More about Probability: Your Study Guide
Hey everyone! Welcome to the study notes for "More about Probability". You might have learned the basics before, but now we're going to level up! We'll learn how to handle more complex situations, like the chances of multiple things happening. It might sound tricky, but don't worry! We'll break it down with simple language and real-life examples. Understanding this chapter is super useful, not just for exams, but for making sense of games, news, and decisions in the real world.
1. The Language of Probability: Sets and Events
To talk about probability accurately, mathematicians use a special language called set notation. Think of it as a grammar for probability. Once you know the symbols, everything becomes much clearer.
What is a Set?
A set is just a collection of distinct items, which we call elements.
- Universal Set (S or U): This is the set of ALL possible outcomes in an experiment. Analogy: If you're rolling a standard six-sided die, the universal set is S = {1, 2, 3, 4, 5, 6}.
- Event (A, B, etc.): An event is a specific outcome or a group of outcomes you are interested in. It's a 'subset' of the universal set. Analogy: The event 'rolling an even number' would be A = {2, 4, 6}.
Visualising Sets: Venn Diagrams
A Venn diagram is a picture that helps us see the relationship between sets. It's a fantastic tool! We use a rectangle for the universal set (S) and circles inside for our events (A, B).
Key Set Operations
These are the three most important actions you'll see. Don't worry, they are based on simple words: AND, OR, and NOT.
Intersection (A ∩ B): The "AND" Operation
This means elements that are in BOTH event A AND event B. It's the overlapping part of the circles in a Venn diagram.
Example: Let A be the event of rolling an even number {2, 4, 6} and B be the event of rolling a number greater than 3 {4, 5, 6}. Then the intersection is A ∩ B = {4, 6}, because these numbers are both even AND greater than 3.
Union (A ∪ B): The "OR" Operation
This means elements that are in event A OR event B (or both). It includes everything in both circles.
Example: Using the same events A = {2, 4, 6} and B = {4, 5, 6}, the union is A ∪ B = {2, 4, 5, 6}. Notice we don't list 4 and 6 twice!
Memory Aid: The Union symbol ∪ looks like a cup that holds everything from both sets together.
Complement (A'): The "NOT" Operation
This means everything in the universal set that is NOT in event A. In a Venn diagram, it's the area outside of circle A.
Example: If A is the event of rolling an even number {2, 4, 6}, its complement is A' = {1, 3, 5}, the set of outcomes that are NOT even.
Key Takeaway
Set notation gives us precise symbols for probability ideas. Master these three:
- Intersection (∩) means AND (the overlap).
- Union (∪) means OR (everything combined).
- Complement (') means NOT (everything else).
2. The Addition Law: Combining "OR" Probabilities
Now that we have the language, let's use it. The addition law helps us find the probability of event A OR event B happening, which is $$P(A \cup B)$$.
The General Addition Law
This is the main formula you need to know:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$Why do we subtract $$P(A \cap B)$$?
Think about the Venn diagram. When we add the probability of circle A and the probability of circle B, we've counted the overlapping part (the intersection) twice! So, we have to subtract it once to get the correct total.
Step-by-Step Example:
In a group of students, the probability of having a cat is 0.4, the probability of having a dog is 0.5, and the probability of having both is 0.15. What is the probability that a randomly chosen student has a cat or a dog?
- Identify the probabilities:
P(Cat) = 0.4
P(Dog) = 0.5
P(Cat ∩ Dog) = 0.15 - Apply the formula:
$$P(Cat \cup Dog) = P(Cat) + P(Dog) - P(Cat \cap Dog)$$
$$P(Cat \cup Dog) = 0.4 + 0.5 - 0.15 = 0.75$$
So, there's a 0.75 (or 75%) chance the student has a cat or a dog.
Special Case 1: Mutually Exclusive Events
Mutually exclusive events are events that CANNOT happen at the same time.
Example: When you roll a die, you can't get a '1' AND a '6' on the same roll.
For these events, the overlap is impossible, so $$P(A \cap B) = 0$$. This makes the addition law much simpler:
Addition Law for Mutually Exclusive Events: $$P(A \cup B) = P(A) + P(B)$$
Special Case 2: Complementary Events
Remember the complement A' (the "NOT A" event)? An event and its complement are always mutually exclusive. Even better, they cover all possibilities!
This gives us a super useful rule:
$$P(A) + P(A') = 1 \quad \text{or} \quad P(A') = 1 - P(A)$$This is a great shortcut! Sometimes it's easier to calculate the probability of an event NOT happening and subtract it from 1.
Common Mistake Alert!
A common mistake is forgetting to subtract $$P(A \cap B)$$ for events that can happen together. Always ask yourself: "Can these two events happen at the same time?". If the answer is yes, you MUST subtract the intersection!
Key Takeaway
To find the probability of A OR B, use the addition law. The full formula is $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$. If the events can't happen together (mutually exclusive), the last part is just zero.
