Exam P — Conditional Probability & Bayes Flashcards

$\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}$, valid only when $\Pr(B)>0$. It rescales the probability of $A$ to the reduced sample space in which $B$ has occurred.

$\Pr(A\cap B)=\Pr(A\mid B)\,\Pr(B)=\Pr(B\mid A)\,\Pr(A)$. It rearranges the definition of conditional probability and holds for any events (no independence required).

$\Pr(A\cap B\cap C)=\Pr(A)\,\Pr(B\mid A)\,\Pr(C\mid A\cap B)$. Each factor conditions on everything to its left; the chain rule generalizes to any number of events.

$A$ and $B$ are independent iff $\Pr(A\cap B)=\Pr(A)\,\Pr(B)$. Equivalently $\Pr(A\mid B)=\Pr(A)$ (when $\Pr(B)>0$) and $\Pr(B\mid A)=\Pr(B)$ (when $\Pr(A)>0$): knowing one event tells you nothing about the other.

No. Mutually exclusive means $\Pr(A\cap B)=0$, while independence requires $\Pr(A\cap B)=\Pr(A)\,\Pr(B)>0$. The two are contradictory for positive-probability events: if $A$ occurs, a mutually exclusive $B$ cannot, so they are maximally dependent, not independent.

Yes. $\Pr(A\cap B^{c})=\Pr(A)-\Pr(A\cap B)=\Pr(A)-\Pr(A)\Pr(B)=\Pr(A)\left(1-\Pr(B)\right)=\Pr(A)\Pr(B^{c})$. Independence is preserved under complementation; likewise $A^{c},B$ and $A^{c},B^{c}$ are independent.

Pairwise: every pair satisfies $\Pr(A\cap B)=\Pr(A)\Pr(B)$, etc. Mutual independence additionally requires $\Pr(A\cap B\cap C)=\Pr(A)\Pr(B)\Pr(C)$ AND all pairwise conditions. Pairwise independence does **not** imply mutual independence — the triple-product condition can fail even when all pairs are independent.

$\Pr(A)=\displaystyle\sum_{i=1}^{n}\Pr(A\mid B_i)\,\Pr(B_i)$. The $B_i$ must be mutually exclusive and exhaustive (a partition); $A$'s probability is the weighted average of its conditional probabilities, weighted by the $\Pr(B_i)$.

$\Pr(B_k\mid A)=\frac{\Pr(A\mid B_k)\,\Pr(B_k)}{\displaystyle\sum_{i=1}^{n}\Pr(A\mid B_i)\,\Pr(B_i)}$. The numerator is the joint $\Pr(A\cap B_k)$; the denominator is $\Pr(A)$ by the law of total probability.

Sensitivity $=\Pr(+\mid D)$ (true-positive rate). Specificity $=\Pr(-\mid D^{c})$ (true-negative rate); the false-positive rate is $\Pr(+\mid D^{c})=1-\text{specificity}$. Prevalence $=\Pr(D)$, the unconditional disease rate in the population.

In general $\Pr(D\mid +)\neq\Pr(+\mid D)$ unless $\Pr(D)=\Pr(+)$. A test can have high sensitivity $\Pr(+\mid D)$ yet a low posterior $\Pr(D\mid +)$ when the disease is rare, because the many healthy people generate a large number of false positives. Bayes' theorem is precisely the tool that converts $\Pr(+\mid D)$ into $\Pr(D\mid +)$.

$\Pr(D)=0.02,\ \Pr(D^{c})=0.98,\ \Pr(+\mid D^{c})=1-0.90=0.10$. Numerator: $\Pr(+\mid D)\Pr(D)=0.95\times0.02=0.019$. Denominator: $0.019+0.10\times0.98=0.019+0.098=0.117$. $\Pr(D\mid +)=\frac{0.019}{0.117}\approx 0.1624$, about $16.2\%$ — low despite high sensitivity, because the disease is rare.

$\Pr(-\mid D)=1-0.95=0.05,\ \Pr(-\mid D^{c})=0.90$. Numerator: $\Pr(-\mid D)\Pr(D)=0.05\times0.02=0.001$. Denominator: $\Pr(-)=0.001+0.90\times0.98=0.001+0.882=0.883$. $\Pr(D\mid -)=\frac{0.001}{0.883}\approx 0.00113$, about $0.11\%$ — a negative result is very reassuring here.

