Download Chapter 15 - The Binomial Formula - PART IV : PROBABILITY

Chapter 15 - The Binomial Formula
PART IV : PROBABILITY
Dr. Joseph Brennan
Math 148, BU
Dr. Joseph Brennan (Math 148, BU)
Chapter 15 - The Binomial Formula
1 / 19
Pascal’s Triangle
In this chapter we explore a mathematical concept which gives rise to
Pascal’s Triangle:
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Chapter 15 - The Binomial Formula
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Pascal’s Triangle
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Chapter 15 - The Binomial Formula
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Pascal’s Triangle
Pascal’s Triangle is connected to simple combinations.
What is a combination?
Assume that we have n distinct objects (say rocks painted with numbers)
in a bucket. We are going to remove k objects from the bucket, without
replacement.
Can we count the number of possible ways such a choice can occur?
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Chapter 15 - The Binomial Formula
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Permutations and Combinations
ORDER
In order to count the number of possible ways to choose, without
replacement, k objects from a collection of n distinct we must be specific
as to we acknowledge order.
A permutation is a choice where order matters.
The choice 1, 3, 4 is different than the choice 4, 1, 3.
A combination is a choice where order does not matter.
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Chapter 15 - The Binomial Formula
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Multiplication Principle
The Multiplication Principle of Counting
If there are A ways to complete a first action and B ways to complete a
second action, then the number of ways to complete both actions is A · B.
How many complete outfits (fashion aside...) can be made when there are
4 shirts and 3 pants from which to choose?
Factorials
A mathematical shorthand notation for the multiplication of descending
integers:
n! = n · (n − 1) · (n − 2) · . . . · 3 · 2 · 1
For example:
1! = 1
3! = 3 · 2 · 1 = 6
5! = 5 · 4 · 3 · 2 · 1 = 120
7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040
Mathematical convention:
0! = 1
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Chapter 15 - The Binomial Formula
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Permutations
The number of permutations that can be constructed with k objects taken
without replacement from n distinct objects is denoted n Pk .
n Pk
=
n!
(n − k)!
Example: Consider the number of permutations from choosing 3 objects
from 10 objects.
10 P3
=
10!
10 · 9 · 8 · 7 · 6 · . . . · 1
=
= 10 · 9 · 8 = 720
(10 − 3)!
7 · 6 · ... · 1
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Chapter 15 - The Binomial Formula
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Example (Permutations)
A social club has 50 members. The club would like to have more
organization and members will be selected to the office of president, vice
president, treasurer, secretary, and webmaster. If no member is able to
hold multiple offices, in how many ways can the club organize itself?
Solution: As no member can hold multiple offices, we are assigning offices
without replacement. Further, as each office has distinct duties order
matters.
We are counting permutations! Specifically
50 P5
=
50 P5 .
50 · 49 · 48 · 47 · 46 · 45 · . . . · 1
50!
=
45 · . . . · 1
(50 − 5)!
= 50 · 49 · 48 · 47 · 46 = 2, 542, 512, 200!
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Chapter 15 - The Binomial Formula
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Combinations
The number of combinations that can be constructed with k objects taken
without replacement from n distinct objects is denoted kn .
n
n!
=
k! · (n − k)!
k
We read
n
k
as ”n choose k”.
The only difference between a permutation and a combination is order.
This leads to very similar counting formulas:
n!
n
n!
=
n Pk =
(n − k)!
k
k! · (n − k)!
Notice that they differ by dividing by k!. Recall that 3! = 6:
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Chapter 15 - The Binomial Formula
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Example (Combinations)
A 12 member jury is to be chosen from a jury pool of 100 individuals. In
how many ways can this jury be formed?
Solution: We are choosing without replacement from our jury pool,
otherwise we won’t have 12 jurors. Further, unless we specify a leader for
the jury, each member holds an indistinguishable office and order does not
matter.
The number of ways to form this jury is denoted 100
12 ; which is read ”100
choose 12”:
100
100!
100!
=
=
= 1.050421 × 1015
12
12! · (100 − 12)!
12! · 88!
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Chapter 15 - The Binomial Formula
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The Binomial Coefficient
n
k
is referred to as the binomial coefficient.
k ≥ 0; you don’t choose a negative number of objects.
k ≤ n; you can’t choose more objects than exists in the bin.
n ≥ 0; there cannot be a negative number of items in the bin.
0
n
n
=
1
and
=
1
and
0
0
n = 1.
