3  Probability

We saw that the probability of an event \(A\), denoted as \(P(A)\), is a number between 0 and 1, where:

3.1 Probability of a Conjunction of Two Events

The probability of the conjunction (intersection) of two independent events \(A\) and \(B\) can be calculated using the formula:

\[ \text{Probability} (A \text{ and } B) = \text{Probability} (A) \times \text{Probability} (B) \]

This formula assumes that events \(A\) and \(B\) are independent, meaning the occurrence of one event does not affect the occurrence of the other.

Example: six-sided die

If you want to find the probability of rolling a 3 on a fair six-sided die (Event \(A\)) and flipping heads on a fair coin (Event \(B\)), and both events are independent, you would first find the individual probabilities of each event:

  • Probability of rolling a 3 on a six-sided die: \(P(A) = \frac{1}{6}\) (1 favorable outcome out of 6 possible outcomes)
  • Probability of flipping heads on a fair coin: \(P(B) = \frac{1}{2}\) (1 favorable outcome out of 2 possible outcomes)

Then, you can calculate the probability of the conjunction of these events:

\[ \text{Probability} (A \text{ and } B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \]

So, the probability of rolling a 3 on a fair six-sided die and flipping heads on a fair coin is \(\frac{1}{12}\) or approximately 0.0833 when both events are independent.

3.2 Probability of a Union of Two Events

The probability of the union of two events \(A\) and \(B\) can be calculated using the formula:

\[ \text{Probability} (A \text{ or } B) = \text{Probability} (A) + \text{Probability} (B) - \text{Probability} (A \text{ and } B) \]

This formula calculates the probability of either event \(A\) or event \(B\) occurring, taking into account the possibility that both events could occur simultaneously. The term \(\text{Probability} (A \text{ and } B)\) represents the probability of the intersection (conjunction) of events \(A\) and \(B\).

If events \(A\) and \(B\) are mutually exclusive (they cannot occur at the same time), meaning \(\text{Probability} (A \text{ and } B) = 0\), the formula simplifies to:

\[ \text{Probability} (A \text{ or } B) = \text{Probability} (A) + \text{Probability} (B) \]

Example: six-sided die

For example, if you want to find the probability of rolling a 3 on a fair six-sided die (Event \(A\)) or flipping heads on a fair coin (Event \(B\)), and these events are mutually exclusive, and \(P(A) = \frac{1}{6}\) and \(P(B) = \frac{1}{2}\), the probability of the union of these events would be:

\[ \text{Probability} (A \text{ or } B) = \frac{1}{6} + \frac{1}{2} = \frac{4}{6} = \frac{2}{3} \]

So, the probability of rolling a 3 on a fair six-sided die or flipping heads on a fair coin (assuming they are mutually exclusive events) is \(\frac{2}{3}\) or approximately 0.6667.

3.3 Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted by \(P(A | B)\), which reads as “the probability of event \(A\) given event \(B\).” The formula for conditional probability is:

\[ P(A | B) = \frac{P(A \text{ and } B)}{P(B)} \]

where: - \(P(A | B)\) is the conditional probability of event \(A\) given event \(B\). - \(P(A \text{ and } B)\) is the joint probability of both events \(A\) and \(B\) occurring. - \(P(B)\) is the probability of event \(B\) occurring.

In words, the conditional probability of \(A\) given \(B\) is the ratio of the probability of both \(A\) and \(B\) occurring to the probability that \(B\) occurs.

3.3.1 Examples of Conditional Probability

Example 1: Medical Test Accuracy

Consider a medical test for a disease, where the test result can be positive (\(+\)) or negative (\(-\)). Let:

  • \(D\): Person has the disease.
  • \(T\): Test result is positive.

The conditional probability here is \(P(D | T)\), the probability that a person has the disease given that the test result is positive. The accuracy of the test can be represented as follows:

  • The probability of a true positive: \(P(T | D) = 0.95\) (the test correctly identifies 95% of those with the disease).
  • The probability of a false positive: \(P(T | \neg D) = 0.10\) (the test incorrectly indicates positive for 10% of those without the disease).

Using Bayes’ theorem, \(P(D | T)\) can be calculated considering both true positives and false positives, making it a key application of conditional probability in real-life scenarios.