Law of total probability: $\Pr(\text{claim})=0.10(0.60)+0.04(0.30)+0.30(0.10)$ $=0.060+0.012+0.030=0.102$. So about $10.2\%$ of drivers file a claim.

From the previous card $\Pr(\text{claim})=0.102$. Bayes: $\Pr(\text{high-risk}\mid\text{claim})=\frac{0.30(0.10)}{0.102}=\frac{0.030}{0.102}\approx 0.2941$. So about $29.4\%$ of claimants are high-risk, far above their $10\%$ population share.

$\Pr(R\mid U_1)=\frac{3}{5},\ \Pr(R\mid U_2)=\frac{1}{5},\ \Pr(U_1)=\Pr(U_2)=\frac12$. $\Pr(R)=\frac12\cdot\frac35+\frac12\cdot\frac15=\frac{3}{10}+\frac{1}{10}=\frac{4}{10}$. $\Pr(U_1\mid R)=\frac{\frac12\cdot\frac35}{\frac{4}{10}}=\frac{3/10}{4/10}=\frac{3}{4}=0.75$.

$\Pr(\text{def})=0.01(0.50)+0.02(0.30)+0.03(0.20)=0.005+0.006+0.006=0.017$. $\Pr(C\mid\text{def})=\frac{0.03(0.20)}{0.017}=\frac{0.006}{0.017}\approx 0.3529$. Machine C, despite making only $20\%$ of output, accounts for about $35.3\%$ of defects.

Build a table whose cells are the joint probabilities $\Pr(A\cap B_i)=\Pr(A\mid B_i)\Pr(B_i)$ for each partition class. The column sum gives $\Pr(A)$ (law of total probability), and each posterior is that cell divided by the column total: $\Pr(B_i\mid A)=\Pr(A\cap B_i)/\Pr(A)$. It avoids re-deriving the denominator and makes the normalization step a single division.

$\Pr(\text{fair})=0.6,\ \Pr(\text{biased})=0.4$; $\Pr(H\mid\text{fair})=0.5,\ \Pr(H\mid\text{biased})=0.8$. Joint: fair $=0.6(0.5)=0.30$; biased $=0.4(0.8)=0.32$. $\Pr(H)=0.30+0.32=0.62$. $\Pr(\text{biased}\mid H)=\frac{0.32}{0.62}\approx 0.5161$.

Conditional on coin type the flips are independent: $\Pr(HH\mid\text{fair})=0.5^2=0.25$, $\Pr(HH\mid\text{biased})=0.8^2=0.64$. Joint: fair $=0.6(0.25)=0.15$; biased $=0.4(0.64)=0.256$. $\Pr(HH)=0.15+0.256=0.406$. $\Pr(\text{biased}\mid HH)=\frac{0.256}{0.406}\approx 0.6305$. The extra head shifts evidence toward the biased coin.

$A,B$ are conditionally independent given $C$ iff $\Pr(A\cap B\mid C)=\Pr(A\mid C)\,\Pr(B\mid C)$. This is independence within the sub-population where $C$ holds. It does not imply unconditional independence (and vice versa): the coin flips above are independent given the coin's type but dependent unconditionally, since shared coin identity links them.

$\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)=0.5+0.4-0.7=0.2$. For independence we'd need $\Pr(A)\Pr(B)=0.5\times0.4=0.20$. Since $0.2=0.2$, the events **are** independent.

First $\Pr(A\cap B)=\Pr(A\mid B)\Pr(B)=0.6\times0.3=0.18$. Then $\Pr(B\mid A)=\frac{\Pr(A\cap B)}{\Pr(A)}=\frac{0.18}{0.4}=0.45$.

$\Pr(H\mid A)=\frac{\Pr(H\cap A)}{\Pr(A)}=\frac{0.5}{0.6}\approx 0.8333$. Note $\Pr(H\mid A)=0.833>\Pr(H)=0.7$, so owning auto is positively associated with owning home (they are dependent).

If $A\subset B$: then $A\cap B=A$, so $\Pr(A\mid B)=\frac{\Pr(A)}{\Pr(B)}$. If $B\subset A$: then $A\cap B=B$, so $\Pr(A\mid B)=\frac{\Pr(B)}{\Pr(B)}=1$ — once $B$ happens, $A$ is certain.

Use the chain rule: $\Pr(R_1)=\frac58$, then $\Pr(R_2\mid R_1)=\frac47$ (one red gone, 7 left). $\Pr(R_1\cap R_2)=\frac58\cdot\frac47=\frac{20}{56}=\frac{5}{14}\approx 0.3571$.

