🟣 Advanced Probability & Combinatorics

So far, we've calculated probabilities of events assuming all outcomes are from a general sample space. But what happens when we get new information that changes the possible outcomes? This is where conditional probability comes into play. It allows us to update probabilities based on prior knowledge or the occurrence of another event.

We will also explore the concept of independent events, which are events where the occurrence of one does not affect the probability of the other. Understanding independence is crucial for simplifying probability calculations in many situations.

Conditional Probability is the probability of an event occurring, given that another event has already occurred. It's about refining our understanding of probability when we have additional information that limits our sample space.

For example, consider drawing cards from a deck. The probability of drawing a King from a full deck is $\frac{4}{52}$. But if we are told that the drawn card is a face card (Jack, Queen, or King), then the probability of it being a King changes, because our sample space is now restricted to just face cards.

Notation: Conditional Probability of event $A$ given that event $B$ has occurred is denoted as $P(A|B)$ and is read as "the probability of $A$ given $B$."

The formula to calculate conditional probability is:

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

Where:

• $P(A|B)$ is the conditional probability of event $A$ occurring given that event $B$ has occurred
• $P(A \cap B)$ is the probability of both events $A$ and $B$ occurring (the probability of their intersection)
• $P(B)$ is the probability of event $B$ occurring (which must be greater than 0, as we are given that $B$ has occurred)

In essence, $P(A|B)$ is asking: out of all the times $B$ occurs, how often does $A$ also occur? We are restricting our focus to the sample space where $B$ has happened.

Example 1: Conditional Probability with Dice

Suppose you roll two fair six-sided dice. What is the probability that the sum is 7, given that the first die shows a 4?

Let $A$ be the event "the sum is 7", and $B$ be the event "the first die shows a 4". We want to find $P(A|B)$.

Solution:

• Event $B$: "The first die shows a 4". Possible outcomes for $B$ are $\{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\}$. So, $P(B) = \frac{6}{36} = \frac{1}{6}$.
• Event $A \cap B$: "The sum is 7 AND the first die shows a 4". The only outcome that satisfies both conditions is $\{(4, 3)\}$. So, $P(A \cap B) = \frac{1}{36}$.

Using the conditional probability formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

$$P(\text{Sum is 7 | First die is 4}) = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{36} \times \frac{6}{1} = \frac{1}{6}$$

Answer: The conditional probability is $\frac{1}{6}$. Notice that given the first die is 4, there are only 6 possible outcomes (the second die can be 1, 2, 3, 4, 5, or 6), and only one of these (when the second die is 3) makes the sum 7. Thus, it intuitively makes sense that the probability is $\frac{1}{6}$.

Example 2: Conditional Probability with Cards

What is the probability that a card drawn from a standard deck of 52 cards is a King, given that it is a face card (Jack, Queen, or King)?

Let $A$ be the event "the card is a King", and $B$ be the event "the card is a face card". We need to find $P(A|B)$.

Solution:

• Event $B$: "The card is a face card". There are 12 face cards in a deck (4 Jacks, 4 Queens, 4 Kings). So, $P(B) = \frac{12}{52}$.
• Event $A \cap B$: "The card is a King AND a face card". If a card is a King, it is automatically a face card. So, $A \cap B$ is simply event $A$, "the card is a King". There are 4 Kings in a deck. So, $P(A \cap B) = P(A) = \frac{4}{52}$.

Using the conditional probability formula:

$$P(\text{King | Face card}) = \frac{P(\text{King and Face card})}{P(\text{Face card})} = \frac{\frac{4}{52}}{\frac{12}{52}} = \frac{4}{12} = \frac{1}{3}$$

Answer: The conditional probability of drawing a King, given that it's a face card, is $\frac{1}{3}$. Initially, the probability of drawing a King from the full deck was $\frac{4}{52} = \frac{1}{13}$. Having the information that it is a face card significantly increased the probability of it being a King to $\frac{1}{3}$.

Example 3: Medical Testing

In a population, 1% of people have a certain disease. A test for this disease is 95% accurate (if you have the disease, the test is positive 95% of the time). What is the probability that a person has the disease given that they tested positive?

Solution:

Let $D$ = has disease, $T$ = tests positive

Given: $P(D) = 0.01$, $P(T|D) = 0.95$

We need: $P(D|T) = \frac{P(T|D) \cdot P(D)}{P(T)}$

Note: This introduces Bayes' Theorem, a powerful application of conditional probability!

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, knowing that one event has happened gives you no new information about whether the other event will happen.

Definition

Independent Events: Two events $A$ and $B$ are independent if and only if the probability of both events occurring is the product of their individual probabilities:

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

Equivalently, event $A$ is independent of event $B$ if $P(A|B) = P(A)$ (provided $P(B) > 0$), and event $B$ is independent of event $A$ if $P(B|A) = P(B)$ (provided $P(A) > 0$).

Example 4: Independent Events - Coin and Die

Tossing a coin and rolling a die are independent events. The result of the coin toss does not affect the result of the die roll, and vice versa.

If event $A$ = "coin shows heads" and event $B$ = "die shows 6":

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

Dependent events are events where the occurrence of one event affects the probability of the other event occurring. They are NOT independent.

For dependent events, we use the Multiplication Rule:

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

or equivalently:

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

Example 5: Dependent Events - Drawing Without Replacement

A bag contains 3 red balls and 2 blue balls. Two balls are drawn without replacement. What is the probability that both balls are red?

Solution:

Let $A$ = "first ball is red" and $B$ = "second ball is red"

• $P(A) = \frac{3}{5}$ (3 red out of 5 total)
• $P(B|A) = \frac{2}{4} = \frac{1}{2}$ (if first was red, 2 red remain out of 4 total)

Using the multiplication rule for dependent events:

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

Answer: The probability of drawing two red balls is $\frac{3}{10}$.

Example 6: Testing Independence

Given $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$. Are events $A$ and $B$ independent?

Solution:

Check if $P(A \cap B) = P(A) \times P(B)$:

$P(A) \times P(B) = 0.4 \times 0.5 = 0.2$

Since $P(A \cap B) = 0.2$ equals $P(A) \times P(B) = 0.2$, the events ARE independent.

1️⃣ Medical Diagnosis

Conditional probability is crucial in interpreting medical test results. The probability that you have a disease given a positive test result depends on the test's accuracy and the disease's prevalence in the population.

2️⃣ Weather Forecasting

Weather forecasts use conditional probability: "There's a 70% chance of rain given current atmospheric conditions."

3️⃣ Quality Control in Manufacturing

If a defective item is found, what's the probability it came from a specific machine? This uses conditional probability to identify problem sources.

4️⃣ Sports Analytics

What's the probability a basketball team wins given they're ahead at halftime? Independent vs. dependent events help analyze game strategies.

5️⃣ Financial Risk Assessment

Banks use conditional probability to assess loan default risk based on credit history and other factors.