3. The Multiplication Law: "AND" Probabilities and Conditions
This section is all about finding the probability of event A AND event B happening, which is $$P(A \cap B)$$. The key question to ask here is: "Does the first event affect the second one?"
Independent Events: No Influence
Independent events are events where the outcome of one has absolutely no effect on the outcome of the other.
Example: Tossing a coin and then rolling a die. The coin landing on Heads doesn't change the probability of the die landing on 6.
For these events, we use the Multiplication Law for Independent Events:
$$P(A \cap B) = P(A) \times P(B)$$Example:
What is the probability of flipping a coin and getting Heads AND rolling a die and getting a 4?
P(Heads ∩ 4) = P(Heads) × P(4) = $$ \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} $$
Dependent Events and Conditional Probability
Dependent events are events where the outcome of the first event CHANGES the probability of the second event.
Example: Drawing two marbles from a bag WITHOUT replacing the first one. The probabilities for the second draw depend on what you drew first.
This introduces a new idea: Conditional Probability. We write it as $$P(B|A)$$, which means "the probability of B happening, GIVEN THAT A has already happened."
This gives us the General Multiplication Law (works for ANY events):
$$P(A \cap B) = P(A) \times P(B|A)$$Step-by-Step Example:
A bag contains 3 red and 7 blue marbles. Two marbles are drawn one by one without replacement. What's the probability that both are red?
- Probability of the 1st event:
Let A = 1st marble is red. $$P(A) = \frac{3}{10}$$ - Probability of the 2nd event (conditional):
Let B = 2nd marble is red. GIVEN the first was red, there are now only 2 red marbles and 9 total marbles left. So, $$P(B|A) = \frac{2}{9}$$. - Apply the formula:
$$P(A \cap B) = P(A) \times P(B|A) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$$
Did you know?
The formal test for independence is checking if $$P(A \cap B) = P(A) \times P(B)$$. You shouldn't just rely on intuition. If the equation holds true, the events are independent. If it doesn't, they are dependent.
Key Takeaway
To find the probability of A AND B, first check if they are independent.
- If YES (independent), just multiply: $$P(A \cap B) = P(A) \times P(B)$$.
- If NO (dependent), use the conditional formula: $$P(A \cap B) = P(A) \times P(B|A)$$.
4. Probability Power-Up: Using Permutations and Combinations
For some problems, there are too many outcomes to list. That's where your counting skills from the "Permutations and Combinations" chapter come in handy! We use them to find the numerator and denominator of our basic probability formula.
Quick Review: The Master Formula
$$ \text{Probability of an event} = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}} $$When to Use Permutation (P) vs. Combination (C)
This is the most important decision. Here's a quick reminder:
- Permutation ($$P_r^n$$): ORDER MATTERS. Use this when arranging items, assigning specific roles (e.g., President, Vice-President), or creating codes. Memory aid: Permutation = Position.
- Combination ($$C_r^n$$): ORDER DOES NOT MATTER. Use this when selecting a group, forming a committee, or choosing items where the order of selection is irrelevant. Memory aid: Combination = Committee.
Step-by-Step Problem Solving
Follow these steps, and you'll solve these problems like a pro!
- Read the question carefully. Decide if order matters (Permutation) or not (Combination).
- Calculate the Total Outcomes (Denominator): Find the total number of ways to select items without any of the question's special conditions.
- Calculate the Favourable Outcomes (Numerator): Find the number of ways to select items that satisfy the specific conditions of the question. You might need to use the multiplication rule of counting here (for "AND").
- Divide and Simplify: Put the favourable number over the total number to get your probability.
Worked Example:
There are 6 boys and 5 girls in a class. A committee of 4 students is to be selected. What is the probability that the committee will have exactly 2 boys and 2 girls?
Step 1: P or C?
It's a committee, so the order of selection doesn't matter. We will use Combination (C).
Step 2: Total Outcomes (Denominator)
There are 11 students in total (6 boys + 5 girls), and we are choosing 4.
Total number of possible committees = $$C_4^{11} = \frac{11!}{4!(11-4)!} = 330$$
Step 3: Favourable Outcomes (Numerator)
We want exactly 2 boys AND 2 girls.
- Ways to choose 2 boys from 6: $$C_2^6 = \frac{6!}{2!4!} = 15$$
- Ways to choose 2 girls from 5: $$C_2^5 = \frac{5!}{2!3!} = 10$$
Since we need both, we multiply: $$15 \times 10 = 150$$ favourable outcomes.
Step 4: Divide
Probability = $$ \frac{\text{Favourable}}{\text{Total}} = \frac{150}{330} = \frac{15}{33} = \frac{5}{11} $$
Key Takeaway
When probabilities involve selections from a group, use Permutations (order matters) or Combinations (order doesn't matter) to quickly count the total and favourable outcomes. This turns a difficult problem into a structured calculation.