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Fundamentals of Business Mathematics and Statistics · Probability

Multiplication Theorem and Conditional Probability Explained

Updated 10 October 2026 · Fact-checked

Conditional probability P(A | B) is the chance of A happening when you already know B has happened: P(A ∩ B) ÷ P(B), for P(B) > 0. The multiplication theorem gives P(A ∩ B) = P(A) × P(B | A). If events are independent, this becomes P(A) × P(B).

Understand Multiplication Theorem and Conditional Probability

Probability often changes when you get new information. Suppose a bag has 3 red and 2 blue balls. The chance the first ball is red is 3/5. If you draw a red ball and do not put it back, the bag now has 2 red and 2 blue. The chance the second ball is red is now 2/4. The second probability depends on the first result.

Conditional probability captures this. P(A | B) means the probability of A given that B has already occurred. You shrink your sample space to only the outcomes where B is true, then ask how many of those also have A. That is why you divide by P(B).

The multiplication theorem answers 'what is the chance that both A and B happen?'. Rearrange the conditional formula and you get P(A ∩ B) = P(A) × P(B | A). Read it as: first A happens, then B happens given A.

Two events are independent if one happening does not change the chance of the other. Typical cases are repeated coin tosses, or draws made with replacement. Then P(B | A) = P(B), and the rule simplifies to P(A ∩ B) = P(A) × P(B). Events that influence each other, such as draws without replacement, are dependent, and you must use the conditional probability.

Do not confuse independent with mutually exclusive. Mutually exclusive events cannot happen together, so P(A ∩ B) = 0. Two such events with non-zero probabilities are never independent.

Key formulas to remember

Conditional probability
P(A | B) = P(A ∩ B) ÷ P(B)
Valid only when P(B) > 0. Divide by the probability of the given event.
Multiplication theorem (general)
P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)
Works for any two events, dependent or independent.
Multiplication rule for independent events
P(A ∩ B) = P(A) × P(B)
Use only when the events are independent.
Test for independence
A and B are independent if P(A ∩ B) = P(A) × P(B), equivalently P(A | B) = P(A)
Check this when the question asks whether events are independent.
Three events (dependent)
P(A ∩ B ∩ C) = P(A) × P(B | A) × P(C | A ∩ B)
For independent events, multiply the three plain probabilities.
Complement of a conditional event
P(A' | B) = 1 − P(A | B)
The condition B stays the same on both sides.

How to solve Multiplication Theorem and Conditional Probability questions

Use this routine for any question on the multiplication theorem or conditional probability.

  1. 1Name the events clearly, for example A = first ball is red, B = second ball is red.
  2. 2Decide what the question wants: a joint probability P(A ∩ B), or a conditional probability P(A | B).
  3. 3Decide whether the events are independent or dependent. Look for 'with replacement', 'without replacement', 'given that' and 'separately'.
  4. 4For a joint probability, multiply: P(A) × P(B | A). If independent, use P(A) × P(B).
  5. 5For a conditional probability, write P(A ∩ B) ÷ P(B). Make sure the denominator is the given event.
  6. 6Update counts after each draw if there is no replacement. Reduce both favourable items and total items as needed.
  7. 7Simplify the fraction and check that the answer lies between 0 and 1.
  8. 8Match the value to the option. Eliminate options greater than 1 or clearly inconsistent.

Quickest way: Shrink the sample space

When to use it: Use when a question gives a table or counts and asks for 'probability of A given B'.

  1. Ignore everything outside the given event B.
  2. Count the total outcomes in B. This is your new denominator.
  3. Count how many of those outcomes also have A. This is your numerator.
  4. Divide: P(A | B) = (count in A and B) ÷ (count in B).
  5. For draws without replacement, just multiply fractions with reduced counts, for example 3/5 × 2/4.

Common mistakes in Multiplication Theorem and Conditional Probability

  • Using P(A) × P(B) when events are dependent, such as draws without replacement.

    Students memorise the simple product rule and apply it everywhere.

    Fix: Ask whether the first result changes the second. If yes, use P(A) × P(B | A) and update the counts.

  • Dividing by the wrong probability in P(A | B), using P(A) instead of P(B).

    The notation is read left to right and the first event feels like the base.

    Fix: The event after the bar is the given one. Always divide by its probability.

  • Confusing independent events with mutually exclusive events.

    Both ideas sound like 'unrelated' events.

    Fix: Mutually exclusive means P(A ∩ B) = 0. Independent means P(A ∩ B) = P(A) × P(B). Non-zero events cannot be both.

  • Not changing the total after a draw without replacement.

    Students rush and reuse the first denominator.

    Fix: After each draw without replacement, reduce the total by 1 and reduce the favourable count if the item drawn was favourable.

  • Mixing up P(A | B) with P(B | A).

    The two look similar and are often different in value.

    Fix: Rewrite the sentence in words: 'given that B has happened, find A'. Then put B in the denominator.

Worked examples

Example 1

A box has 5 red and 3 green pens. Two pens are drawn one after another without replacement. Find the probability that both are red.

Show the solution
  1. Let A = first pen is red and B = second pen is red.
  2. P(A) = 5/8.
  3. After one red pen is removed, 4 red pens remain out of 7 pens, so P(B | A) = 4/7.
  4. P(A ∩ B) = P(A) × P(B | A) = 5/8 × 4/7 = 20/56 = 5/14.

Answer: 5/14

Example 2

In a class, the probability that a student passes Mathematics is 0.6 and the probability that a student passes both Mathematics and Statistics is 0.45. Find the probability that a student passes Statistics given that the student passed Mathematics.

Show the solution
  1. Let M = passes Mathematics and S = passes Statistics.
  2. Given P(M) = 0.6 and P(M ∩ S) = 0.45.
  3. P(S | M) = P(M ∩ S) ÷ P(M).
  4. P(S | M) = 0.45 ÷ 0.6 = 0.75.

Answer: 0.75

Exam tips

  • Look for trigger words: 'given that', 'if it is known that' and 'provided that' signal conditional probability.
  • Check for 'with replacement' or 'without replacement' before choosing between the independent and dependent rule.
  • With no negative marking, always attempt every question. Eliminate any option above 1 and then guess among the rest.
  • In table-based questions, use the shrink-the-sample-space method. It is faster than the formula.
  • If a question asks whether two events are independent, test whether P(A ∩ B) equals P(A) × P(B).

Practice questions from Probability

Multiplication Theorem and Conditional Probability in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Multiplication Theorem and Conditional Probability: frequently asked questions

What is the multiplication theorem of probability?

It says the probability that both A and B occur is P(A ∩ B) = P(A) × P(B | A). If A and B are independent, it reduces to P(A) × P(B). It extends to three or more events by chaining conditional probabilities.

How do I find conditional probability quickly?

Divide the probability of both events by the probability of the given event: P(A | B) = P(A ∩ B) ÷ P(B). With counts, treat B as your whole sample space and count how many outcomes in it also have A.

What is the difference between independent and dependent events?

Independent events do not affect each other, so P(B | A) = P(B). Dependent events do affect each other, so you must use P(B | A) in the product. Draws without replacement are the usual dependent case.

Can mutually exclusive events be independent?

Not when both have non-zero probability. Mutually exclusive events have P(A ∩ B) = 0, while independence needs P(A ∩ B) = P(A) × P(B), which is greater than 0 in that case.