STAT 211 Topic 2

Lecture 3

Lecture Notes

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Probability

Sample Space

All possible outcomes of an experiment:

EX flipping a coin: $S=\{H,T\}$

EX rolling a die: $S=\{1,2,3,4,5,6\}$

Event

Any subset of outcomes contained in sample space

EX flipping a coin: $E=\{H\},\{T\},\{H,T\},...$

Set Theory

For two events $A$ and $B$

Union ($A\cup B$)

all outcomes that are in A, B, or both

Intersection($A\cap B$)

all outcomes that are in A and B

Complement ($A^{\prime }$)

all outcomes that are not in A

Mutually Exclusive

$A\cap B=\{\}$

Properties

• $0\le P(A)\le 1$
• $P()=0,P(S)=1$
• $P(A^{\prime })=1-P(A)$
• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
• If A and B are mutually exclusive, $P(A\cup B)=P(A)+P(B)$

Example

Probability of stopping at first light is 0.4. Probability of stopping at second light is 0.5. Probability of stopping at at least one light is 0.6:

$\begin{array}{rl}P(A)& =0.4\\ P(B)& =0.5\\ P(A\cup B)& =0.6\\ P(A\cap B)& =0.3\\ P(A\cup B^{\prime })& =0.1\end{array}$

Equally Likely Outcomes

If $N$ is number of possible outcomes and $A$ is event, then $P(A)=\frac{N(A)}{N}$

Conditional Probability

Probability of event $A$ given event $B$ has occurred:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

Lecture 4

Lecture 4 Notes

Thursday, January 27, 2011

Sample Warm-up Problem

Jane is taking a flight. Chance that she will sit in front row is .4; chance that her flights will be delayed is .3; chance that she willl sit in front row and flight is on time is .2:

(Draw a venn diagram)

We know:

• A: sits in front row = .4
• B: flight is delayed = .3
• A∩B': front and not delayed = .2

We can find:

1. P(A ∩ B) = P(A) − P(A∩B') = .2
2. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = .5
3. P(A' ∩ B') = P( (A ∪ B)' ) = 1 - P(A ∪ B) = .5
4. P(B' | A) = P(B' ∩ A) ÷ P(A) = 0.2 ÷ 0.4 = 0.5
5. P(A' | B) = P(A' ∩ B) ÷ P(B) = 0.1 ÷ 0.3 = 1/3

Independent Events

$\begin{array}{rl}P(A\cap B)& =P(A)\times P(B)\\ P(A\cap B\cap \dots \cap Z)& =P(A)\times P(B)\times \dots \times P(Z)\end{array}$

B has no effect on A when independent:

$P(A|B)=P(A)$

• A: student is Male
• B: student is an engineer
• P(A) = .513
• P(B) = .518
• P(A ∩ B) = .414

.513 × .518 ≠ .414 ∴ Not independent.

Mutually Exhaustive

NOT TO BE CONFUSED WITH MUTUALLY EXCLUSIVE

(Mutually Exclusive means that P(A ∩ B) = 0)

Mutually Exhaustive means that P(A ∪ B) = Sample Space

Events that are Mutually Exclusive AND Mutually Exhaustive, then the sample space can be partitioned based on each event.

For any event D that is partitioned by Mutually exhaustive and exclusive events A, B, and C, then

$P(D)=P(A\cap D)+P(B\cap D)+P(C\cap D)$

$P(A\cap D)=P(D|A)\times P(A)$

…which brings us to…

Bayes' Theorem

Suppose $A_1,\dots ,A_n$ are mutually exclusive and exhaustive events:

$P(A_i|B)=\frac{P(A_i\cap B)}{P(B)}=\frac{P(B|A_i)P(A_i)}{\sum _{j=1}^{n}P(B|A_j)P(A_j)}$

Example 1

Suppose a colon cancer test is 95% accurate (i.e. if a patient has colon cancer, the test will detect it 95% of the time)

If there is no cancer, the test will give a false positive 1% of the time.

If 5% of the population has colon cancer, what is the probability that a randomly selected patient does not have cancer given a positive test result.

Events:

$A_1$: Patient does not have cancer

$A_2={A_1}^{\prime }$: Patient has cancer

$B$: Test returns positive

A₁ and A₂ are mutually exclusive and mutually exhaustive, therefore:

We know:

$P(A_2)=0.05$ 5% of population has cancer, so $P(A_1)=1-P(A_2)=0.95$

Since the test is 95% accurate, $P(B|A_2)=0.95$

Since the test returns a false positive 1% of the time, $P(B|A_1)=0.01$

We want to find $P(A_1|B)$

Solution

$P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)}=\frac{0.01\times 0.95}{0.01\times 0.95+0.95\times 0.05}=0.167$

Example 2

Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn’t rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie’s wedding?

• Let A be the event that it rains
• Let B be the event that the weatherman predicts it will rain

Mutually Exhaustive and Exclusive Events

The two events A and A' are mutually exclusive and exhaustive.

$\begin{array}{rl}P(A)& =5/365\\ P(B|A)& =\frac{P(A\cap B)}{P(A)}=0.9\\ P(B|A^{\prime })& =\frac{P(A^{\prime }\cap B)}{P(A^{\prime })}=\frac{P(A^{\prime }\cap B)}{1-P(A)}=0.1\\ P(A|B)& =\frac{P(B\cap A)}{P(B)}=?\end{array}$

Using the given information, we can solve for P(A ∩ B) and P(B):

$P(A\cap B)=P(B|A)\times P(A)=0.9\times 5/365$

$P(B)=P(A\cap B)+P(A^{\prime }\cap B)=0.9\times 5/365+0.1\times (1-5/365)$

Plug these values into the equation:

$P(A|B)=\frac{0.9\times 5/365}{0.9\times 3/365+0.1\times (1-5/365)}\approx 0.111$