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225 LearnersLast updated on November 27, 2025

Conditional probability is when an event occurs only if certain conditions are met. This concept can be used to determine the likelihood of real-life events occurring under specific conditions. For example, in weather forecasting, we can calculate the probability of rain. In this topic, students will learn how conditional probability can be applied easily.
Conditional probability is an essential idea in probability and statistics. It tells us the chance that event A occurs after we already know that event B has occurred. We can write this as P(A | B), P(A / B), or PB(A).
In simple words, P(A | B) means “the probability of A occurring given that B has happened.” Because B gives us new information, the probability of A may change.
This idea is often explained with a conditional probability example: knowing one condition affects the final answer.
Example:
A card is drawn from a standard deck of 52 cards. If it is known that the card is red, find the probability that the card is a king.
Solution:
Let:
A = event that the card drawn is a king
B = event that the card drawn is red
We want that: P(A∣B)
Step 1: First, we need to determine the total number of red cards (event B).
There are 26 red cards (13 hearts + 13 diamonds).
Step 2: Next, we find the number of red kings (A ∩ B).
There are two red kings (the King of Hearts and the King of Diamonds).
Step 3: Then apply the conditional probability formula:
P(A∣B) = \(\frac{\text{Number of red kings}}{\text{Total red cards}} = \frac{2}{26} = \frac{1}{13}\)
In the card example, we found that:
P(A∣B) = \(\frac{1}{13}\)
Here:
The 1 represents the one specific event that satisfies both A and B. The card drawn is a red king. The 13 represents the total number of possible outcomes in event B. There are 13 red cards in each suit, but we treat the total as 26, so the simplified probability becomes 113. This reasoning helps us to understand how the conditional probability equation is formed.
General Formula
Probability of A given B:
P(A∣B) = P(A∩B)P(B), (where P(B) = 0)
Probability of B given A:
P(B∣A) = P(A∩B)P(A), (where P(A) = 0).
These formulas are part of the Kolmogorov definition of conditional probability.
Here is the meaning of each term:
The three types of probability are conditional, joint, and marginal. Here, we will discuss a few key points that differentiate them:
| Features | Conditional Probability | Joint Probability | Marginal Probability |
| Meaning | The probability of event A happening, given that event B has been already occurred. Formula: P(A∣B) = P(A∩B)P(B) |
The probability that two events occur at the same time, A and B. Formula: P(A∩B) |
The probability that an event will occur, regardless of other events. |
| Formula | \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) | P(A∩B) | \(P(A) = \sum P(A \cap B)\) |
| Variables Involved | Two or more related events | Two or more events | A single event |


To find the conditional probability, we use the following steps:
1. Identify the events
2. Calculate the joint probability P(A ∩ B)
3. Determine the probability of the stated condition P(B).
4. Now, we use the formula: P(A|B) = P(A ∩ B) / P(B)
The probability of independent events calculates the probability of events whose occurrences do not depend on each other. Here, the conditional probability of two independent events, A and B, can be written as:
P(A|B) = P(A)
P(B|A) = P(B)
These events satisfy the multiplication rule: P(A ∩ B) = P(A) × P(B)
Mutually exclusive events are those that cannot occur simultaneously. For mutually exclusive events, the conditional probability will always be equal to zero.
P(A|B) = 0
P(B|A) = 0
Bayes’ theorem helps us to find the probability of an event when we have new information about another related event. It allows us to update or revise our earlier probabilities based on this new evidence. In simple terms, Bayes’ theorem tells us how likely a cause is, given that we have observed a particular result.
A tree diagram is used to understand the conditional probabilities. It visually shows each possible outcome step by step, helping us to see how the probabilities split and change at every stage.
Conditional probability has a few rules that help us understand how probabilities change when we already know that one event has occurred. These rules make it easier to calculate and compare the probabilities. Here are some properties that are explained in simple terms:
Property 1: Whole Sample Space Has Probability 1
If E and F are events in the sample space S, then:
P(S∣F) = P(F∣F) = 1
Meaning:
If we already know that F has happened, then we are completely sure, 100% certain, that the outcome lies in F. So the probability is 1.
Property 2: Conditional Probability of A or B
If A, B, and F are events, and P(F) ≠ 0, then:
P(A∪B∣F) = P(A∣F) + P(B∣F) − P(A∩B∣F)
Meaning:
When we want the probability of A or B happening under the condition F, we:
Add the conditional probabilities of A and B. Subtract the part where both A and B occur together (to avoid double-counting). This rule is the same as the usual addition rule, just applied under the condition F.
Property 3: Conditional Probability of Complements
The probability of A not happening, given B, is:
P(A′∣B) = 1 − P(A∣B)
Meaning:
Once B has happened, only two outcomes are possible:
A happens, or A does not occur.
Since these two cover all possibilities under B, their conditional probabilities must add up to 1. So we can easily find one by subtracting the other from 1.
