Conditional Probability — Hard Practice Quiz
A Probability cheat sheet for Conditional Probability — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Conditional Probability
Where: \(P(B) \neq 0\)
Bayes' Theorem
Law of Total Probability
Where: \(B_i\) partition the sample space
Practice quiz
Given that events $B_1$ and $B_2$ form a partition of the sample space, with $P(B_1) = 0.6$ and $P(B_2) = 0.4$. If $P(A|B_1) = 0.3$ and $P(A|B_2) = 0.7$, what is the value of $P(B_1|A)$?
- $\frac{9}{23}$
- $\frac{18}{23}$
- $0.46$
- $0.3$
Answer: $\frac{9}{23}$
Given $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$. Which of the following is a correct conclusion?
- $A$ and $B$ are independent, and $P(A|B) = 0.5$.
- $A$ and $B$ are not independent, because $P(A|B) \neq P(A)$.
- $P(B|A) = 0.5$, which implies $A$ and $B$ are independent.
- $P(A|B) = 0.4$, which implies $A$ and $B$ are not independent.
Answer: $A$ and $B$ are independent, and $P(A|B) = 0.5$.
Events $B_1$ and $B_2$ form a partition of the sample space. If $P(A) = 0.6$, $P(B_1) = 0.7$, and $P(A|B_1) = 0.8$, what is the value of $P(A|B_2)$?
- $\frac{2}{15}$
- $0.04$
- $0.3$
- $0.56$
Answer: $\frac{2}{15}$
A rare disease affects $1\%$ of the population. A diagnostic test for this disease has a sensitivity of $95\%$ (i.e., $P(\text{positive test}|\text{disease}) = 0.95$) and a specificity of $90\%$ (i.e., $P(\text{negative test}|\text{no disease}) = 0.90$). If a randomly selected person tests positive, what is the probability that they actually have the disease?
- $\frac{19}{217}$
- $0.95$
- $0.1085$
- $0.0095$
Answer: $\frac{19}{217}$
Given $P(A|B)$, $P(B|A)$, and $P(A)$, which of the following expressions correctly represents $P(B)$?
- $P(B) = \frac{P(B|A) \cdot P(A)}{P(A|B)}$
- $P(B) = \frac{P(A|B) \cdot P(A)}{P(B|A)}$
- $P(B) = P(A|B) \cdot P(A) \cdot P(B|A)$
- $P(B) = \frac{P(A|B)}{P(B|A) \cdot P(A)}$
Answer: $P(B) = \frac{P(B|A) \cdot P(A)}{P(A|B)}$
Under what condition is $P(A|B) = P(B|A)$ always true, assuming $P(A) > 0$ and $P(B) > 0$?
- When $A$ and $B$ are independent.
- When $P(A) = P(B)$.
- When $A$ and $B$ are mutually exclusive.
- When $P(A \cap B) = 1$.
Answer: When $P(A) = P(B)$.
Given $P(A) = 0.5$, $P(B) = 0.4$, and $P(A|B) = 0.6$. What is the value of $P(A|B^c)$?
- $\frac{13}{30}$
- $0.26$
- $0.6$
- $0.24$
Answer: $\frac{13}{30}$
If events $B_1, B_2, \dots, B_n$ form a partition of the sample space, and for a given event $A$, it is known that $P(A|B_i) = k$ for all $i$, where $k$ is a constant. What is $P(A)$?
- $k$
- $k \cdot P(B_1)$
- $k \cdot n$
- $1 - k$
Answer: $k$
Suppose $P(A|B) = 2 P(A|B^c)$ and $P(B) = 0.3$. What is the ratio $P(B|A) / P(B^c|A)$?
- $\frac{6}{7}$
- $\frac{2}{3}$
- $\frac{3}{7}$
- $2$
Answer: $\frac{6}{7}$
A factory has three machines, $M_1, M_2$, and $M_3$, which produce $50\%$, $30\%$, and $20\%$ of the total output, respectively. The defect rates for these machines are $1\%$, $2\%$, and $3\%$, respectively. If a randomly selected item is found to be defective, what is the probability that it was produced by machine $M_2$?
- $\frac{6}{17}$
- $0.017$
- $0.006$
- $0.3$
Answer: $\frac{6}{17}$
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