Multiplication Law of Probability

The multiplication law of probability is used to determine the probability that two events will both occur.

Key Concepts

Events: An event is a specific outcome or a set of outcomes of a random experiment.
Independent Events: Two events are independent if the occurrence of one does not affect the occurrence of the other.
Dependent Events: Two events are dependent if the occurrence of one affects the occurrence of the other.

Multiplication Law for Independent Events:

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

This formula applies when events A and B are independent of each other.

Example:

Suppose event $A$ is rolling a 3 on a die $P(A) = \frac{1}{6}$ , and event $B$ is flipping a head on a coin $P(B) = \frac{1}{2}$ . Since these events are independent:

$P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$

Multiplication Law for Dependent Events:

If events A and B are dependent, the probability of both events occurring is given by:

$P(A \cap B) = P(A) \times P(B|A)$

Where $P(B|A)$ is the conditional probability of event B given that event A has occurred.

Example:

Suppose event $A$ is drawing an ace from a deck of cards $P(A) = \frac{4}{52}$ , and event $B$ is drawing a king from the remaining deck (without replacement). The probability of drawing a king after drawing an ace:

$P(B|A) = \frac{4}{51}$

Therefore:

$P(A \cap B) = P(A) \times P(B|A) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}$

Practice Question

Question 1. A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?

Solution:

Let A be the event that the first ball drawn is red, and B be the event that the second ball drawn is red.

$P(A) = \frac{5}{8}$

If the first ball drawn is red, there are now 4 red balls left out of 7 remaining balls.

$P(B|A) = \frac{4}{7}$

Therefore, the probability that both balls are red:

$P(A \cap B) = P(A) \times P(B|A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Question 2. A box contains 10 pens: 4 are blue, 3 are black, and 3 are red. Two pens are selected at random with replacement. What is the probability that both pens are black?
Solution:

Since the pens are selected with replacement, the events are independent.

$P(A) = \frac{3}{10}$

The probability of drawing a black pen on the second draw is the same:

$P(B) = \frac{3}{10}$

Therefore, the probability that both pens are black:

$P(A \cap B) = P(A) \times P(B) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}$