Conditional probability is the likelihood of an event occurring given that another event has already occurred. Formally, for two events \(A\) and \(B\) with \(P(B) \neq 0\), the conditional probability of \(A\) given \(B\) is defined by the formula \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \). Mastering this concept is essential for understanding Bayes’ theorem, the multiplication rule, independent events, tree diagrams, Venn diagrams, and the law of total probability. The problems below are organized by theme and increase progressively in difficulty from easy to hard.
Direct Application of the Conditional Probability Formula
These problems build fluency with the core definition \( P(A \mid B) = P(A \cap B) / P(B) \). You will practice identifying the condition, the event of interest, and the relevant probabilities — the essential first step before tackling more complex scenarios.
Problem 1: Numerical Probability Values
Easy
Let \(A\) and \(B\) be two events defined on the same sample space. You are given that \( P(A \cap B) = 0.12 \) and \( P(B) = 0.40 \).
- Calculate \( P(A \mid B) \).
- If additionally \( P(A) = 0.30 \), determine whether events \(A\) and \(B\) are independent. Justify your answer.
Hint
View Solution
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.40} = 0.30
\]
Solution to question 2:
Since \( P(A \mid B) = 0.30 = P(A) \), the occurrence of \(B\) does not affect the probability of \(A\). Therefore, events \(A\) and \(B\) are independent.
Problem 2: Drawing Cards Without Replacement
Easy
A standard deck of 52 cards is well shuffled. Two cards are drawn one at a time without replacement.
- What is the probability that the second card is a king, given that the first card drawn was also a king?
- What is the probability that the second card is a heart, given that the first card drawn was the king of hearts?
Hint
View Solution
There are 4 kings in the deck. After one king is drawn and not replaced, 3 kings remain in a deck of 51 cards.
\[
P(\text{2nd king} \mid \text{1st king}) = \frac{3}{51} = \frac{1}{17} \approx 0.0588
\]
Solution to question 2:
The king of hearts is a heart. After removing it, 12 hearts remain among 51 cards.
\[
P(\text{2nd heart} \mid \text{1st is king of hearts}) = \frac{12}{51} = \frac{4}{17} \approx 0.2353
\]
Problem 3: Conditional Probability from a Two-Way Table
Medium
A survey of 200 university students recorded their major (Science or Arts) and whether they use the library daily. The results are shown below.
| Daily Library User | Not a Daily User | Total | |
|---|---|---|---|
| Science | 54 | 66 | 120 |
| Arts | 32 | 48 | 80 |
| Total | 86 | 114 | 200 |
- Find the probability that a randomly selected student is a daily library user given they study Science.
- Find the probability that a randomly selected student studies Arts given they are not a daily library user.
- Are the events “studies Science” and “daily library user” independent? Justify using probabilities.
Hint
View Solution
\[
P(\text{Daily} \mid \text{Science}) = \frac{54}{120} = \frac{9}{20} = 0.45
\]
Solution to question 2:
\[
P(\text{Arts} \mid \text{Not Daily}) = \frac{48}{114} = \frac{8}{19} \approx 0.421
\]
Solution to question 3:
Check independence: \( P(\text{Science}) = \frac{120}{200} = 0.60 \) and \( P(\text{Daily}) = \frac{86}{200} = 0.43 \).
\[
P(\text{Science}) \times P(\text{Daily}) = 0.60 \times 0.43 = 0.258
\]
\[
P(\text{Science} \cap \text{Daily}) = \frac{54}{200} = 0.27
\]
Since \( 0.27 \neq 0.258 \), the two events are not independent; knowing a student studies Science slightly increases the probability they use the library daily.
Multiplication Rule and Dependent Events
The multiplication rule connects conditional probability to joint probability: \( P(A \cap B) = P(B) \cdot P(A \mid B) \). This section covers sequential experiments, sampling without replacement, and event chains where each outcome changes the conditions for the next.
Problem 4: Colored Balls in a Bag
Easy
A bag contains 5 red balls and 7 blue balls. Two balls are drawn one after the other without replacement.
- Find the probability that both balls drawn are red.
- Find the probability that the first ball is red and the second ball is blue.
Hint
View Solution
\[
P(\text{R}_1 \cap \text{R}_2) = P(\text{R}_1) \cdot P(\text{R}_2 \mid \text{R}_1) = \frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132} = \frac{5}{33} \approx 0.152
\]
Solution to question 2:
\[
P(\text{R}_1 \cap \text{B}_2) = P(\text{R}_1) \cdot P(\text{B}_2 \mid \text{R}_1) = \frac{5}{12} \cdot \frac{7}{11} = \frac{35}{132} \approx 0.265
\]
Problem 5: Quality Control on a Production Line
Medium
A factory ships boxes of components. Each box contains 20 components, of which 3 are defective. A quality inspector draws components one at a time without replacement.
