Conditional Probability
In short: Conditional probability is the chance that one event happens when you already know another event has happened, found by restricting attention to the known group.
GeometryProbabilityS-CP.A.3
What is Conditional Probability?
Learning that a randomly picked student is in the band changes the odds that the student also plays a sport, and that updated chance is a conditional probability, written \(P(A|B)\) and read the probability of \(A\) given \(B\).
Knowing \(B\) shrinks the sample space to just the outcomes in \(B\). Then \(P(A|B) = \frac{P(A \text{ and } B)}{P(B)}\), or, with counts, the number in both groups divided by the number in the given group. It requires \(P(B) > 0\).
How to spot it
- A problem uses the phrase given that, if, among those who, or knowing that.
- A two-way table or Venn diagram is provided, and the question narrows to one row, column or circle.
- Cards or marbles are drawn one after another without replacement.
- A percentage is asked of a subgroup, such as the share of band members who also play a sport.
The method
- For any conditional probability, identify the condition \(B\), the group you already know you are in.
- Count or find the probability of the overlap, \(A \text{ and } B\).
- Divide the overlap by the size or probability of \(B\), not by the full total.
- Interpret the result in words, and compare it to \(P(A)\) to see whether knowing \(B\) changed the odds.
Conditional Probability: solved example
Problem
Of 150 students at a school in Ohio, 90 play a sport, 60 are in the band and 25 do both. A student is chosen at random and is known to be in the band. What is the probability that the student also plays a sport?
Solution
The condition is band, so the sample space shrinks to the 60 band members. Of those, 25 play a sport.
\(P(\text{sport}|\text{band}) = \frac{25}{60} = \frac{5}{12} \approx 0.417\). In comparison, \(P(\text{sport}) = \frac{90}{150} = 0.6\), so being in the band lowers the chance. The answer is \(\frac{5}{12}\).
Try it
Question 1
Two cards are drawn from a standard 52-card deck without replacement. What is the probability that the second is an ace, given that the first was an ace?
Show the answer
Three aces remain among 51 cards: \(\frac{3}{51} = \frac{1}{17}\).
Question 2
Using the school data above, find the probability that a student is in the band, given that the student plays a sport.
Show the answer
\(\frac{25}{90} = \frac{5}{18} \approx 0.278\).
Conditional Probability vs. Independent Events
Here the known event changes the odds, which is why the symbol \(P(A|B)\) differs from \(P(A)\). With independent events, learning the first outcome changes nothing, so \(P(A|B) = P(A)\). Comparing the two numbers is the test: 0.417 against 0.6 shows the sport and band groups are not independent.
Keep going
Lessons
Worksheets and quizzes
- Conditional Probability and Independence – Worksheet
- Conditional Probability and Independence – Quiz
- Conditional Probability and Independence – Worksheet
- Conditional Probability and Independence – Quiz
Related words
Whole dictionary: A–Z · Geometry · Probability