By symmetry $\Pr(R_2)=\frac58$. The joint $\Pr(R_1\cap R_2)=\frac{5}{14}$ from the previous card. $\Pr(R_1\mid R_2)=\frac{\Pr(R_1\cap R_2)}{\Pr(R_2)}=\frac{5/14}{5/8}=\frac{5}{14}\cdot\frac{8}{5}=\frac{8}{14}=\frac{4}{7}\approx 0.5714$.

$\Pr(\text{ace}_1)=\frac{4}{52}=\frac{1}{13}$ and $\Pr(\text{ace}_2\mid\text{ace}_1)=\frac{3}{51}=\frac{1}{17}$. $\Pr(\text{both aces})=\frac{1}{13}\cdot\frac{1}{17}=\frac{1}{221}\approx 0.004525$.

Independence across days lets the daily survival probabilities multiply: $\Pr(\text{survive 5 days})=(1-0.05)^5=0.95^5$. $0.95^5\approx 0.7738$.

Use the complement of 'none works': $\Pr(\text{none})=(1-0.9)^3=0.1^3=0.001$. $\Pr(\text{at least one})=1-0.001=0.999$. Computing 'at least one' via $1-\Pr(\text{none})$ is far easier than summing the individual cases.

$\Pr(A)=\sum_i\Pr(A\mid B_i)\Pr(B_i)$ requires the $B_i$ to be exhaustive. Dropping a cell — e.g. counting only true positives and omitting false positives — understates $\Pr(A)$ and inflates the posterior. Every person who could produce the event $A$ (a positive test) must appear in the denominator.

$\Pr(S)=0.20,\ \Pr(\text{flag}\mid S)=0.98,\ \Pr(\text{flag}\mid S^{c})=0.03$. Joint: spam $=0.20(0.98)=0.196$; ham $=0.80(0.03)=0.024$. $\Pr(\text{flag})=0.196+0.024=0.220$. $\Pr(S\mid\text{flag})=\frac{0.196}{0.220}\approx 0.8909$.

$\frac{\Pr(D\mid E)}{\Pr(D^{c}\mid E)}=\frac{\Pr(E\mid D)}{\Pr(E\mid D^{c})}\cdot\frac{\Pr(D)}{\Pr(D^{c})}$. Posterior odds $=$ likelihood ratio $\times$ prior odds. The shared denominator $\Pr(E)$ cancels, so no normalization is needed; convert the final odds back to a probability if required.

Prior odds $=\frac{0.001}{0.999}\approx 0.0010010$. Posterior odds $=50\times0.0010010\approx 0.050050$. Convert: $\Pr(D\mid +)=\frac{\text{odds}}{1+\text{odds}}=\frac{0.050050}{1.050050}\approx 0.04766$, about $4.8\%$ — still low because the prior is tiny.

$\Pr(\text{no claim}\mid\text{low})=0.95,\ \Pr(\text{no claim}\mid\text{high})=0.75$. Joint: low $=0.80(0.95)=0.760$; high $=0.20(0.75)=0.150$. $\Pr(\text{no claim})=0.760+0.150=0.910$. $\Pr(\text{high}\mid\text{no claim})=\frac{0.150}{0.910}\approx 0.1648$ — slightly below the $0.20$ prior, as a clean year is mild evidence of being low-risk.

Use the updated prior $\Pr(\text{high})=0.1648,\ \Pr(\text{low})=0.8352$ (Bayes is sequential). Joint: low $=0.8352(0.95)=0.79344$; high $=0.1648(0.75)=0.12360$. Total $=0.79344+0.12360=0.91704$. $\Pr(\text{high}\mid\text{2nd no claim})=\frac{0.12360}{0.91704}\approx 0.1348$. Each clean year further lowers the high-risk probability.

Compute the complement directly: $\Pr(S^{c}\mid\text{flag})=\frac{0.024}{0.220}\approx 0.1091$. Then $0.8909+0.1091=1.000$. Posteriors over a partition always sum to 1 because they share the denominator $\Pr(A)$ and the numerators are the partition cells of $\Pr(A)$ — a quick exam sanity check.

Joint products: $B_1{:}\ 0.5(0.2)=0.10$; $B_2{:}\ 0.3(0.5)=0.15$; $B_3{:}\ 0.2(0.8)=0.16$. $\Pr(E)=0.10+0.15+0.16=0.41$. $\Pr(B_2\mid E)=\frac{0.15}{0.41}\approx 0.3659$.