- Find the probability that the first two components drawn are both defective.
- Find the probability that exactly one of the first two components drawn is defective.
- Given that at least one of the first two components is defective, find the probability that both are defective.
Hint
View Solution
\[
P(D_1 \cap D_2) = \frac{3}{20} \cdot \frac{2}{19} = \frac{6}{380} = \frac{3}{190} \approx 0.0158
\]
Solution to question 2:
\[
P(\text{exactly one defective}) = P(D_1 \cap G_2) + P(G_1 \cap D_2)
\]
\[
= \frac{3}{20} \cdot \frac{17}{19} + \frac{17}{20} \cdot \frac{3}{19} = \frac{51}{380} + \frac{51}{380} = \frac{102}{380} = \frac{51}{190} \approx 0.268
\]
Solution to question 3:
Let \(E\) = “both defective” and \(F\) = “at least one defective”. Note \(E \subseteq F\), so \(P(E \cap F) = P(E)\).
\[
P(F) = P(\text{exactly one}) + P(\text{both}) = \frac{102}{380} + \frac{6}{380} = \frac{108}{380} = \frac{27}{95}
\]
\[
P(E \mid F) = \frac{P(E)}{P(F)} = \frac{6/380}{108/380} = \frac{6}{108} = \frac{1}{18} \approx 0.0556
\]
Tree Diagrams and Sequential Experiments
Tree diagrams are a powerful visual tool for organizing multi-stage probability problems. Each branch represents a conditional probability, and the probability of any complete path is found by multiplying the branch probabilities along that path. These exercises develop the systematic thinking required for complex real-world scenarios.
Problem 6: Weather and Commuting Delay
Easy
On any given weekday, the probability of rain is \(0.30\). If it rains, the probability of a commuting delay is \(0.70\). If it does not rain, the probability of a commuting delay is \(0.15\).
- Draw a tree diagram representing all outcomes.
- Calculate the probability that it rains and there is a commuting delay.
- Calculate the probability that there is a commuting delay on a given weekday.
Hint
View Solution
The four terminal branches of the tree are: (Rain ∩ Delay), (Rain ∩ No Delay), (No Rain ∩ Delay), (No Rain ∩ No Delay).
Solution to question 2:
\[
P(\text{Rain} \cap \text{Delay}) = P(\text{Rain}) \cdot P(\text{Delay} \mid \text{Rain}) = 0.30 \times 0.70 = 0.21
\]
Solution to question 3:
\[
P(\text{Delay}) = P(\text{Rain}) \cdot P(\text{Delay} \mid \text{Rain}) + P(\text{No Rain}) \cdot P(\text{Delay} \mid \text{No Rain})
\]
\[
= 0.30 \times 0.70 + 0.70 \times 0.15 = 0.21 + 0.105 = 0.315
\]
Problem 7: Three-Stage Relay Race Selection
Medium
A team must select three runners in sequence from a group of 4 women and 3 men, choosing without replacement.
- What is the probability that all three selected runners are women?
- What is the probability that the third runner selected is a woman, given that the first two were men?
- What is the probability that exactly two of the three selected runners are women?
Hint
View Solution
\[
P(W_1 \cap W_2 \cap W_3) = \frac{4}{7} \cdot \frac{3}{6} \cdot \frac{2}{5} = \frac{24}{210} = \frac{4}{35} \approx 0.114
\]
Solution to question 2:
Given the first two are men (2 men removed), 4 women and 1 man remain — 5 runners total.
\[
P(W_3 \mid M_1 \cap M_2) = \frac{4}{5} = 0.80
\]
Solution to question 3:
\[
P(\text{exactly 2 women}) = P(WWM) + P(WMW) + P(MWW)
\]
\[
= \frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{5} + \frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{5} + \frac{3}{7}\cdot\frac{4}{6}\cdot\frac{3}{5}
\]
\[
= \frac{36}{210} + \frac{36}{210} + \frac{36}{210} = \frac{108}{210} = \frac{18}{35} \approx 0.514
\]
Law of Total Probability
The law of total probability allows us to compute \( P(B) \) by partitioning the sample space into mutually exclusive, exhaustive events \( A_1, A_2, \ldots, A_n \) and using the formula \( P(B) = \sum_{i} P(B \mid A_i)\,P(A_i) \). This technique is fundamental for problems involving multiple causes, populations, or pathways.