Understanding conditional probability becomes much easier when you know a few smart strategies. These tips and tricks help you break problems into simpler parts, avoid common mistakes, and solve questions faster and more accurately.
Conditional probability is a significant concept in probability theory. Students can often make mistakes when determining conditional probability. These errors can be avoided with a proper understanding of the concept. Here are a few common mistakes along with some tricks to help you avoid them.
Conditional probability plays a vital role in the prediction of various real-life situations based on conditions. Whether predicting the weather or solving complex real-life problems, conditional probability has numerous uses. Let’s look at a few examples:
Aden has 60% chance of qualifying for an exam if they prepare and 30% chance if they don’t. If the probability of studying is 80%, what is the probability that Aden qualifies for the exam?
The probability of Aden qualifying for the exam is 54% or 0.54.
We have:
A = qualifying for the exam
B = Preparing for the exam
The given probabilities are:
Probability of qualifying for the exam if they prepare
P(A|B) = 0.6
Probability of qualifying even without preparation:
P(A|Bc) = 0.3
Probability of preparing for the exam:
P(B) = 0.8
Probability of not preparing for the exam:
\(P(Bc) = 1 – P(B) = 1 – 0.8 = 0.2\)
Here, we use the law of total probability:
\(P(A) = P(A \mid B)\,P(B) + P(A \mid B^{c})\,P(B^{c})\)
Now, substitute the values into the equation:
\(P(A) = (0.6 × 0.8) + (0.3 × 0.2)\)
\(= 0.48 + 0.06\)
\(= 0.54\)
Therefore, we get 0.54 as the probability of Aden qualifying for the exam, which is equal to 54%.
In a class of 80 pupils, 50 learn Mathematics, 30 learn English, and 15 learn both Mathematics and English. If a pupil is selected randomly and is found to be learning English, what is the probability that they also learn Mathematics?
The probability of a pupil learning both English and Math at the same time is 0.5 or 50%.
Here,
A = Pupil learning mathematics
B = Pupil learning English
Now, we determine P(A|B):
\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
We will now change the given values into probabilities:
\(P(A \cap B) = \frac{15}{80}\)
\(P(B) = \frac{30}{80}\)
Using the formula: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
= \(\frac{\frac{15}{80}}{\frac{30}{80}} = \frac{15}{30} = 0.5\)
Therefore, the probability of a pupil who learns English and Math at the same time is 0.5 or 50%.
On the turf, 70 children play badminton, and 30 play both badminton and basketball. If a child plays badminton, what is the probability that they also play basketball?
The probability of a child playing both badminton and basketball is 42.86% or 0.43.
We use the formula:
\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
Here, we have:
P(A ∩ B) = 30 (children playing both badminton and basketball)
P(B) = 70 (children playing badminton)
Substituting the values:
\(P(A \mid B) = \frac{30}{70} = 0.4286\)
Therefore, the probability of a child playing both badminton and basketball is 42.86% or 0.43.
A company manufactures 2,000 items daily. 600 come from Machine A, and 1400 from Machine B. Machine A makes 60 defective items, while Machine B makes 20 defective items. If an item is defective, what is the chance it came from Machine A?
The probability of a defective item from Machine A is 0.75 or 75%.
We use the formula:
\(P(A \mid D) = \frac{P(A \cap D)}{P(D)}\)
P(A) = Probability of an item from Machine A = \(\frac{600}{2000} = 0.3\)
P(B) = Probability of an item from Machine B = \(\frac{1400}{2000} = 0.7\)
P(D∣A) = Probability of a defective item from Machine A = \(\frac{60}{600} = 0.1\)
P(D∣B) = Probability of a defective item from Machine B = \(\frac{20}{1400} = 0.0143\)
Here, we use the total probability formula:
\(P(D) = P(D|A) P(A) + P(D|B) P(B)\)
\(P(D) = (0.1 × 0.3) + (0.0134 × 0.7)\)
\(= 0.03 + 0.01001\)
\(= 0.04001\)
\(≈ 0.04\)
Here, we apply Bayes’ theorem:
\(P(A \mid D) = \frac{P(D \mid A)\, P(A)}{P(D)}\)
\(= \frac{0.1 \times 0.3}{0.04} = \frac{0.03}{0.04} = 0.75\)
Therefore, the probability of a defective item from machine A is 0.75 or 75%.
A city records that 60% of youngsters drive above the speed limit. Among them, 25% are caught violating traffic rules. If a youngster is found violating traffic rules, what is the probability that they were also speeding?
0.75 is the probability.
Use the formula:
\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
A: The youngster was speeding
B: The youngster was caught
\(P(A) = 0.60\)
\(P(B|A) = 0.25\)
\(P(B) = 0.20\)
\(P (A ∩ B) = P(B|A) P(A) = 0.25 × 0.60 = 0.15\)
\(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)
\(= \frac{0.15}{0.20} = 0.75\)
Therefore, if a youngster violates the traffic rules, the probability that they were speeding is 75%.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






