Lecture #4 (Class Notes + Video ...)
Class Notes:
Conditional Probability Law
What is it? Conditional probability tells us the chance of an event happening given that another event has already happened.
We write it as P(A\,||\,B), which means "the probability of event A happening, given that event B has already happened."
The formula is: P(A\,||\,B) = \frac{P(A \text{ and } B)}{P(B)}.
This means: (Probability of both A and B happening) divided by (Probability of B happening).
Think of it like this: When you know event B has happened, your entire possible world (sample space) shrinks from all outcomes (\Omega) to just the outcomes where B happened.
So, we only consider the outcomes within B, and event A inside this new world becomes "A and B".
We then calculate the probability of "A and B" relative to the size of "B".
The rules still apply: Conditional probabilities still follow the basic rules of probability:
It's never negative: The chance of an event happening can't be less than zero. P(A||B) \ge 0.
The whole space is certain: If B has happened, then the probability of any outcome within B is 100%. P(\text{all outcomes}||B) = 1.
Adding probabilities for separate events: If two events, A1 and A2, can't happen at the same time (they are "disjoint"), then the probability of either A1 or A2 happening, given B, is simply the sum of their individual conditional probabilities.
P(A1||B) + P(A2||B) = P((A1 \text{ or } A2) || B).
Multiplication Rule
What does it do? This rule helps us find the probability that multiple events all happen one after another.
The formula for events A1, A2, \ldots, An all happening is: P(A1 \text{ and } A2 \text{ and } \ldots \text{ and } An) = P(A1) \cdot P(A2 \mid A1) \cdot P(A3 \mid A1 \text{ and } A2) \cdot \ldots \cdot P(An \mid A1 \text{ and } \ldots \text{ and } A_{n-1}).
In simpler words: It's the probability of the first event, multiplied by the probability of the second event given the first one happened, multiplied by the probability of the third event given the first two happened, and so on.
How it works (A quick look at why):
You can break down the probability of multiple events happening step-by-step using conditional probability.
For example, P(A1 \text{ and } A2) = P(A1) \cdot P(A2 \mid A_1).
Then, P(A1 \text{ and } A2 \text{ and } A3) = P(A1 \text{ and } A2) \cdot P(A3 \mid A1 \text{ and } A2).
If you keep substituting like this, you get the long multiplication chain shown in the rule.
Example: Drawing 3 cards (without putting them back) and none are hearts.
Let's say:
A_1 = "The first card is not a heart."
A_2 = "The second card is not a heart."
A_3 = "The third card is not a heart."
Step 1: Probability the first card isn't a heart.
There are 39 non-heart cards out of 52 total. So, P(A_1) = \frac{39}{52}.
Step 2: Probability the second card isn't a heart, given the first wasn't a heart.
Now there are 51 cards left, and 38 of them are not hearts. So, P(A2 \mid A1) = \frac{38}{51}.
Step 3: Probability the third card isn't a heart, given the first two weren't hearts.
Now there are 50 cards left, and 37 of them are not hearts. So, P(A3 \mid A1 \text{ and } A_2) = \frac{37}{50}.
Final Result: To get the probability that none of the three cards are hearts, we multiply these probabilities:
P(A1 \text{ and } A2 \text{ and } A_3) = \frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50}.
Examples Using the Multiplication Rule
Example 1: Drawing 3 cards (no hearts)
This is the same as the example in the "Multiplication Rule" section.
We want the probability that the 1st card isn't a heart, AND the 2nd isn't a heart, AND the 3rd isn't a heart.
The calculation is:
P(\text{1st not heart}) = \frac{39}{52}
P(\text{2nd not heart} \mid \text{1st not heart}) = \frac{38}{51}
P(\text{3rd not heart} \mid \text{1st and 2nd not heart}) = \frac{37}{50}
Total Probability: P(\text{none are hearts}) = \frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50}.
Example 2: Grouping Graduates and Undergraduates
Imagine you have 4 graduates and 12 undergraduates (total 16 people). You divide them into 4 groups of 4 people each.
Question: What's the chance that each of the 4 groups ends up with exactly one graduate?
Let's call the graduates G1, G2, G3, G4.
We look at this step-by-step:
Place the first graduate (G1): It doesn't matter where G1 goes.
Place the second graduate (G2): For G2 to be in a different group from G1, there are 15 spots left in total, and 12 of them lead to a different group (since G1 took up one spot in their group, leaving 3 spots in that group, and 12 spots in the other 3 groups). So, P(G2 \text{ in different group from } G1) = \frac{12}{15}.
Place the third graduate (G3): Now, G1 and G2 are in different groups. There are 14 spots remaining. For G3 to be in a different group from both G1 and G2, there are 8 valid spots left (since G1's group has 3 spots left, G2's group has 3 spots left, leaving 14 - 3 - 3 = 8 spots in the remaining two empty groups). So, P(G3 \text{ in different group from } G1 \text{ and } G_2) = \frac{8}{14}.
Place the fourth graduate (G4): G1, G2, G3 are all in different groups. There are 13 spots left. For G4 to be in a different group from all three, it must go into the last remaining group that doesn't have a graduate yet. There are 4 valid spots in that last group. So, P(G4 \text{ in different group from } G1, G2 \text{ and } G_3) = \frac{4}{13}.