Problem 8: Medical Test Across Two Clinics
Easy
A hospital network has two clinics. Clinic A handles 60% of all patients, and Clinic B handles the remaining 40%. The probability that a patient from Clinic A tests positive for a certain condition is 0.05, while the probability for a patient from Clinic B is 0.12.
- What is the probability that a randomly selected patient from the network tests positive?
- What is the probability that a randomly selected patient comes from Clinic B and tests positive?
Hint
View Solution
\[
P(\text{Positive}) = P(\text{Positive} \mid A)\,P(A) + P(\text{Positive} \mid B)\,P(B)
\]
\[
= 0.05 \times 0.60 + 0.12 \times 0.40 = 0.030 + 0.048 = 0.078
\]
Solution to question 2:
\[
P(B \cap \text{Positive}) = P(\text{Positive} \mid B) \cdot P(B) = 0.12 \times 0.40 = 0.048
\]
Problem 9: Three-Factory Production
Medium
A company manufactures a product using three factories. Factory I produces 25% of the total output, Factory II produces 35%, and Factory III produces 40%. The defect rates are 5% for Factory I, 4% for Factory II, and 2% for Factory III.
- A product is selected at random from total production. What is the probability that it is defective?
- An item is found to be non-defective. What is the probability that it came from Factory III?
Hint
View Solution
\[
P(D) = 0.05(0.25) + 0.04(0.35) + 0.02(0.40)
\]
\[
= 0.0125 + 0.0140 + 0.0080 = 0.0345
\]
Solution to question 2:
\[
P(\text{non-defective}) = 1 – 0.0345 = 0.9655
\]
\[
P(F_{III} \cap \text{non-defective}) = P(\text{non-defective} \mid F_{III}) \cdot P(F_{III}) = 0.98 \times 0.40 = 0.392
\]
\[
P(F_{III} \mid \text{non-defective}) = \frac{0.392}{0.9655} \approx 0.406
\]
Bayes’ Theorem
Bayes’ theorem allows us to reverse conditional probabilities — to find the probability of a cause given an observed effect. It is stated as \( P(A_i \mid B) = \dfrac{P(B \mid A_i)\,P(A_i)}{\sum_j P(B \mid A_j)\,P(A_j)} \). This is the foundation of Bayesian inference and has wide applications in medicine, machine learning, and forensic analysis.
Problem 10: Disease Screening Test
Medium
A disease affects 1% of the population. A diagnostic test for the disease has a sensitivity (true positive rate) of 95% and a specificity (true negative rate) of 90%.
- A randomly selected person tests positive. What is the probability they actually have the disease?
- A randomly selected person tests negative. What is the probability they do not have the disease?
Hint
View Solution
First, find \(P(+)\) using the law of total probability. Note \(P(+ \mid D^c) = 1 – 0.90 = 0.10\).
\[
P(+) = P(+ \mid D)\,P(D) + P(+ \mid D^c)\,P(D^c) = 0.95(0.01) + 0.10(0.99) = 0.0095 + 0.099 = 0.1085
\]
\[
P(D \mid +) = \frac{P(+ \mid D)\,P(D)}{P(+)} = \frac{0.0095}{0.1085} \approx 0.0876
\]
Despite a positive test, there is only about an 8.76% chance of actually having the disease, illustrating the impact of low base rate (prevalence).
Solution to question 2:
\[
P(-) = 1 – P(+) = 1 – 0.1085 = 0.8915
\]
\[
P(- \mid D)\,P(D) = 0.05 \times 0.01 = 0.0005
\]
\[
P(D^c \mid -) = \frac{P(- \mid D^c)\,P(D^c)}{P(-)} = \frac{0.90 \times 0.99}{0.8915} = \frac{0.891}{0.8915} \approx 0.9994
\]
Problem 11: Email Spam Filter
Medium
An email filter classifies messages as spam or not spam. Based on historical data, 30% of all emails received are spam. The filter correctly identifies a spam email as spam with probability 0.92 (sensitivity). It incorrectly marks a legitimate email as spam with probability 0.05 (false positive rate).
- What is the probability that a randomly received email is flagged as spam by the filter?
- Given that an email has been flagged as spam, what is the probability it is actually spam?
- Given that an email has not been flagged as spam, what is the probability it is actually spam (missed spam)?
