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 ($A'$)

all outcomes that are not in A

Mutually Exclusive

$A\cap B=\{\}$

Properties

• $0\leq P(A)\leq 1$
• $P({})=0,P(S)=1$
• $P(A')=1-P(A)$
• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
• 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:

${\begin{aligned}P(A)&=0.4\\P(B)&=0.5\\P(A\cup B)&=0.6\\P(A\cap B)&=0.3\\P(A\cup B')&=0.1\end{aligned}}$

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

${\begin{aligned}P(A\cap B)&=P(A)\times P(B)\\P(A\cap B\cap \ldots \cap Z)&=P(A)\times P(B)\times \ldots \times P(Z)\end{aligned}}$

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 $A_{1},\ldots ,A_{n}$ 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

$A_{2}=A_{1}'$: 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.

${\begin{aligned}P(A)&=5/365\\P(B|A)&={\tfrac {P(A\cap B)}{P(A)}}=0.9\\P(B|A')&={\tfrac {P(A'\cap B)}{P(A')}}={\tfrac {P(A'\cap B)}{1-P(A)}}=0.1\\P(A|B)&={\tfrac {P(B\cap A)}{P(B)}}=?\end{aligned}}$

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

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

$P(B)=P(A\cap B)+P(A'\cap B)=0.9\times 5/365+0.1\times (1-5/365)$

Plug these values into the equation:

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