Multiplication Theorem on Probability 

In probability theory, the Multiplication Theorem plays a crucial role in determining the likelihood of two or more events happening together. It is especially useful when dealing with dependent or independent events. While the addition theorem calculates the probability of either event occurring, the multiplication theorem is used to find the probability of both events occurring. 

1.0What is the Multiplication Theorem on Probability?

The Multiplication Theorem on Probability states that the probability of the intersection of two events (i.e., both events occurring) is equal to the product of the probability of one event and the conditional probability of the second event given the first has occurred.

2.0Multiplication Theorem on Probability Formula

Let A and B be two events.

For any two events:

$P(A\cap B)=P(A)\cdot P(B\mid A)$

For independent events:

$P(A\cap B)=P(A)\cdot P(B)$

This is also called the Multiplication Rule of Probability.

3.0What is the Multiplication Principle of Probability?

The multiplication principle helps us find the probability of a sequence of events occurring. If multiple independent actions are performed, then the total number of outcomes is the product of the number of outcomes for each action.

It’s closely related to the Multiplication Function of Probability, which refers to evaluating probabilities over combinations of events.

4.0Multiplication Theorem on Probability Examples

Example 1: A coin is tossed and a die is rolled. What is the probability of getting a head and a 6?

Solution:

• $P(\text{ Head })=\frac{1}{2}$
• $P(6\text{ on die })=\frac{1}{6}$

Using multiplication theorem for independent events:

$P(\text{ Head }\cap 6)=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$

Example 2: Two cards are drawn without replacement from a standard deck. Find the probability that both are kings.

Solution:

• $P(1\text{ st king })=\frac{4}{52}$
• After one king is drawn, 3 kings left: $P(2\text{ nd king }\mid 1\text{ st })=\frac{3}{51}$
• $P(\text{ both kings })=\frac{4}{52}\cdot \frac{3}{51}=\frac{1}{221}$

Example 3: Given P(A) = 0.6, P(B|A) = 0.5, find P(A \cap B).

Solution:

$P(A\cap B)=P(A)\cdot P(B\mid A)=0.6\cdot 0.5=0.3$

Example 4: A coin is tossed and a die is rolled. What is the probability of getting a head and a number greater than 4?

Solution:
Let:

• Event A: Getting a head → $P(A)=\frac{1}{2}$
• Event B: Getting number > 4 → Possible outcomes: {5, 6} →$P(B)=\frac{2}{6}=\frac{1}{3}$

Since tossing a coin and rolling a die are independent:

$P(A\cap B)=P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}$

Answer: $\frac{1}{6}$

Example 5: Two cards are drawn without replacement from a standard deck. What is the probability that both are aces?

Solution:

• First card is an ace: $P(A)=\frac{4}{52}$
• Second ace: $P(B\mid A)=\frac{3}{51}$

Using the Multiplication Theorem (dependent events):

$P(A\cap B)=\frac{4}{52}\cdot \frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$

Answer: $\frac{1}{221}$

Example 6: Let P(A) = 0.5, and P(B|A) = 0.8. Find P(A \cap B).

Solution:

$P(A\cap B)=P(A)\cdot P(B\mid A)=0.5\cdot 0.8=0.4$

Answer: 0.4

Example 7: Two coins are tossed. What is the probability that both show heads?

Solution:
Let A = First coin shows head → $P(A)=\frac{1}{2}$

Let B = Second coin shows head → $P(B)=\frac{1}{2}$

$P(A\cap B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

Answer: $\frac{1}{4}$

Example 8: A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Find the probability that both are red.

Solution:

• First red:$P(A)=\frac{3}{5}$
• Second red: $P(B\mid A)=\frac{2}{4}=\frac{1}{2}$

$P(A\cap B)=\frac{3}{5}\cdot \frac{1}{2}=\frac{3}{10}$

Answer: $\frac{3}{10}$

Example 9: A bag contains 4 red balls and 6 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball was blue?

Solution:

Let:

• A: First ball is blue
• B: Second ball is red

We are asked to find P(B|A)

After first blue is drawn (6/10), remaining red = 4, total balls = 9.