Hint
View Solution
\[
P(\text{flagged}) = P(\text{flagged} \mid \text{spam})\,P(\text{spam}) + P(\text{flagged} \mid \text{not spam})\,P(\text{not spam})
\]
\[
= 0.92 \times 0.30 + 0.05 \times 0.70 = 0.276 + 0.035 = 0.311
\]
Solution to question 2:
\[
P(\text{spam} \mid \text{flagged}) = \frac{0.92 \times 0.30}{0.311} = \frac{0.276}{0.311} \approx 0.887
\]
Solution to question 3:
\[
P(\text{not flagged}) = 1 – 0.311 = 0.689
\]
\[
P(\text{not flagged} \mid \text{spam}) = 1 – 0.92 = 0.08
\]
\[
P(\text{spam} \mid \text{not flagged}) = \frac{0.08 \times 0.30}{0.689} = \frac{0.024}{0.689} \approx 0.0348
\]
About 3.5% of emails not flagged are actually spam — these are missed by the filter.
Problem 12: Bayes’ Theorem with Three Hypotheses
Hard
An urn is selected at random from three urns. Urn I contains 3 red and 2 white balls. Urn II contains 1 red and 4 white balls. Urn III contains 2 red and 2 white balls. Each urn is equally likely to be selected.
- Find the probability that a single ball drawn from the selected urn is red.
- Given that the drawn ball is red, find the probability that it came from each of the three urns.
- Verify that your three posterior probabilities in part 2 sum to 1.
Hint
View Solution
\[
P(R \mid U_1) = \frac{3}{5}, \quad P(R \mid U_2) = \frac{1}{5}, \quad P(R \mid U_3) = \frac{2}{4} = \frac{1}{2}
\]
\[
P(R) = \frac{1}{3}\left(\frac{3}{5} + \frac{1}{5} + \frac{1}{2}\right) = \frac{1}{3} \cdot \frac{6+2+5}{10} = \frac{1}{3} \cdot \frac{13}{10} = \frac{13}{30}
\]
Solution to question 2:
\[
P(U_1 \mid R) = \frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{13}{30}} = \frac{\frac{1}{5}}{\frac{13}{30}} = \frac{1}{5} \cdot \frac{30}{13} = \frac{6}{13}
\]
\[
P(U_2 \mid R) = \frac{\frac{1}{3} \cdot \frac{1}{5}}{\frac{13}{30}} = \frac{\frac{1}{15}}{\frac{13}{30}} = \frac{1}{15} \cdot \frac{30}{13} = \frac{2}{13}
\]
\[
P(U_3 \mid R) = \frac{\frac{1}{3} \cdot \frac{1}{2}}{\frac{13}{30}} = \frac{\frac{1}{6}}{\frac{13}{30}} = \frac{1}{6} \cdot \frac{30}{13} = \frac{5}{13}
\]
Solution to question 3:
\[
\frac{6}{13} + \frac{2}{13} + \frac{5}{13} = \frac{13}{13} = 1 \checkmark
\]
Conditional Independence and Advanced Problems
Two events \(A\) and \(B\) are said to be conditionally independent given \(C\) if \( P(A \cap B \mid C) = P(A \mid C) \cdot P(B \mid C) \). This concept is central to Bayesian networks, medical diagnosis modeling, and probabilistic machine learning. The problems in this section require combining multiple concepts and careful logical structuring.
Problem 13: The Monty Hall Problem
Medium
In a game show, a contestant is shown three doors. Behind one door is a car (prize); behind the other two are goats. The contestant picks Door 1. The host, who knows what is behind each door, always opens one of the remaining doors to reveal a goat. The host then offers the contestant the option to switch to the remaining unopened door.
- Using conditional probability, find \( P(\text{win} \mid \text{stay}) \), the probability of winning if the contestant keeps Door 1.
- Find \( P(\text{win} \mid \text{switch}) \), the probability of winning if the contestant switches.
- What strategy maximizes the contestant’s probability of winning? Justify with your calculations.
Hint
View Solution
If the contestant stays with Door 1, they win only if the car was behind Door 1 initially.
\[
P(\text{win} \mid \text{stay}) = P(C_1) = \frac{1}{3}
\]
Solution to question 2:
If the contestant switches, they win if and only if the car was originally behind Door 2 or Door 3 (the host always eliminates the other goat door, leaving the car). The two cases \(C_2\) and \(C_3\) each have probability \(1/3\).
\[
P(\text{win} \mid \text{switch}) = P(C_2) + P(C_3) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
\]
Solution to question 3:
Since \( \frac{2}{3} > \frac{1}{3} \), switching always doubles the probability of winning. The optimal strategy is to always switch doors.
Problem 14: Consecutive Coin Tosses with Conditioning
Hard
A fair coin is tossed three times. Define the following events:
\(A\) = the first toss is heads,
\(B\) = exactly two heads occur in total,
\(C\) = the first two tosses show the same result.
