Lesson 2.3: The Laws of Probability

Introduction

Welcome to Lesson 2.3 of AS-Level Statistics, where we will explore the laws of probability. Understanding probability is essential for quantifying uncertainty in various scenarios, and this lesson will help you build a solid foundation in the language and laws that govern probability.

Learning Objectives

By the end of this lesson, you will be able to:

Understand and apply the addition law for the union of two events, including the general form and the case of mutually exclusive events.
Understand and apply the multiplication law for the intersection of two events.
Recognize and use mutually exclusive events.
Use the addition law, including for mutually exclusive events, to find the probability of a union.
Utilize the multiplication law to find the probability of an intersection.

Hook

Have you ever asked yourself how likely it is to rain on your picnic day or if you can get a number you want in a game of dice? Probability helps you answer these questions by quantifying uncertainty. Let’s delve into how we can calculate these probabilities through well-defined laws.

The Addition Law of Probability

The addition law of probability allows us to calculate the probability of either of two events occurring. This is especially important when both events can occur at the same time, referred to as the union of events.

General Addition Law

The general addition law states that for two events $ A $ and $ B $, the probability of either event occurring can be expressed as:

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

This formula accounts for the overlap between events $ A $ and $ B $, ensuring we do not double-count the scenarios where both events take place.

Example 1: General Addition Law

Problem: Suppose the probability of event $ A $ (it rains today) is $ \mathbb{P}(A) = 0.4 $ and the probability of event $ B $ (it snows today) is $ \mathbb{P}(B) = 0.2 $. The probability that it rains and snows at the same time is $ \mathbb{P}(A \cap B) = 0.1 $. What is the probability that it either rains or snows?

Solution: Using the general addition law,

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

Substituting the values into the equation, we get:

$$\mathbb{P}(A \cup B) = 0.4 + 0.2 - 0.1 = 0.5$$

Thus, the probability that it either rains or snows today is $ 0.5 $.

Mutually Exclusive Events

Events $ A $ and $ B $ are said to be mutually exclusive if they cannot occur at the same time. In this scenario, the intersection probability $ \mathbb{P}(A \cap B) $ is 0. Therefore, the addition law for mutually exclusive events simplifies to:

$$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$$

Example 2: Mutually Exclusive Events

Problem: Let’s say you roll a single die. What is the probability of rolling a $ 1 $ or a $ 2 $? Since these events cannot occur simultaneously (you cannot roll both at once), they are mutually exclusive.

Solution: Here, we have:

$ \mathbb{P}(1) = \frac{1}{6} $
$ \mathbb{P}(2) = \frac{1}{6} $

Using the addition law for mutually exclusive events,

$$\mathbb{P}(1 \cup 2) = \mathbb{P}(1) + \mathbb{P}(2) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

Thus, the probability of rolling either a $ 1 $ or a $ 2 $ is $ \frac{1}{3} $.

The Multiplication Law of Probability

The multiplication law of probability describes the likelihood of both events occurring simultaneously, or the intersection of two events. For two events $ A $ and $ B $, this is given as:

$$\mathbb{P}(A \cap B) = \mathbb{P}(A) \times \mathbb{P}(B | A)$$

where $ \mathbb{P}(B | A) $ is the conditional probability that event $ B $ occurs given that event $ A $ has occurred.

Independent Events

Events $ A $ and $ B $ are independent if the occurrence of one does not affect the probability of the other occurring. In this case, the multiplication law simplifies to:

$$\mathbb{P}(A \cap B) = \mathbb{P}(A) \times \mathbb{P}(B)$$

Example 3: Independent Events

Problem: Consider two independent events: flipping a fair coin (event $ A $, where heads is $ H $) with probability $ \mathbb{P}(H) = \frac{1}{2} $ and rolling a die (event $ B $, where rolling a $ 4 $ is $ \mathbb{P}(4) = \frac{1}{6} $). What is the probability that both the coin shows heads and the die shows a $ 4 $?

Solution: Since both events are independent,

$$\mathbb{P}(H \cap 4) = \mathbb{P}(H) \times \mathbb{P}(4) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

Thus, the probability of getting heads on the coin and a $ 4 $ on the die is $ \frac{1}{12} $.

Conclusion

In this lesson, we have explored the laws of probability, focusing on the addition law for unions of events and the multiplication law for intersections. Understanding these concepts helps us quantify the likelihood of various outcomes in a variety of situations.

Key Takeaways

The addition law accounts for the probability of either event occurring and can be modified for mutually exclusive events.
The multiplication law determines the probability of both events occurring simultaneously, taking into account whether they are independent.
Mutual exclusivity simplifies calculations of the addition law.
Recognizing whether events are mutually exclusive or independent is vital to applying the appropriate laws correctly.

Study Notes

The addition law: $ \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B) $
For mutually exclusive events: $ \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) $
The multiplication law: $ \mathbb{P}(A \cap B) = \mathbb{P}(A) \times \mathbb{P}(B | A) $
For independent events: $ \mathbb{P}(A \cap B) = \mathbb{P}(A) \times \mathbb{P}(B) $
Always clarify whether events are mutual exclusive or independent before applying the laws.