Example 2: Weather Forecast

Consider two weather events:

  • \(S\): It will be sunny tomorrow.
  • \(R\): It will rain tomorrow.

Let’s say meteorologists have found that:

  • The probability of a sunny day given that it rained today is \(P(S | R) = 0.20\).
  • The probability of a rainy day given that it was sunny today is \(P(R | S) = 0.15\).

These probabilities represent how weather conditions are interrelated. For instance, \(P(S | R) = 0.20\) means that there is a 20% chance of a sunny day tomorrow if it rains today. These conditional probabilities are crucial for weather forecasting, helping meteorologists make predictions based on current weather conditions.

Example 2: Six-Sided Die

Consider the experiment of rolling a fair six-sided die. Let’s define two events:

  • \(A\): The outcome is an even number \(\{2, 4, 6\}\).
  • \(B\): The outcome is greater than 3 \(\{4, 5, 6\}\).

We want to find \(P(A | B)\), which represents the probability that the outcome is an even number given that it is greater than 3.

First, let’s find \(P(A \cap B)\), the probability that the outcome is both an even number and greater than 3. The outcomes in the intersection of \(A\) and \(B\) are \(\{4, 6\}\), so:

\[ P(A \cap B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3} \]

Next, let’s find \(P(B)\), the probability that the outcome is greater than 3. There are 3 outcomes in \(B\), so:

\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2} \]

Using the formula for conditional probability:

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} \]

So, the probability that the outcome is an even number given that it is greater than 3 (\(P(A | B)\)) is \(\frac{2}{3}\) or approximately \(0.6667\).

3.4 Independence in Probability

Two events, \(A\) and \(B\), are considered independent if the occurrence of one event does not affect the probability of the other event. In other words, \(A\) and \(B\) are independent if and only if:

\[ P(A \text{ and } B) = P(A) \times P(B) \]

In this case, the probability of both events \(A\) and \(B\) occurring is equal to the product of the probabilities of each event occurring individually.

Alternatively, events \(A\) and \(B\) are independent if either of the following conditions (equivalent statements) holds:

  1. \(P(A | B) = P(A)\) (the probability of \(A\) given \(B\) is the same as the probability of \(A\)).
  2. \(P(B | A) = P(B)\) (the probability of \(B\) given \(A\) is the same as the probability of \(B\)).

If events \(A\) and \(B\) are independent, knowing whether \(A\) occurs or not provides no information about the occurrence of \(B\), and vice versa.

Independence is a fundamental concept in probability theory and plays a crucial role in various applications, including statistics, engineering, and decision-making processes. It allows for simpler calculations and modeling of complex systems by assuming that certain events do not influence each other.

3.4.1 Examples of Independence in Probability

Example 1: Coin Toss and Dice Roll

Consider the events:

  • \(A\): Tossing a fair coin and getting heads.
  • \(B\): Rolling a fair six-sided die and getting a 4.

The outcome of the coin toss does not affect the outcome of the die roll, and vice versa. The probability of getting heads on the coin (\(P(A) = \frac{1}{2}\)) and the probability of rolling a 4 on the die (\(P(B) = \frac{1}{6}\)) are independent events. The joint probability \(P(A \text{ and } B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\).

Example 2: Drawing Cards

Consider a standard deck of 52 playing cards. Let:

  • \(C\): Drawing a spade from the deck.
  • \(D\): Drawing a face card (jack, queen, or king) from the deck.

The event of drawing a spade (\(P(C) = \frac{13}{52} = \frac{1}{4}\)) is independent of the event of drawing a face card (\(P(D) = \frac{12}{52} = \frac{3}{13}\)). The joint probability \(P(C \text{ and } D) = P(C) \times P(D) = \frac{1}{4} \times \frac{3}{13} = \frac{3}{52}\).

Example 3: Weather Events

Consider two weather events:

  • \(S\): It will be sunny tomorrow.
  • \(R\): It will rain tomorrow.

If the occurrence of rain tomorrow does not influence the probability of it being sunny tomorrow (and vice versa), events \(S\) and \(R\) are independent. For instance, if \(P(S) = 0.7\) (70% chance of sun) and \(P(R) = 0.3\) (30% chance of rain), these events are independent if \(P(S \text{ and } R) = P(S) \times P(R) = 0.7 \times 0.3 = 0.21\).