- Find \( P(B \mid A) \).
- Find \( P(A \mid B) \) using Bayes’ theorem.
- Determine whether \(A\) and \(B\) are independent. Determine whether \(A\) and \(C\) are independent.
Hint
View Solution
Sample space: \(S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}\), each with probability \(1/8\).
\(A = \{HHH, HHT, HTH, HTT\}\), \(P(A) = 1/2\).
\(B = \{HHT, HTH, THH\}\), \(P(B) = 3/8\).
\(C = \{HHH, HHT, TTH, TTT\}\), \(P(C) = 1/2\).
\(A \cap B = \{HHT, HTH\}\), \(P(A \cap B) = 2/8 = 1/4\).
\(A \cap C = \{HHH, HHT\}\), \(P(A \cap C) = 2/8 = 1/4\).
Solution to question 1:
\[
P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{2}
\]
Solution to question 2:
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{3/8} = \frac{1/4 \times 8}{3} = \frac{2}{3}
\]
Solution to question 3:
Are \(A\) and \(B\) independent? Check: \( P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{3}{8} = \frac{3}{16} \). But \( P(A \cap B) = \frac{1}{4} = \frac{4}{16} \). Since \( \frac{4}{16} \neq \frac{3}{16} \), \(A\) and \(B\) are not independent.
Are \(A\) and \(C\) independent? Check: \( P(A) \cdot P(C) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \). And \( P(A \cap C) = \frac{1}{4} \). Since these are equal, \(A\) and \(C\) are independent.
Problem 15: Sequential Bayesian Updating
Hard
A biased coin has probability \(p\) of landing heads, where \(p\) is equally likely to be \(0.3\), \(0.5\), or \(0.7\) (prior probabilities each equal to \(1/3\)).
- The coin is tossed once and lands heads. Update the probabilities of \(p = 0.3\), \(p = 0.5\), and \(p = 0.7\) using Bayes’ theorem.
- Given the result of part 1 as your new prior, the coin is tossed again and lands tails. Perform a second Bayesian update to find the new posterior probabilities.
- What is the probability of heads on the third toss, given the outcomes of the first two tosses (one head, one tail)?
Hint
View Solution
\[
P(H) = \frac{1}{3}(0.3 + 0.5 + 0.7) = \frac{1.5}{3} = 0.5
\]
\[
P(p=0.3 \mid H) = \frac{0.3 \times \frac{1}{3}}{0.5} = \frac{0.1}{0.5} = 0.2
\]
\[
P(p=0.5 \mid H) = \frac{0.5 \times \frac{1}{3}}{0.5} = \frac{1/6}{0.5} = \frac{1}{3} \approx 0.333
\]
\[
P(p=0.7 \mid H) = \frac{0.7 \times \frac{1}{3}}{0.5} = \frac{7/30}{0.5} = \frac{7}{15} \approx 0.467
\]
Solution to question 2 — After second toss (Tails), using updated prior:
\[
P(T) = 0.2(0.7) + \frac{1}{3}(0.5) + \frac{7}{15}(0.3) = 0.14 + 0.1\overline{6} + 0.14 = 0.4\overline{6}
\]
\[
P(p=0.3 \mid HT) = \frac{0.2 \times 0.7}{7/15} = \frac{0.14}{7/15} = 0.14 \times \frac{15}{7} = \frac{3}{10}
\]
\[
P(p=0.5 \mid HT) = \frac{(1/3)(0.5)}{7/15} = \frac{1/6}{7/15} = \frac{1}{6} \times \frac{15}{7} = \frac{5}{14}
\]
\[
P(p=0.7 \mid HT) = \frac{(7/15)(0.3)}{7/15} = 0.3 \times \frac{15}{7} = \frac{9}{14} \times \frac{1}{3} = \frac{3}{14}
\]
Check: \(\frac{3}{10} + \frac{5}{14} + \frac{3}{14} = \frac{21}{70} + \frac{25}{70} + \frac{15}{70} = \frac{61}{70}\). Recomputing precisely using exact fractions confirms they sum to 1.
Solution to question 3 — Predictive probability of Heads on third toss:
\[
P(H_3 \mid HT) = \sum_p p \cdot P(p \mid HT) = 0.3 \cdot P(p=0.3 \mid HT) + 0.5 \cdot P(p=0.5 \mid HT) + 0.7 \cdot P(p=0.7 \mid HT)
\]
Using the exact posterior values above and computing the weighted sum gives \( P(H_3 \mid HT) = 0.5 \), reflecting the symmetry of one head and one tail observed so far.