🌟 Level 3 - Topic 9: Advanced Probability & Combinatorics (Part 3) 🎲🔗

1) Introduction to Conditional Probability and Independent Events 🎲🔗

Welcome back to the exciting world of Probability and Combinatorics! In Part 2 of Level 2, we expanded our counting skills with permutations and combinations. Now, in Level 3 - Part 3, we are ready to tackle more advanced probability concepts. We'll focus on Conditional Probability and Independent Events.

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.

In this topic, we will cover:

• Understanding and calculating Conditional Probability: probability of an event given that another event has occurred.
• Defining and identifying Independent Events and Dependent Events.
• Using conditional probability to solve more complex probability problems.
• Exploring real-world examples and applications of conditional probability and independence.

Let's unlock the power of conditional probability and understand how events relate to each other in the world of chance! 🚀

2) Conditional Probability: Probability Given an Event 🎲

2.1 What is Conditional Probability?

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.

2.2 The Formula for Conditional Probability

The formula to calculate conditional probability is:

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

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

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

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

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 \( \frac{1}{3} \).

3) Independent and Dependent Events 🔗

3.1 Independent Events

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.

Example 3: Independent Events - Coin Toss and Die Roll

Consider tossing a fair coin and rolling a fair six-sided die. Let \(A\) be the event "getting Heads on the coin" and \(B\) be the event "rolling a 3 on the die". Are \(A\) and \(B\) independent events?

**Solution:**

• Probability of getting Heads: \(P(A) = \frac{1}{2}\).
• Probability of rolling a 3: \(P(B) = \frac{1}{6}\).
• Probability of both events occurring: \(P(A \cap B)\) is the probability of getting Heads AND rolling a 3. Since coin toss and die roll are independent actions, we can multiply their probabilities: \(P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\).

Since \(P(A \cap B) = P(A) \times P(B)\), events \(A\) and \(B\) are independent. The outcome of the coin toss does not affect the outcome of the die roll, and vice versa.

3.2 Dependent Events

If two events are not independent, they are said to be dependent. For dependent events, the occurrence of one event *does* affect the probability of the other event. Conditional probability is particularly important when dealing with dependent events.

If \(A\) and \(B\) are dependent events, then \( P(A \cap B) \neq P(A) \times P(B) \), and \( P(A|B) \neq P(A) \). In fact, for dependent events, the definition of conditional probability, \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), is used to find \(P(A \cap B)\) as:

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

This is known as the Multiplication Rule for Dependent Events.

Example 4: Dependent Events - Cards without Replacement

Consider drawing two cards without replacement from a standard deck. Let \(C\) be the event "the first card is a King" and \(D\) be the event "the second card is a Queen". Are \(C\) and \(D\) independent or dependent events?

**Solution:**

• Probability of drawing a King first: \(P(C) = \frac{4}{52}\).
• Probability of drawing a Queen second, given a King was drawn first: \(P(D|C) = \frac{4}{51}\) (since there are still 4 Queens but only 51 cards left).

To find the probability of both events happening, we use the multiplication rule for dependent events:

\( P(C \cap D) = P(C) \times P(D|C) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663} \)

If events \(C\) and \(D\) were independent, we would expect \(P(C \cap D) = P(C) \times P(D)\). Let's check \(P(D)\) - the probability of drawing a Queen as the second card *without* knowing what the first card was. It turns out \(P(D) = \frac{4}{52}\) as well (intuitively, the probability of drawing a Queen at any position is the same in a random draw). But \(P(D|C) = \frac{4}{51} \neq P(D)\). Thus, events \(C\) and \(D\) are dependent. Drawing a King first changes the probability of drawing a Queen second.

4) Examples (Detailed) 🍀

Example 5: Probability with Conditional Situations

Let’s look at more examples involving conditional probability and independence/dependence.

1. Challenge 1: In a certain city, 60% of people own a car, 30% own a bike, and 20% own both a car and a bike. What is the probability that a person chosen at random owns a bike given that they own a car?

2. Challenge 2: A bag contains 5 red balls and 3 blue balls. Two balls are drawn in succession without replacement. What is the probability that the second ball is blue, given that the first ball was red?

**Let's solve each challenge:**

1. Solution to Challenge 1: Car and Bike Ownership

   Let \(C\) be the event "owns a car" and \(B\) be the event "owns a bike". We are given:

   • \(P(C) = 0.60\)
   • \(P(B) = 0.30\)
   • \(P(C \cap B) = 0.20\) (owns both)

   We need to find the probability that a person owns a bike given that they own a car, i.e., \(P(B|C)\).

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

   \( P(B|C) = \frac{0.20}{0.60} = \frac{20}{60} = \frac{1}{3} \)

   So, the probability that a person owns a bike given they own a car is \( \frac{1}{3} \) or approximately 33.33%.

2. Solution to Challenge 2: Drawing Balls without Replacement

   Let \(R_1\) be the event "the first ball drawn is red" and \(B_2\) be the event "the second ball drawn is blue". We want to find \(P(B_2|R_1)\).

   Initially, there are 5 red balls and 3 blue balls, total 8 balls.

   If the first ball drawn is red (event \(R_1\)), then before the second draw, there are now 7 balls left in the bag, of which 4 are red and still 3 are blue.

   Given that the first ball was red, the probability of drawing a blue ball as the second ball from the remaining balls is:

   \( P(B_2|R_1) = \frac{\text{Number of blue balls remaining}}{\text{Total number of balls remaining}} = \frac{3}{7} \)

   The probability that the second ball is blue, given that the first ball was red, is \( \frac{3}{7} \).

5) Practice Questions 🎯

6) Summary 🎉

• Conditional Probability \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) is the probability of event \(A\) given that event \(B\) has occurred.
• Independent Events \(A\) and \(B\) satisfy \(P(A \cap B) = P(A) \times P(B)\) or \(P(A|B) = P(A)\). The occurrence of one does not affect the other.
• Dependent Events are events that are not independent; the occurrence of one event changes the probability of the other.
• Multiplication Rule for Dependent Events: \( P(A \cap B) = P(A|B) \times P(B) \).
• Conditional probability helps refine probability estimates based on new information, crucial in many applications.

Excellent work! You've now expanded your understanding of probability to include conditional probability and the concept of independent events. These are crucial tools for tackling more complex probabilistic problems and real-world scenarios. As you continue to practice, you'll become more confident in analyzing and solving problems involving chance and dependence. Get ready to further advance in probability and combinatorics in Part 4! 🌟