Probability — independent and conditional events (KCSE Mathematics Form 4)

TL;DR

Independent events don't affect each other's probabilities, while conditional events do. You'll learn how to calculate probabilities for both scenarios using specific formulas. Mastering these concepts helps predict outcomes in various real-life situations.

1. The Mental Model

Think of probability as the chance of something happening. Independent events are like two separate coin flips – one doesn't change the other. Conditional events are like drawing cards without replacement – what you draw first affects what's left.

2. The Core Material

What is Probability?

Probability is a measure of how likely an event is to occur. It's always a number between 0 and 1 (or 0% and 100%).
* P(Event) = (Number of favourable outcomes) / (Total number of possible outcomes)

For example, if you roll a standard six-sided die, the probability of rolling a 4 is 1/6 because there's one '4' and six possible outcomes (1, 2, 3, 4, 5, 6).

Independent Events

Two events, A and B, are independent if the occurrence of one does not affect the probability of the other occurring.
* Key idea: The outcome of event A doesn't change the likelihood of event B, and vice-versa.
* Formula for P(A and B): P(A ∩ B) = P(A) × P(B)
This means the probability of both A and B happening is the product of their individual probabilities.

Example:
You flip a coin and roll a die.
* Event A: Getting a Head on the coin. P(A) = 1/2.
* Event B: Rolling a 6 on the die. P(B) = 1/6.
Are these independent? Yes, the coin flip doesn't influence the die roll.
P(Head and 6) = P(Head) × P(6) = (1/2) × (1/6) = 1/12.

Conditional Events

Two events, A and B, are conditional (or dependent) if the occurrence of one does affect the probability of the other occurring.
* Key idea: The probability of event B happening changes based on whether event A has already happened.
* Notation: P(B|A) means "the probability of event B happening given that event A has already happened."
* Formula for P(B|A): P(B|A) = P(A ∩ B) / P(A) (provided P(A) > 0)
* Rearranged formula for P(A and B): P(A ∩ B) = P(A) × P(B|A)
This is often called the Multiplication Rule for Dependent Events.

Example:
You have a bag with 5 red balls and 3 blue balls (total 8 balls). You draw two balls without replacement.
* Event A: Drawing a red ball first. P(A) = 5/8.
* Event B: Drawing another red ball second, given that the first was red.
If the first was red, there are now 4 red balls left and 7 total balls.
So, P(B|A) = 4/7.
P(Drawing two red balls) = P(A) × P(B|A) = (5/8) × (4/7) = 20/56 = 5/14.

Using Tree Diagrams for Conditional Probability

Tree diagrams are super helpful for visualising sequences of events, especially when they are conditional. Each branch represents a possible outcome, and probabilities are written along the branches.

graph TD
    A[Start] --> B{First Draw};
    B --> C(Red Ball 1: 5/8);
    B --> D(Blue Ball 1: 3/8);
    C --> E(Red Ball 2: 4/7);
    C --> F(Blue Ball 2: 3/7);
    D --> G(Red Ball 2: 5/7);
    D --> H(Blue Ball 2: 2/7);
    E --> I(P(R1 and R2) = 5/8 * 4/7 = 20/56);
    F --> J(P(R1 and B2) = 5/8 * 3/7 = 15/56);
    G --> K(P(B1 and R2) = 3/8 * 5/7 = 15/56);
    H --> L(P(B1 and B2) = 3/8 * 2/7 = 6/56);

• To find the probability of a sequence of events (like "Red then Red"), you multiply the probabilities along the branches.
• To find the probability of an event that can happen in multiple ways (like "one red and one blue"), you add the probabilities of those different paths.

Mutually Exclusive Events (Quick Distinction)

Don't confuse independent events with mutually exclusive events.
* Mutually Exclusive: Events that cannot happen at the same time (e.g., rolling a 1 and a 6 on a single die roll). If A and B are mutually exclusive, P(A ∩ B) = 0.
* Independent: Events where one doesn't affect the other (e.g., rolling a 1 on one die and a 6 on another die). P(A ∩ B) = P(A) × P(B).
If two events are mutually exclusive and have non-zero probabilities, they cannot be independent.

3. Worked Example

A box contains 7 green pens and 5 blue pens. Two pens are picked at random, one after the other, without replacement.

a) Draw a tree diagram to represent all possible outcomes.
b) Find the probability that:
i) Both pens are green.
ii) The first pen is green and the second is blue.
iii) The pens are of different colours.

Solution:

Total pens = 7 (Green) + 5 (Blue) = 12 pens.

a) Tree Diagram:

graph TD
    A[Start] --> B{First Pick};
    B --> C(Green 1: 7/12);
    B --> D(Blue 1: 5/12);
    C --> E(Green 2: 6/11);
    C --> F(Blue 2: 5/11);
    D --> G(Green 2: 7/11);
    D --> H(Blue 2: 4/11);
    E --> I(P(G1 and G2) = 7/12 * 6/11 = 42/132);
    F --> J(P(G1 and B2) = 7/12 * 5/11 = 35/132);
    G --> K(P(B1 and G2) = 5/12 * 7/11 = 35/132);
    H --> L(P(B1 and B2) = 5/12 * 4/11 = 20/132);

b) Probabilities:

i) Both pens are green (G1 and G2):
This is the path Green then Green.
P(G1 and G2) = P(G1) × P(G2|G1) = (7/12) × (6/11) = 42/132 = 7/22.

ii) The first pen is green and the second is blue (G1 and B2):
This is the path Green then Blue.
P(G1 and B2) = P(G1) × P(B2|G1) = (7/12) × (5/11) = 35/132.

iii) The pens are of different colours:
This can happen in two ways: (Green then Blue) OR (Blue then Green).
P(Different colours) = P(G1 and B2) + P(B1 and G2)
P(B1 and G2) = P(B1) × P(G2|B1) = (5/12) × (7/11) = 35/132.
P(Different colours) = 35/132 + 35/132 = 70/132 = 35/66.

4. Key Takeaways

• Probability is a number between 0 and 1 representing the likelihood of an event.
• Independent events don't influence each other; P(A and B) = P(A) × P(B).
• Conditional events depend on previous outcomes; P(A and B) = P(A) × P(B|A).
• P(B|A) is the probability of B given that A has already occurred.
• Tree diagrams are excellent tools for visualising and calculating probabilities for sequential events.
• The sum of probabilities of all possible outcomes for an event must always equal 1.

Common Mistakes to Avoid:
- Confusing independent events with mutually exclusive events.
- Forgetting to adjust the total number of outcomes (and favourable outcomes) when dealing with "without replacement" problems.
- Incorrectly multiplying probabilities for events that are not independent.
- Not simplifying fractions to their lowest terms in your final answer.

5. Now Try It

A bag contains 4 red marbles and 6 blue marbles. You pick two marbles at random.

a) If you pick the marbles with replacement, what is the probability that both marbles are blue?
b) If you pick the marbles without replacement, what is the probability that both marbles are blue?
c) If you pick the marbles without replacement, what is the probability that at least one marble is red?

What success looks like:
For (a), you'll use the independent events formula. For (b), you'll use the conditional probability formula (or a tree diagram) and adjust the probabilities for the second pick. For (c), you'll either sum the probabilities of (Red then Blue), (Blue then Red), and (Red then Red), or use the complement rule (1 - P(Both Blue)). Your answers should be simplified fractions.