7 fully solved examples and 5 practice problems on \(P(A \mid B)\)
Conditional probability measures the probability of event \( A \) occurring given that event \( B \) has already occurred. The idea is to restrict the sample space from all of \( S \) down to only the outcomes where \( B \) happens. We denote this \( P(A \mid B) \) — read "the probability of \( A \) given \( B \)."
| Rule | Formula |
|---|---|
| Conditional probability | \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} \) |
| Multiplication rule (rearranged) | \( P(A \cap B) = P(A \mid B) \cdot P(B) \) |
Validity: the conditional formula requires \( P(B) \ne 0 \).
A fair 6-sided die is rolled.
(a) Simple probability. The sample space is \( S = \{1,2,3,4,5,6\} \). Let \( A = \{1,2,3\} \). Then:
\[ P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \](b) Conditional probability. Let \( B = \{1,3,5\} \) (odd). The two events overlap on \( A \cap B = \{1,3\} \).
Given \( B \), our universe shrinks to just \( \{1, 3, 5\} \). Within that restricted space, 2 of the 3 outcomes are also in \( A \):
\[ P(A \mid B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{3} \]
Deriving the formula. Divide numerator and denominator by \( n(S) \):
\[ P(A \mid B) = \frac{n(A \cap B)/n(S)}{n(B)/n(S)} = \frac{P(A \cap B)}{P(B)} \]In a group of children, the probability that a randomly selected child likes oranges is 0.6. The probability they like both oranges and apples is 0.3. If a child who likes oranges is selected, what is the probability they also like apples?
Let \( O \) = likes oranges, \( A \) = likes apples. Given \( P(O) = 0.6 \) and \( P(A \cap O) = 0.3 \):
\[ P(A \mid O) = \frac{P(A \cap O)}{P(O)} = \frac{0.3}{0.6} = 0.5 \]Conclusion: 50%.
A survey of 100 people who recently bought mobile phones or tablets:
| Mobile Phone | Tablet | Total | |
|---|---|---|---|
| Brand A | 20 | 10 | 30 |
| Brand B | 30 | 40 | 70 |
| Total | 50 | 50 | 100 |
If a person is selected at random, find:
Restricted-space shortcut: within the 70 brand-B buyers, 30 bought a mobile phone, giving \( 30/70 = 3/7 \). ✓
A single card is drawn from a standard 52-card deck. Find:
A dealership's vehicle inventory by type and color:
| SUV | Sport Car | Van | Coupe | Total | |
|---|---|---|---|---|---|
| Black | 35 | 10 | 25 | 15 | 85 |
| White | 10 | 15 | 20 | 5 | 50 |
| Blue | 15 | 15 | 5 | 30 | 65 |
| Total | 60 | 40 | 50 | 50 | 200 |
If a car is selected at random, find:
Shortcut: each answer is just the ratio of "row/column cell" to "row/column total" — the formula agrees with the table's marginal counts.
A survey of 100 people showed that 40% have both home and car insurance with a specific company. The probability of having car insurance is 0.7. What is the probability that a person has home insurance given they have car insurance?
Let \( H \) = home insurance, \( C \) = car insurance. Given \( P(C) = 0.7 \) and \( P(H \cap C) = 0.4 \):
\[ P(H \mid C) = \frac{P(H \cap C)}{P(C)} = \frac{0.4}{0.7} = \frac{4}{7} \approx 0.571 \]Conclusion: 57.1% of car-insurance holders also hold home insurance.
Out of 200 students surveyed, 120 play football, 50 play basketball, and 20 play both.
Let \( F \) = plays football, \( B \) = plays basketball. Then \( P(F) = 120/200 = 0.6 \), \( P(B) = 50/200 = 0.25 \), \( P(F \cap B) = 20/200 = 0.1 \).
Let \( O \) = exactly one sport. Then:
\[ n(O) = (120 - 20) + (50 - 20) = 100 + 30 = 130 \] \[ P(O) = \frac{130}{200} = 0.65 \]The intersection \( F \cap O \) is "football only" = 100 students:
\[ P(F \cap O) = \frac{100}{200} = 0.5 \] \[ P(F \mid O) = \frac{0.5}{0.65} = \frac{10}{13} \approx 0.769 \]
A fair die is rolled. Find the probability of rolling an even number given that the number rolled is greater than 3.
\( E = \{2,4,6\} \), \( G = \{4,5,6\} \), \( E \cap G = \{4,6\} \).
\[ P(E \mid G) = \frac{2/6}{3/6} = \frac{2}{3} \]A card is drawn from a standard 52-card deck. Find the probability of getting a 3 given that the card is red.
Out of 120 people: 90 own cars, 40 own bicycles, 10 own neither.
Let \( x \) = owns both. Then \( (90-x) + (40-x) + x + 10 = 120 \Rightarrow x = 20 \).
Company A produces 70 products (40 software, 30 hardware). Company B produces 50 products (x hardware, y software).
The probability the market goes down given the economy deteriorates is 0.8. The probability the economy deteriorates is 0.5. What is the probability the economy deteriorates and the market goes down?
\(P(A \mid B)\) is the probability that event \(A\) occurs given that \(B\) has already occurred. It is defined as \(P(A \mid B) = P(A \cap B)/P(B)\) whenever \(P(B) > 0\). Conceptually, it restricts the sample space from \(S\) down to \(B\).
Rearranging the definition gives \(P(A \cap B) = P(A \mid B) \cdot P(B)\). This computes the probability of both events occurring when the conditional probability is known.
Events \(A\) and \(B\) are independent iff \(P(A \mid B) = P(A)\), equivalently \(P(A \cap B) = P(A) \cdot P(B)\). Independence means knowing \(B\) gives no information about whether \(A\) occurs.
Bayes' Theorem is built from two applications of conditional probability: \(P(A \mid B) = P(A \cap B)/P(B)\) and \(P(B \mid A) = P(A \cap B)/P(A)\). Eliminating \(P(A \cap B)\) gives \(P(A \mid B) = P(B \mid A) \cdot P(A)/P(B)\), which inverts the direction of conditioning.
Last updated: October 20, 2026