Quantitative Aptitude · Probability

Multiplication Theorem and Independent Events (CA Foundation Probability)

Updated 1 October 2026 · Fact-checked

The multiplication theorem gives the probability that two events both happen: P(A ∩ B) = P(A) × P(B | A). If A and B are independent, this becomes P(A) × P(B). To solve a question, decide whether the events are independent, then multiply the right probabilities.

Understand Multiplication Theorem and Independent Events

The multiplication theorem answers the question "what is the chance that A and B both happen?" You find the chance of A first. Then you find the chance of B, given that A has already happened. You multiply the two.

The word "given" matters. If A happening changes the chance of B, the events are dependent. Drawing two cards from a pack without replacement is an example. After the first card is gone, the pack has changed.

If A happening does not change the chance of B, the events are independent. Tossing a coin twice is an example. The first toss does not affect the second. Then P(B | A) = P(B), and the rule becomes a simple product.

The test for independence is: A and B are independent if and only if P(A ∩ B) = P(A) × P(B). Compute both sides and compare. If they match, the events are independent.

Do not confuse independent with mutually exclusive. Mutually exclusive events cannot happen together, so P(A ∩ B) = 0. Independent events can happen together, and their joint probability is the product. If both events have non-zero probability, they cannot be both independent and mutually exclusive.

Key formulas to remember

General multiplication theorem: P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B); Works for any two events. Use it when events are dependent.
Multiplication rule for independent events: P(A ∩ B) = P(A) × P(B); Use only when A and B are independent.
Test for independence: A and B are independent ⇔ P(A ∩ B) = P(A) × P(B); Equivalent to P(B | A) = P(B) when P(A) is non-zero.
Three events (independent): P(A ∩ B ∩ C) = P(A) × P(B) × P(C); All events must be independent of each other.
At least one of independent events: P(at least one) = 1 − P(A′) × P(B′); Complements of independent events are also independent.
Mutually exclusive events: P(A ∩ B) = 0; If both P(A) and P(B) are above 0, such events are not independent.

How to solve Multiplication Theorem and Independent Events questions

Use this method for any multiplication-theorem or independence question.

1Write down the events clearly, for example A = first card is an ace, B = second card is a king.
2Look for the words "and", "both", "all" or "then". These signal multiplication.
3Decide if the events are independent. Check for replacement, separate trials or the phrase "independently". Without replacement means dependent.
4If independent, find each probability and multiply. If dependent, find P(A), then P(B | A) using the changed counts, and multiply.
5For "at least one" questions, find the probability that none happens and subtract from 1.
6To test independence from given values, compute P(A) × P(B) and compare it with P(A ∩ B).
7Match your answer with the options and check it lies between 0 and 1.

Quickest way: Fast route for MCQs

When to use it: Use this in the exam when the question has numbers and four options and time is short.

Spot the key word: "without replacement" means update the counts, "independent" means just multiply.
Write the product as a fraction and cancel before multiplying.
For "at least one", go straight to 1 − (product of failure probabilities).
Use elimination: a joint probability can never exceed the smaller single probability. Reject any option that does.
If an independence test is asked, multiply P(A) and P(B) once and compare with P(A ∩ B). Do not do anything else.

Common mistakes in Multiplication Theorem and Independent Events

Treating mutually exclusive events as independent.
Both ideas sound like "unrelated" events.
Fix: Mutually exclusive means they cannot occur together, so P(A ∩ B) = 0. Independent means P(A ∩ B) = P(A) × P(B).
Multiplying P(A) × P(B) when the draw is without replacement.
Students use the short formula by habit.
Fix: Reduce the total and the favourable count after the first draw, then use P(B | A).
Adding probabilities for "and" questions.
Confusion with the addition theorem used for "or".
Fix: "And" means multiply. "Or" means add, then subtract the overlap.
Using P(A) × P(B) = P(A ∩ B) for any events.
The rule is remembered without its condition.
Fix: This holds only for independent events. For others, use P(A) × P(B | A).
Computing "at least one" by adding the individual probabilities.
It looks like an "or" situation.
Fix: Use 1 − P(none). It is faster and avoids double counting.

Worked examples

Example 1

Two cards are drawn one after another, without replacement, from a well-shuffled pack of 52 cards. The probability that both are aces is: (a) 1/169 (b) 1/221 (c) 4/663 (d) 1/13

Show the solution

Let A = first card is an ace and B = second card is an ace.
P(A) = 4/52 = 1/13.
After one ace is removed, 3 aces remain among 51 cards, so P(B | A) = 3/51 = 1/17.
P(A ∩ B) = (1/13) × (1/17) = 1/221.

Answer: (b) 1/221

Example 2

A and B solve a problem independently. The chance that A solves it is 2/3 and that B solves it is 3/5. The probability that the problem is solved by at least one of them is: (a) 2/5 (b) 13/15 (c) 4/5 (d) 1/15

Show the solution

P(A fails) = 1 − 2/3 = 1/3.
P(B fails) = 1 − 3/5 = 2/5.
The events are independent, so P(both fail) = (1/3) × (2/5) = 2/15.
P(at least one solves) = 1 − 2/15 = 13/15.

Answer: (b) 13/15

Example 3

For two events A and B, P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Which statement is correct? (a) A and B are independent (b) A and B are mutually exclusive (c) P(B | A) = 0.5 (d) P(A ∪ B) = 0.9

Show the solution

Compute P(A) × P(B) = 0.5 × 0.4 = 0.20.
This equals P(A ∩ B) = 0.2, so A and B are independent. Option (a) is true.
Option (b) is false because P(A ∩ B) = 0.2, which is not 0.
P(B | A) = 0.2 ÷ 0.5 = 0.4, not 0.5, so (c) is false.
P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7, not 0.9, so (d) is false.

Answer: (a) A and B are independent

Exam tips

Read for "with replacement" or "without replacement" first. It decides the whole method.
For "at least one" and "none", the complement method saves time.
Questions often give P(A), P(B) and P(A ∩ B) and ask if the events are independent. Just compare the product.
Do not spend long on three-stage dependent draws. If the fractions become messy, mark the question and return later, because wrong answers cost 0.25 marks.
Remember that complements of independent events are also independent. It speeds up many problems.

Practice questions from Probability

Multiplication Theorem and Independent Events: frequently asked questions

What is the multiplication theorem of probability?

It says P(A ∩ B) = P(A) × P(B | A). It gives the chance that both A and B occur. For independent events it reduces to P(A) × P(B).

What is the difference between independent and mutually exclusive events?

Mutually exclusive events cannot happen together, so P(A ∩ B) = 0. Independent events do not affect each other, so P(A ∩ B) = P(A) × P(B). If both have non-zero probability, they cannot be both.

How do I check if two events are independent?

Find P(A) × P(B) and compare it with P(A ∩ B). If they are equal, the events are independent. You can also check whether P(B | A) equals P(B).

Does drawing without replacement give independent events?

No. Removing an item changes the counts for the next draw, so the events are dependent. Use P(A) × P(B | A) with the updated numbers.