Multiplication Rule for Probability Formula | Solved Examples, FAQs, Important Questions

Class 11 Mathematics | Written by Neeraj Anand
Published by ANAND TECHNICAL PUBLISHERS
For CBSE Board Exams, JEE Mains & Advanced Preparation

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📘 Introduction

The Multiplication Rule of Probability is used to determine the probability of two or more events occurring simultaneously. It is especially useful when the events are either independent or dependent.

In probability theory, the Multiplication Rule helps in finding the likelihood of two or more events occurring together. This rule is applied differently depending on whether the events are independent or dependent.

• Independent Events: The occurrence of one event does not influence the occurrence of another.
• Dependent Events: The occurrence of one event affects the probability of another event.

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🔍 1. Multiplication Rule for Independent Events

If two events A and B are independent, the occurrence of one does not affect the occurrence of the other. The probability that both A and B occur is given by:

\begin{array}{l} \mathbf{P \left( A \cap B \right ) = P \left( A \right ) \times P \left( B \right )} \end{array}

Example: Suppose you flip a fair coin and roll a fair die.
If the probability of getting a head when flipping a coin is 1/2 and the probability of rolling a 4 on a die is 1/6​,

• Probability of getting a head on the coin:

\begin{array}{l} \mathbf{P \left( \text{Head} \right ) = \frac{1}{2}} \end{array}

• Probability of rolling a 4 on a die:

\begin{array}{l} \mathbf{P \left( 4 \right ) = \frac{1}{6}} \end{array}

Since both events are independent:

The probability of both happening together is :

\begin{array}{l} \mathbf{P \left( \text{Head and 4} \right ) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}} \end{array}

This rule can be extended to more than two events. For example, if A, B, and C are three independent events, then the probability of all three events occurring is:

P(A∩B∩C) = P(A) × P(B) × P(C)

The multiplication rule is based on the assumption that the events are independent, meaning that the occurrence of one event does not affect the occurrence of the other events. If the events are dependent, the multiplication rule does not apply directly, and conditional probability may be used instead.

🔍 Extended Multiplication Rule for Multiple Events

For multiple independent events A₁,A₂,A₃,…,A_n

\begin{array}{l} \mathbf{P \left( A_1 \cap A_2 \cap … \cap A_n \right ) = P \left( A_1 \right ) \times P \left( A_2 \right ) \times … \times P \left( A_n \right )} \end{array}

Example : What is the probability of getting 3 heads when tossing a fair coin 3 times?

Each flip is independent:

\begin{array}{l} \mathbf{P \left( \text{3 Heads} \right ) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}} \end{array}

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🔍 2. Multiplication Rule for Dependent Events

If two events A and B are dependent, the occurrence of event A affects the occurrence of event B. The probability that both A and B occur is:

\begin{array}{l} \mathbf{P \left( A \cap B \right ) = P \left( A \right ) \times P \left( B | A \right )} \end{array}

Here, P(B∣A) represents the probability of B occurring given that A has already occurred.

Example: Suppose you draw two cards from a deck of 52 cards without replacement.

Probability of drawing an Ace on the first draw:

\begin{array}{l} \mathbf{P \left( A_1 \right ) = \frac{4}{52} = \frac{1}{13}} \end{array}

If an Ace is already drawn, the probability of drawing another Ace:

\begin{array}{l} \mathbf{P \left( A_2 | A_1 \right ) = \frac{3}{51}} \end{array}

If a card is drawn from a deck and not replaced, the probability of drawing two aces consecutively is:

\begin{array}{l} \mathbf{P \left( A_1 \cap A_2 \right ) = \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}} \end{array}

Proof

We know that the conditional probability of event A given that B has occurred is denoted by P(A|B) and is given by:

\(\begin{array}{l}P(A|B) = \frac{P(A∩B)}{P(B)}\end{array} \)

Where, P(B)≠0

P(A∩B) = P(B)×P(A|B) ……………………………………..(1)

\(\begin{array}{l}P(B|A)~ = ~\frac{P(B∩A)}{P(A)}\end{array} \)

Where, P(A) ≠ 0.

P(B∩A) = P(A)×P(B|A)

Since, P(A∩B) = P(B∩A)

P(A∩B) = P(A)×P(B|A) ………………………………………(2)

From (1) and (2), we get:

P(A∩B) = P(B)×P(A|B) = P(A)×P(B|A) where,

P(A) ≠ 0,P(B) ≠ 0.

The above result is known as the multiplication rule of probability.

For independent events A and B, P(B|A) = P(B). The equation (2) can be modified into,

P(A∩B) = P(B) × P(A)

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Probability Multiplication Rule for n Events

Here, to get the probability multiplication rule for n events, we must extend the probability multiplication theorem to n events for n events.A1, A2, … , An, we get

P(A₁ ∩ A₂ ∩ … ∩ A_n) = P(A₁) P(A₂ | A₁) P(A₃ | A₁ ∩ A₂) … × P(A_n |A₁ ∩ A₂ ∩ … ∩ A_n-1)

Now for the case of n independent events, the multiplication theorem get reduced to P(A₁ ∩ A₂ ∩ … ∩ A_n) = P(A₁) P(A₂) … P(A_n).

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Multiplication Theorem of Probability

We have already learned the multiplication rules we follow in probability, such as;

P(A∩B) = P(A)×P(B|A) ; if P(A) ≠ 0

P(A∩B) = P(B)×P(A|B) ; if P(B) ≠ 0

Let us learn here the multiplication theorems for independent events A and B.

If A and B are two independent events for a random experiment, then the probability of simultaneous occurrence of two independent events will be equal to the product of their probabilities. Hence,

P(A∩B) = P(A).P(B)

Now, from multiplication rule we know;

P(A∩B) = P(A)×P(B|A)

Since A and B are independent, therefore;

P(B|A) = P(B)

Therefore, again we get;

P(A∩B) = P(A).P(B)

Hence, proved.

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Solved Example of Multiplication Rule of Probability

Example 1: An urn contains 20 red and 10 blue balls. Two balls are drawn from a bag one after the other without replacement. What is the probability that both the balls are drawn are red?

Solution: Let A and B denote the events that the first and the second balls are drawn are red balls. We have to find P(A∩B) or P(AB).

P(A) = P(red balls in first draw) = 20/30

Now, only 19 red balls and 10 blue balls are left in the bag. The probability of drawing a red ball in the second draw too is an example of conditional probability where the drawing of the second ball depends on the drawing of the first ball.

Hence Conditional probability of B on A will be,

P(B|A) = 19/29

By multiplication rule of probability,

P(A∩B) = P(A) × P(B|A)

\(\begin{array}{l}P(A∩B)~ =~ \frac{20}{30} ~× ~\frac{19}{29} ~=~ \frac{38}{87}\end{array} \)

Addition Rule of Probability

The addition rule states the probability of two events is the sum of the probabilities of two events that will happen minus the probability of both the events that will happen.

Mathematically, the addition rule of probability is expressed as:

P(A ∪ B) = P (A) + P(B) – P(A ∩ B)

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🎯 Key Points to Remember

• Use the independent events formula when events do not affect each other.
• Use the dependent events formula when the outcome of one event affects the other.
• Always reduce fractions to their simplest form.

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