Transcript Probability

Probability
We love Section 9.3a and b!
Most people have an intuitive sense of
probability, but that intuition is often
incorrect…
Let’s test your intuition
about probability!!!
Intuition about Probability?
Find the probability of each of the following events.
1. Tossing a head on one toss of a fair coin.
Two equally likely outcomes: { T, H }. Probability is 1/2.
2. Tossing two heads in a row on two tosses of a fair coin.
Four equally likely outcomes: { TT, TH, HT, HH }.
Probability is 1/4.
Intuition about Probability?
Find the probability of each of the following events.
3. Drawing a queen from a standard deck of 52 cards.
There are 52 equally likely outcomes, 4 of which are
queens. Probability is 4/52, or 1/13.
4. Rolling a sum of 4 on a single roll of two fair dice.
Multiplication Principle of Counting  6 x 6 = 36 equally
likely outcomes. Of these, three { (1, 3), (3, 1), (2, 2) }
yield a sum of 4. Probability is 3/36, or 1/12.
Intuition about Probability?
Find the probability of each of the following events.
5. Guessing all 6 numbers in a state lottery that requires
you to pick 6 numbers between 1 and 46, inclusive.
Number of equally likely ways to choose 6 numbers
from 46 numbers without regard to order?
C

9,366,819
46 6
 Probability is 1/9366819.
Terminology of Probability
Sample Space – the set of all possible outcomes
of an experiment
Event – a subset of the sample space
Each sample space consists of a finite number of
equally likely outcomes…
Probability of an Event
(Equally Likely Outcomes)
If E is an event in a finite, nonempty sample space
S of equally likely outcomes, then the probability
of the event E is
the number of outcomes in E
P(E) =
the number of outcomes in S
A key part of the hypothesis!!!
Rolling Dice
Possible outcomes for the sum on two fair dice:
{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }
So the probability of rolling a sum of 4 is 1/11, right???
NO!!!, because these sums are not equally likely…
Possible outcomes when rolling a BLUE and Aqua die:
1-1 Sum = 2
Probability Distribution
Outcome
Probability
1-2, 2-1 Sum = 3
2
1/36
1-3, 2-2, 3-1 Sum = 4
3
2/36
1-4, 2-3, 3-2, 4-1 Sum = 5
4
3/36
1-5, 2-4, 3-3, 4-2, 5-1 Sum = 6
5
4/36
1-6, 2-5, 3-4, 4-3, 5-2, 6-1 Sum = 7
6
5/36
2-6, 3-5, 4-4, 5-3, 6-2 Sum = 8
7
6/36
3-6, 4-5, 5-4, 6-3 Sum = 9
8
5/36
4-6, 5-5, 6-4 Sum = 10
9
4/36
5-6, 6-5 Sum = 11
10
3/36
6-6 Sum = 12
Note: The sum of the probabilities of
any prob. dist. is always 1!!!
11
12
2/36
1/36
Definition: Probability Function
A probability function is a function P that assigns a real
number to each outcome in a sample space S subject to
the following conditions:
1.
0  P O  1
for every outcome O.
2. The sum of the probabilities of all outcomes
in S is 1.
The empty set…
3. P   0
 
Probability of an Event
(Outcomes not Equally Likely)
Let S be a finite, nonempty sample space in which
every outcome has a probability assigned to it by
a probability function P. If E is any event in S, the
probability of the event E is the sum of the
probabilities of all the outcomes in E.
Returning to our dice…
Suppose that two fair dice have been rolled. State the
probability of each of the following events.
1. The sum is 7. P(sum 7) = 1/6
2. The same number is rolled on both dice.P(doubles) = 1/6
3. The sum is 2 or 3. P(sum 2 or 3) = 1/12
4. The sum is a multiple of 3.
P(multiple of 3) = 1/3
Strategy for Determining Probabilities
1. Determine the sample space of all possible outcomes. Whe
possible, choose outcomes that are equally likely.
2. If the sample space has equally likely outcomes, the
Probability of an event E is determined by counting:
the number of outcomes in E
P( E ) =
the number of outcomes in S
3. If the sample space does not have equally likely outcomes,
determine the probability function. (This is not always easy
to do.) Check to be sure that the conditions of a probability
function are satisfied. Then the probability of an event E is
determined by adding up the probabilities of all the
outcomes contained in E.
Multiplication Principle of Probability
1
Suppose an event A has probability p and an event B
has probability p 2 under the assumption that A occurs.
then the probability that both A and B occur is p p .
1
2
If the events A and B are independent, we omit the
phrase “under the assumption that A occurs.”
Practice Problems
Sal opens a box of a dozen chocolate covered doughnuts and
offers two of them to Val. Val likes the vanilla-filled doughnut
the best, but all of the doughnuts look alike on the outside. If
four of the twelve doughnuts are vanilla-filled, what is the
probability that both of Val’s picks turn out to be vanilla?
Here, we are choosing two doughnuts (without regard to orde
from a box of 12:
12 C2  66
outcomes of this experiment
Are they all equally likely?  YES!!!
Practice Problems
Sal opens a box of a dozen chocolate covered doughnuts and
offers two of them to Val. Val likes the vanilla-filled doughnuts
the best, but all of the doughnuts look alike on the outside. If
four of the twelve doughnuts are vanilla-filled, what is the
probability that both of Val’s picks turn out to be vanilla?
The event E consists of all possible pairs of 2 vanilla-filled
doughnuts (without regard to order) from the 4 vanilla-filled
doughnuts available:
ways to form such pairs
4 2
C 6
 P  E   6 66  1 11
Practice Problems
Another way to solve the same problem:
There are two types of doughnuts: vanilla (V) and unvanilla (U
When choosing two doughnuts, there are four possible
outcomes: { VV, VU, UV, UU }
Are they all equally likely?  NO WAY!!!
 We want the probability of outcome VV.
Practice Problems
There are two types of doughnuts: vanilla (V) and unvanilla (U
When choosing two doughnuts, there4/12
are four possible
outcomes: { VV, VU, UV, UU }
Probability of picking a V on the first draw:
3/11
Probability of picking a V on the second draw, under the
assumption that a V was drawn on the first:
Probability of drawing a V on both draws
(use the Mult Prop!!!):
4 3 1

