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stats1-week8

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Deep study for Quiz 2 week 8. Conditioning narrows the sample space; Bayes flips the direction of a conditional statement.

Week map

Given information →

P(A|B)

→ multiplication rule → independence → law of total probability → Bayes theorem.

Conditional notation

$P(A|B)$ → “probability of $A$ given $B$ ” → probability $A$ occurs knowing $B$ occurred.
Formula: $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$ — requires $P(B) > 0$ .
Narrow sample space: restrict to outcomes where $B$ is true, renormalize.

Mini-example: die roll,

B

= “even,”

A

= “greater than 3.” Given even

\{2,4,6\}

, favorable

\{4,6\}

→

P(A|B) = 2/3

Multiplication rule

P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A)

Chain events sequentially: multiply conditionals.

Independence

$A$ and $B$ independent if $P(A|B) = P(A)$ — knowing $B$ does not change $A$ .
Equivalent: $P(A \cap B) = P(A) \cdot P(B)$ .
Independent ≠ mutually exclusive (disjoint events with positive probability are dependent).

Mini-example: fair coin tosses.

A

= first head,

B

= second head.

P(A \cap B) = 1/4 = P(A)P(B)

Bayes and total probability

Law of total probability (partition

B_1, \ldots, B_k

P(A) = \sum_{i=1}^{k} P(A|B_i) P(B_i)

Bayes theorem:

P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}

Flip from

P(A|B)

P(B|A)

using prior

P(B)

and evidence

P(A)

Mini-example (screening): disease rate 1%, test sensitivity

P(+|D)=0.99

, specificity

P(-|\text{no }D)=0.95

. Positive test does not imply 99% sick — denominator

P(+)

mixes true and false positives via total probability.

Pattern families

Easy — Read conditional from table or story

Identify

P(A|B)

numerator and denominator. Compute from counts in restricted row.

Medium — Multiply chain

P(A \cap B)

from

P(A|B)P(B)

. Two-step tree: multiply along branch.

Hard — Bayes setup

Identify hypothesis

H

and evidence

E

. Plug

P(E|H)

P(H)

, compute

P(E)

via total probability, then

P(H|E)

Worked mini-examples

Example 1 — Conditional from counts.

100 students: 40 take stats, 25 take both CS and stats.

P(\text{CS}|\text{stats}) = 25/40

Example 2 — Multiplication.

P(\text{rain})=0.2

P(\text{late}|\text{rain})=0.5

→

P(\text{rain and late})=0.1

Example 3 — Independence check.

Roll die:

A

odd,

B

greater than 4.

P(A)=3/6

P(A|B)

\{5,6\}

is 1/2 — not equal → dependent.

Example 4 — Complement conditional.

P(A^c|B) = 1 - P(A|B)

Example 5 — Bayes skeleton.

Prior

P(H)=0.01

, likelihood

P(E|H)=0.9

P(E|H^c)=0.1

. Compute

P(H|E)

using

P(E)=0.9(0.01)+0.1(0.99)

Traps

Swapping $P(A|B)$ and $P(B|A)$ without Bayes.
Using $P(A)+P(B)$ instead of conditional formula.
Treating independent as disjoint.
Forgetting to renormalize when conditioning.
Base rate neglect: ignoring small $P(H)$ in medical/test stories.

Diagnostic (try yourself)

$P(A \cap B)=0.12$ , $P(B)=0.4$ . Find $P(A|B)$ .
Events independent with $P(A)=0.3$ , $P(B)=0.5$ . Find $P(A \cap B)$ .
Two cards without replacement: $P(\text{second ace}|\text{first ace})$ on standard deck — qualitative: larger or smaller than $P(\text{second ace})$ unconditional?
Write Bayes formula for $P(H|E)$ in terms of $P(E|H)$ , $P(H)$ , $P(E)$ .
Table row “Test +” split into disease / no disease counts. What probability does $P(\text{disease}|\text{test}+)$ use as denominator?

stats1-week8

Week map

Conditional notation

Multiplication rule

Independence

Bayes and total probability

Pattern families

Easy — Read conditional from table or story

Medium — Multiply chain

Hard — Bayes setup

Worked mini-examples

Traps

Diagnostic (try yourself)

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