Total Probability: Multiply these chances together:
P(\text{each group has a graduate}) = \frac{12}{15} \cdot \frac{8}{14} \cdot \frac{4}{13}.
Example 3: The Prize Ticket Draw
You have 5 tickets, and only one is a prize winner. 5 people take turns drawing one ticket each, without putting them back.
Question: Is this game fair? Does the person who draws first have a better chance to win than someone who draws later?
Challenge for you: Try to use the conditional probability and multiplication rule ideas to figure out the probability of winning for each position (1st person, 2nd person, etc.). You'll find it's surprisingly fair!
________________________________________________________
Video Notes:
Understanding Probability Rules Easily
1. The "OR" Rule (Addition Rule)
This rule helps you find the chance of event A OR event B happening. We write it as P(A \cup B).
The basic idea: Add the chances of A and B, then subtract the chance of both happening at the same time so you don't count it twice.
The formula: P(A \cup B) = P(A) + P(B) - P(A \cap B)
Key thing to remember: The "- P(A \cap B) part is important! It stops you from double-counting the outcomes that are in both A and B.
What if events can't happen at the same time? (Mutually Exclusive)
If A and B cannot happen together (like flipping a coin and getting both heads AND tails), they are mutually exclusive.
In this special case, the chance of both happening (the intersection P(A \cap B)) is 0.
So, the rule becomes simpler: P(A \cup B) = P(A) + P(B). No need to subtract anything if there's no overlap!
2. The "AND" Rule (Multiplication Rule & Conditional Probability)
This rule helps you find the chance of event A AND event B happening together. We write it as P(A \cap B).
What is Conditional Probability? (Probability Given Something Else)
P(A|B) means "the probability of A happening, given that B has already happened."
Think of it as: "If we already know B happened, what's the chance of A?"
The formula for conditional probability: P(A|B) = \frac{P(A \cap B)}{P(B)}
How to find P(A and B) using this: You can rearrange the formula to find P(A \cap B).
If you know P(A|B) and P(B): P(A \cap B) = P(A|B) \cdot P(B)
Or, if you know P(B|A) and P(A): P(A \cap B) = P(B|A) \cdot P(A)
3. Are Events Connected? (Independent vs. Dependent)
Independent Events: If what happens in event B does not change the probability of event A, then A and B are independent.
This means P(A|B) = P(A). (The chance of A is the same, even if B happened).
If two events are independent, the "AND" rule is very simple: P(A \cap B) = P(A) \cdot P(B)
Dependent Events: If what happens in event B does change the probability of event A, then they are dependent.
In this case, P(A|B) \neq P(A).
You must use the conditional form: P(A \cap B) = P(A|B) \cdot P(B)
4. Order Doesn't Matter for "AND"
When we talk about A and B happening "at the same time" (their intersection), the order doesn't change the outcome.
So, the probability of A and B is the same as B and A: P(A \cap B) = P(B \cap A)
Practical Example: Sarah's Courses (Let's walk through it!)
We know: P(A) = 0.3 (chance of algebra), P(B) = 0.7 (chance of biology), and P(A|B) = 0.4 (chance of algebra GIVEN biology).
Part A: Chance of BOTH Algebra AND Biology (P(A \cap B))
Use the conditional form: P(A \cap B) = P(A|B) \cdot P(B) = 0.4 \times 0.7 = 0.28.
Answer: There's a 28% chance Sarah enrolls in both.
Part B: Chance of Algebra OR Biology (P(A \cup B))
Use the addition rule: P(A \cup B) = P(A) + P(B) - P(A \cap B)
We already found P(A \cap B) = 0.28
So: 0.3 + 0.7 - 0.28 = 1.0 - 0.28 = 0.72.
Answer: There's a 72% chance Sarah enrolls in at least one of the two.
Part C: Are Algebra and Biology Enrollment Independent?
To check, compare P(A|B) with P(A) (Does knowing she took biology change algebra's chance?)
Here, P(A|B) = 0.4 but P(A) = 0.3. Since 0.4 \neq 0.3, they are NOT independent.
The chance of algebra changed when we knew she took biology.
Part D: Are Algebra and Biology Enrollment Mutually Exclusive?
To check, see if P(A \cap B) = 0 (Can both happen at the same time?)
We found P(A \cap B) = 0.28. Since 0.28 \neq 0, they are NOT mutually exclusive.
Sarah can take both courses.
Key Things to Remember for Your Test!
When to use which rule?
If the problem asks for "A OR B" (P(A \cup B)), think Addition Rule. Remember to subtract the overlap! (-P(A \cap B)) unless they're mutually exclusive.
If the problem asks for "A AND B" (P(A \cap B)), think Multiplication Rule. You'll likely need conditional probability (P(A|B) \cdot P(B)) unless they're independent.
How to check for Independence?
Compare P(A|B) with P(A). If they are the same, they're independent. If different, they're dependent.
Or check if P(A \cap B) = P(A) \cdot P(B). If this holds true, they are independent.
How to check for Mutually Exclusive?
See if P(A \cap B) = 0. If they can't happen together, their intersection is zero.
Don't double-count! That's why the P(A \cap B) subtraction is in the addition rule. It handles the overlap.
These rules help you understand chances in the real world, from games to decisions about uncertainty!