12 11 11
Practice Problems
Now, I want you to create a probability function for this
experiment…
Outcome
VV
Probability
1/11
VU
(4/12)(8/11) = 8/33
UV
(8/12)(4/11) = 8/33
UU
(8/12)(7/11) = 14/33
Does this probability function “check out?”
Practice Problems
Skittles candy comes in the following color proportions:
Red Green Yellow Orange Purple
Color
Proportion 0.25
0.2
0.15
0.3
0.1
A single Skittle is selected at random from a newly-opened ba
What is the probability that the candy has the given color(s)?
1. Green or Orange P(G or O) = P(G) + P(O) = 0.2 + 0.3 = 0.5
2. Purple or Orange or Red
P(P or O or R) = P(P) + P(O) + P(R) = 0.1 + 0.3 + 0.25 = 0.65
Practice Problems
Skittles candy comes in the following color proportions:
Color
Proportion
Red
0.25
Green Yellow Orange Purple
0.2
0.15
0.3
0.1
A single Skittle is selected at random from a newly-opened ba
What is the probability that the candy has the given color(s)?
3. Not Yellow P(not Y) = 1 – P(Y) = 1 – 0.15 = 0.85
4. Neither Green nor Red
P[not (G or R)] = 1 – P(G or R) = 1 – (0.2 + 0.25) = 0.55
Practice Problems
Skittles candy comes in the following color proportions:
Color
Proportion
Red Green Yellow Orange Purple
0.2
0.15
0.3
0.25
0.1
Now, a Skittle is selected at random from each of three newly
opened bags. What is the probability that the three Skittles
have the given color(s)?
5. All three are Orange
P(O1 and O2 and O3) = P(O1) x P(O2) x P(O3)
= (0.3)(0.3)(0.3) = 0.027
Practice Problems
Skittles candy comes in the following color proportions:
Color
Proportion
Red
0.25
Green Yellow Orange Purple
0.2
0.15
0.3
0.1
Now, a Skittle is selected at random from each of three newlyopened bags. What is the probability that the three Skittles
have the given color(s)?
6. All three are Purple
P(P1 and P2 and P3) = P(P1) x P(P2) x P(P3)
= (0.1)(0.1)(0.1) = 0.001
Practice Problems
Skittles candy comes in the following color proportions:
Color
Red Green Yellow Orange Purple
Proportion 0.25
0.2
0.15
0.3
0.1
Now, a Skittle is selected at random from each of three newly
opened bags. What is the probability that the three Skittles
have the given color(s)?
7. None are Red
P(none R) = P(not R1 and not R2 and not R3)
= P(not R1) x P(not R2) x P(not R3)
= (0.75)(0.75)(0.75) = 0.421875
Practice Problems
Skittles candy comes in the following color proportions:
Color
Proportion
Red
0.25
Green Yellow Orange Purple
0.2
0.15
0.3
0.1
Now, a Skittle is selected at random from each of three newlyopened bags. What is the probability that the three Skittles
have the given color(s)?
8. The 1st is Red, the 2nd is Orange, and the 3rd is not Purple
P(R1 and O2 and not P3) = P(R1) x P(O2) x P(not P3)
= (0.25)(0.3)(0.9) = 0.0675