
672180180!6!7!2!15 zyx
672180,180 zyx
What is the coefficient of the term ____ x2y7z6 ?
Jeff Bivin -- LZHS
Given: (2x + 3y - z)9
In how many ways can you arrange the
letters in the non-word
x x x y y y y z z ?243 )()3()2(1260
!2!4!3!9 zyx
243480,816 zyx
What is the coefficient of the term ____ x3y4z2 ?
243 8181260 zyx
Jeff Bivin -- LZHS
In how many ways can you arrange the
letters in the non-word
a a a a a b b b b b b c c c c c c c d d ?
Given: (a + b + c + d)20
2765720,510,793,2!2!7!6!5
!20 dcba
2765720,510,793,2 dcba
What is the coefficient of the term ____ a5b6c7d2 ?
Jeff Bivin -- LZHS
Jeff Bivin -- LZHS
Can we think of combinations as permutations with repetitions?
In a group of 7 people, how many different committees of 3 people can we select?
Two choices….
A member of the 3 person committee
A member of the 4 person non-committee
Jeff Bivin -- LZHS
In a group of 7 people, how many different committees of 3 people can we select?
Alex Betty Chuck Deb Ed Fiona Gabe
N C N C N N C
How many ways can you arrange 3 C’s and 4 N’s?
7! 353! 4!
Jeff Bivin -- LZHS
In a group of 7 people, a pair of 2 people are needed for one committee and 3 different people are need for a second committee. How many possibilities exist?
Alex Betty Chuck Deb Ed Fiona Gabe
C1 N C2 C1 N C2 N
How many ways can you arrange 2 C1’s and 2 C’s and 3 N’s?
7! 2102! 2! 3!
Jeff Bivin -- LZHS
In a group of 7 people, how may ways can a president, a vice-president and a secretary be selected? Alex Betty Chuck Deb Ed Fiona Gabe
N VP N P N N S
How many ways can you arrange 1 P and 1 VP and 1 S and 4 N’s?
7! 2101! 1! 1! 4!
Can we think of most permutations as permutations with repetitions?
Jeff Bivin -- LZHS
Jeff Bivin -- LZHS
PROBABILITY Defined
number of successtotal number of outcomes
The ratio
Jeff Bivin -- LZHS
Probability
A coin is tossed, what is the probability that you will obtain a heads?
Look at the sample space/possible outcomes:
{ H , T }
number of successtotal number of outcomesPr(H) = 1
2=
Jeff Bivin -- LZHS
number of success
Probability
A die is tossed, what is the probability that you will obtain a number greater than 4?
Look at the sample space/possible outcomes:
total number of outcomesPr(>4) = 26= 1
3=
{ 1 , 2 , 3 , 4 , 5 , 6 }
Jeff Bivin -- LZHS
number of failurestotal number of outcomes
total number of outcomesnumber of success
Probability – Success & Failure
A die is tossed, what is the probability that you will obtain a number greater than 4?
Pr(>4) = 26= 1
3=
What is the probability that you fail to obtain a number greater than 4?
Pr(>4) = 46
23= =
TOTAL = Pr(success) + Pr(failure) = 1
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red?
number of successtotal number of outcomes
Probability
Pr(3R) = 5C3
13C3=
5 red
8 blue
have want
3 red
Total: 13 3
1435
28610
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are blue?
number of successtotal number of outcomes
Probability
Pr(3B) = 8C3
13C3=
5 red
8 blue
have want
3 blue
Total: 13 3
14328
28656
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one is red and two are blue?
number of successtotal number of outcomes
Probability – “and”
Pr(1R2B) = 5C1 ● 8C2
13C3=
5 red
8 blue
have want
1 red
Total: 13 3
14370
286140
286285
2 blue
multiply
Jeff Bivin -- LZHS
A jar contains 5 red, 8 blue and 7 white marbles. If 3 marbles are selected at random, what is the probability that one of each color is selected?
# of successtotal # of outcomesPr(1R,1B,1W) = 5C1●8C1●7C1
20C3=
5 red8 blue7 white
have want
1 red
Total: 20 3
5714
1140280
1140785
1 blue1 white
1 red, 1 blue, & 1 white
Jeff Bivin -- LZHS
A jar contains 7 red, 5 blue and 3 white marbles. If 4 marbles are selected at random, what is the probability that 2 red and 2 white marbles are selected?
# of successtotal # of outcomesPr(2R,2W) = 7C2 ● 3C2
15C4=
7 red5 blue3 white
have want
2 red
Total: 15 4
653
136563
1365321
2 white
Jeff Bivin -- LZHS
Five cards are dealt from a standard deck of cards. What is the probability that 3 hearts and 2 clubs are obtained?