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Chapter 15 - The Binomial Formula
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Binomial Probabilities
Recall: An event E in the sample space S has probability
number of outcomes in E
P(E ) =
number of outcomes in S
Example: There are 22 men and 8 women in a jury pool. What is the
probability of forming a 7 person jury where 4 members are women.
Solution: As order doesn’t matter, the outcomes MMMMWWWW and
WWWWMMM are the same, we are dealing with combinations!
Let E be the event where 4 women are chosen. The number of
combinations with 4 women and 3 men from the pool of 22 men and 8
women:
8
22
Size of E =
·
4
3
30
The sample space S is 7 .
22
8
70 · 1, 540
4 · 3
P(E ) =
= 0.053
=
30
2, 035, 800
7
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Chapter 15 - The Binomial Formula
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Example (Poker)
A deck of cards consists of 52 cards divided into:
Two Colors (26 of each):
Red and Black.
Four Suits (13 of each):
Clubs, Diamonds, Hearts, and Spades.
Cards (4 of each, one per suit):
Numbers and Face Cards.
The standard ”hand” in poker consists of 5 cards, dealt to a player
without replacement. As the order of a hand is irrelevant, we focus on a
hand’s composition, the act of being dealt a hand is a combination.
How many five card hands are possible?
52
52!
= 2, 598, 960
=
5! · 47!
5
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Chapter 15 - The Binomial Formula
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Example (Poker)
Using combinations what is the probability of being dealt 5 cards and
obtaining:
A pair of 2’s
48
4
2 · 3
= 0.04
52
5
3 Kings
4
3
5
48
2
·
52
5
A full house; 5’s full and
2’s
4
4
3 · 2
= 0.000009
52
= 0.0017
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A flush of diamonds
13
Chapter 15 - The Binomial Formula
5
52
5
= 0.0005
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The Binomial Formula
Let S be the sample space for a single occurrence and let E be an event
in S. Let p = P(E ).
The probability that E will occur k times in n trials is given by the
Binomial Formula:
n
P(n occurences of E ) =
· p k · (1 − p)n−k
k
Be careful to note:
n is predetermined.
p is P(E) while (1-p) is P(not E).
Trials are independent; p is unchanged trial to trial.
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Chapter 15 - The Binomial Formula
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Example (Binomial Formula)
Example:
A box contains 3 red, 4 green, and 2 blue marbles. What is the probability
of choosing with replacement in 5 trials:
(a) Two red marbles.
(b) Three green marbles.
Solution: As we are choosing with replacement, the trials are
independent. We may use the Binomial Formula with n = 5.
(a) Here p =
3
9
(b) Here p =
4
9
and k = 2. The probability can be found:
2 3
6
5
3
P(2 red marbles) =
·
·
= 0.33
2
9
9
and k = 3. The probability can be found:
3 2
5
5
4
P(3 green marbles) =
·
·
= 0.27
3
9
9
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Example (Blood Disorder)
The probability to recover from a certain blood disorder is 0.3. Five
patients are observed. Compute the probabilities of the following events.
(a) Two out of 5 patients will recover.
(b) Four or more patients will recover.
(c) No one will recover.
(d) All 5 patients will recover.
(e) Not exactly 2 patients will recover.
Note that trials are independent and we may use the Binomial Formula
with n = 5. The probability p is universal for patients and p = 0.3.
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Chapter 15 - The Binomial Formula
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Example (Blood disorder)
(a) We proceed with k = 2.
5
P(2 patients recover) =
(0.3)2 (1 − 0.3)5−2 = 0.3087
2
(b) We need to sum two calculations involving the Binomial Formula; one
with k=4 the other with k=5.
P(4+ patients recover) = P(4 patients recover) + P(5 patients recover)
5
5
4
5−4
=
· (0.3) (1 − 0.3)
+
· (0.3)5 (1 − 0.3)5−5
4
5
= 0.03078
(c) We proceed with k = 0.
P(no patients recover) =
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5
· (0.3)0 · (1 − 0.3)5−0 = 0.16807
0
Chapter 15 - The Binomial Formula
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Example (Blood disorder)
(d) We proceed with k = 5.
P(5 patients recover) =
5
(0.3)5 (1 − 0.3)5−5 = 0.35 = 0.00243
5
Notice that the chance that nobody will recover is much
greater than the chance that everyone will recover ...
(e) The probability that exactly two patients will recover is calculated
using the Binomial Formula with k = 2; by part (a):
P(2 patients recover) = 0.3087
Recall that P(A) = 1 − P(not A), therefore:
P(not exactly 2 patients recover) = 1 − P(2 patients recover) = 0.6913
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Chapter 15 - The Binomial Formula
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