So:

$P(B|A)=\frac{4}{9}$

Answer: $\frac{4}{9}$

Example 10: Two cards are drawn one after the other without replacement. What is the probability that at least one card is an ace?

Solution:

Let A = First card is an ace

B = Second card is an ace

We need:

$P(\text{ At least one ace })=1-P(\text{ No ace in both draws })$

Probability of not getting an ace in first draw: $\frac{48}{52}$, second draw: 

$\frac{47}{51}$

$P(\text{ no ace })=\frac{48}{52}\cdot \frac{47}{51}=\frac{2256}{2652}$

$P(\text{ at least one ace })=1-\frac{2256}{2652}=\frac{396}{2652}=\frac{33}{221}$ 

Answer: $\frac{33}{221}$

Example 11:  Three balls are drawn without replacement from a bag of 4 red and 2 green balls. What is the probability that all three balls are red?

Solution:

P(3 red)=46⋅35⋅24=24120=15

$P(3\mathrm{red})=\frac{4}{6}\cdot \frac{3}{5}\cdot \frac{2}{4}=\frac{24}{120}=\frac{1}{5}$

Answer: $\frac{1}{5}$

Example 12: In a class of 10 students, 4 are boys and 6 are girls. A committee of 2 is chosen randomly. What is the probability that the first is a boy given that at least one is a boy?

Solution:

Let:

• A: First person is a boy
• B: At least one of the two is a boy

We use:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

• $P(A\cap B)$ : First is boy, second can be anyone → $\frac{4}{10}\cdot \frac{9}{9}=\frac{4}{10}$
• $P(B)=1-P(\text{ both girls })$

$\begin{array}{rl}& P(B)=1-\frac{6}{10}\cdot \frac{5}{9}\\ & P(B)=1-\frac{1}{3}=\frac{2}{3}\end{array}$

$P(A\mid B)=\frac{4/10}{2/3}=\frac{4}{10}\cdot \frac{3}{2}=\frac{6}{10}=\frac{3}{5}$

Answer: $\frac{3}{5}$

Example 13: A box contains 6 white balls and 4 black balls. 3 balls are drawn one after another without replacement. Find the probability that exactly one black ball is drawn.

Solution:

Case 1: Black comes first → BWW:

$\frac{4}{10}\cdot \frac{6}{9}\cdot \frac{5}{8}=\frac{120}{720}$

Case 2: Black in second draw → WBW:

$\frac{6}{10}\cdot \frac{4}{9}\cdot \frac{5}{8}=\frac{120}{720}$

Case 3: Black in third draw → WWB:

$\frac{6}{10}\cdot \frac{5}{9}\cdot \frac{4}{8}=\frac{120}{720}$

Add all three cases:

$\frac{120+120+120}{720}=\frac{360}{720}=\frac{1}{2}$

Answer: $\frac{1}{2}$

5.0Addition and Multiplication Theorems on Probability with Examples

• Addition Theorem:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

• Multiplication Theorem:

$P(A\cap B)=P(A)\cdot P(B\mid A)$

Combined Example:

Let P(A) = 0.4, P(B) = 0.5, $P(A\cap B)=0.2$

Then:

$P(A\cup B)=0.4+0.5-0.2=0.7$

What is the multiplication theorem on probability with example?

Ans: It states $P(A\cap B)=P(A)\cdot P(B\mid A)$.

Example: Drawing two cards without replacement.

6.0Applications of the Multiplication Theorem in Real Life

• Medical Testing: Probability of having a disease and testing positive.
• Genetics: Inheritance of two traits occurring together.
• Finance: Risk calculation when combining two financial events.
• Game Theory: Finding compound probabilities in strategies.

7.0Practice Questions on Multiplication Rule of Probability

1. A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability both are red.
2. A coin is flipped twice. What is the probability of getting heads both times?
3. P(A) = 0.7, P(B|A) = 0.4. Find $P(A\cap B)$.
4. Two events A and B are independent. P(A) = 0.6, P(B) = 0.5. Find $P(A\cap B)$.

What’s the probability of drawing an ace and then a king from a deck without replacement?