# of successtotal # of outcomesPr(3H,2C) = 13C3 ● 13C2
52C5=
13 diamonds13 hearts13 clubs
13 spades
have want
3 hearts
Total: 52 5
6497405577
259896022308
259896078286
2 clubs
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red or all three are blue?
# of successtotal # of outcomes
Probability – “or”
Pr(3R or 3B) = 5C3 + 8C3 13C3
=
5 red
8 blue
have want
3 red
Total: 13 3
133
28666
2865610
3 blue
want
OR
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles and 7 yellow marbles. If 3 marbles are selected at random, what is the probability that all three are the same color?
5C3 + 8C3 + 7C3 # of success
total # of outcomesPr(3R or 3B or 3w) = 20C3
=
5 red8 blue
7 yellow
have want
3 red
Total: 20 3
286101
286355610
3 blue
want
OR
3 red or 3 blue or 3 yellow ?
want
3 yellowOR
Jeff Bivin -- LZHS
26C2 + 4C2 – 2C2 # of successtotal # of outcomes
Probability – “or” with overlap
Pr(2R or 2B) = Pr(2R) + Pr(2K) – Pr(2RK)
52C2=
26 red26 black
have want 2 red
Total: 52 2
2 kings
want
OR 2 red
kings
overlap 132616325
22155
1326330
If two cards are selected from a standard deck of cards, what is the probability that both are red or both are kings?
4 kings48 other
Jeff Bivin -- LZHS
5C2● 8C1 + 5C1 ● 8C2# of success
total # of outcomes
Probability – “and” with “or”
Pr(2R1B or 1R2B) = 13C3
=
5 red
8 blue
have want
2 red
Total: 13 3
1310
286220
286285810
1 blue
want
OR1 red
2 blue
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that two are red and one is blue or that one is red and two are blue?
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least two red marbles are selected?
5C2● 8C1 + 5C3# of success
total # of outcomes
Probability – “at least”
Pr(at least 2Red) = 13C3
=
5 red
8 blue
have want
2 red
Total: 13 3
14345
28690
28610810
1 blue
want
OR3 red
2 red or 3 red2 red and 1 blue or 3 red
Pr(2R1B or 3R) =
Jeff Bivin -- LZHS
5C1● 8C2 + 5C2 ● 8C1 + 5C3
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected?
Probability – “at least”
Pr(at least 1Red) = 13C3
5 red
8 blue
have want
1 red
Total:13 3
28610810285
2 blue
want
OR2 red
Pr(1R2B or 2R1B or 3R) =
want
OR3 red
1 blue
143115
286230
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that NO red marbles are selected?
8C3
Probability – “at least”
Pr(0R3B) = 13C3
5 red
8 blue
have want
Total:13 3
14328
28656
3 blue
In the previous example we found
1431151Pr red
Pr(success) + Pr(failure) = 1
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected?
Probability – “at least”
Pr(>1 red) = 1 – Pr( 0 red )
143115
286230
2865611
313
38 CC
Pr(success) + Pr(failure) = 1
Pr(success) = 1 - Pr(failure)
Pr(3 blue)
Jeff Bivin -- LZHS
A jar contains 8 red and 9 blue marbles. If 7 marbles are selected at random, what is the probability that at least one red marbles is selected?
Probability – “at least”
Pr(at least 1Red)
Pr(1R6B or 2R5B or 3R4B or 4R3B or 5R2B or 6R1B or 7R)
Pr(0Red) Pr( 0R7B )
success
failure
FASTEST
Pr(at least 1Red) = 1 - Pr(0R7B) = 717
791CC
48624853
19448361
Jeff Bivin -- LZHS
720
51228612187121C
CCCCC
A jar contains 8 red, 9 blue and 3 white marbles. If 7 marbles are selected at random, what is the probability that at least three red marbles are selected?
Probability – “at least”
Pr(> 3Red) Pr(3-7 red)
Pr(< 3Red) Pr(0-2 red)
success
failure
FASTEST
1 - Pr(0R7NR or 1R6NR or 2R5NR)
77520303601
646393
Jeff Bivin -- LZHS
Probability – “with replacement”
2197320
138
138
135
138
138
135
Must use fractions! R B B
Note: In this example an order is specified
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red followed by two blue marbles are selected if each marble is replaced after each selection?
Jeff Bivin -- LZHS
A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red and two blue marbles are selected if each marble is replaced after each selection?
Probability – “with replacement”
2197960
138
138
135
138
138
135
23 3 C
Must use fractions!
Must account of any order!
Problem: Fractions imply order!
R B B
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