Mathematics (9709) | Probability

AS & A Level Mathematics 9709 (Paper 5) Study Notes: Probability (5.3)

Hello future mathematician! This chapter is all about quantifying chance—giving a precise numerical value to how likely something is to happen. Probability is central to Statistics, and a strong understanding here is crucial for tackling discrete and continuous variables later on. Don't worry if counting outcomes seems tricky; we’ll use the Permutations and Combinations skills you mastered in the previous section (5.2) to make this much easier!

1. The Basics of Probability

Probability measures the likelihood of an event occurring. The result is always a value between 0 (impossible) and 1 (certainty).

Key Definitions

• Experiment: A process that leads to well-defined results or outcomes. (e.g., rolling a dice, choosing a card)
• Outcome: A single result of an experiment. (e.g., rolling a '4', drawing the Ace of Spades)
• Sample Space (\(S\)): The set of all possible outcomes. (e.g., for a standard die, \(S = \{1, 2, 3, 4, 5, 6\}\))
• Event (\(A\)): A subset of the sample space (one or more outcomes). (e.g., Event A = rolling an even number, so \(A = \{2, 4, 6\}\))

The Classical Probability Formula

If all outcomes are equally likely (equiprobable elementary events), the probability of an event \(A\) is calculated as:

\[P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes in S}}\]

The Complementary Event (\(A'\))

The complement of event \(A\) (denoted \(A'\) or \(A^c\)) is the event that \(A\) does not happen.

\[P(A') = 1 - P(A)\]

This is incredibly useful! If calculating \(P(A)\) is hard (e.g., "at least one 6 in three rolls"), calculate \(P(A')\) instead ("no 6s in three rolls") and subtract from 1.

2. Evaluating Probabilities (Counting Outcomes)

In most exam questions, especially those involving selections or arrangements, you cannot simply list all outcomes (enumeration). This is where your knowledge of Permutations and Combinations (P&C) becomes vital.

Method: Use P&C to Count

To calculate \(P(A)\), follow these steps:

1. Use P&C formulas (or the Fundamental Principle of Counting) to find the Total Number of possible outcomes (\(|S|\)).
2. Use P&C formulas to find the Number of ways event \(A\) can occur (\(|A|\)).
3. Calculate the ratio: \(P(A) = \frac{|A|}{|S|}\).

Example: A bag has 5 red and 3 blue balls. If 2 balls are drawn at random, what is the probability that both are red?

• Total outcomes (\(|S|\)): Selecting 2 from 8 balls: \(\binom{8}{2} = 28\).
• Favourable outcomes (\(|A|\)): Selecting 2 red from 5 red: \(\binom{5}{2} = 10\).
• Probability: \(P(\text{2 Red}) = \frac{10}{28} = \frac{5}{14}\).

Key Takeaway: Probability calculations often depend entirely on correctly counting the number of ways things can happen, so ensure your P&C knowledge is solid!

3. Combining Events: Addition (OR) and Multiplication (AND)

When you deal with two or more events, \(A\) and \(B\), you need rules to combine their probabilities.

3.1. The Addition Rule (OR) - Mutually Exclusive Events

If you want the probability that event \(A\) or event \(B\) occurs, you are looking for \(P(A \cup B)\) (A union B).

Mutually Exclusive (or Exclusive) Events:

Two events \(A\) and \(B\) are mutually exclusive if they cannot happen at the same time. This means their intersection is empty, \(P(A \cap B) = 0\).

• Analogy: Rolling a 4 OR a 5 on a single die roll. You can’t do both.

For mutually exclusive events:

\[P(A \cup B) = P(A) + P(B)\]

Common Mistake to Avoid: Only use the simple addition rule above if the events are *definitely* mutually exclusive. If they overlap (i.e., \(P(A \cap B) \ne 0\)), you cannot simply add them, as you would double-count the overlap.

3.2. The Multiplication Rule (AND) - Independent Events

If you want the probability that event \(A\) and event \(B\) both occur, you are looking for \(P(A \cap B)\) (A intersection B).

Independent Events:

Two events \(A\) and \(B\) are independent if the occurrence of one does not affect the probability of the other.

• Analogy: Rolling a die and flipping a coin. Getting a Head doesn't change the chance of rolling a 6.

For independent events:

\[P(A \cap B) = P(A) \times P(B)\]

Did you know? Independence is tested directly using the formula above. If you calculate \(P(A) \times P(B)\) and it is equal to the calculated probability of the intersection \(P(A \cap B)\), the events are independent. If they are unequal, they are dependent.

Key Takeaway: "OR" usually means add (if exclusive). "AND" usually means multiply (if independent).

4. Conditional Probability (The 'Given That' Scenario)

Conditional probability deals with situations where the occurrence of one event directly impacts the probability of a subsequent event. These events are dependent.

Definition and Notation

The conditional probability of event \(A\) occurring, given that event \(B\) has already occurred, is written as \(P(A|B)\).

The formula is:

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

This formula essentially restricts your original sample space (\(S\)) to only include outcomes where \(B\) happened. \(P(B)\) becomes the new "total probability."

Using Conditional Probability with Dependent Events

When dealing with dependent events (like drawing without replacement), the joint probability \(P(A \cap B)\) must use the conditional probability:

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

Example: Drawing two cards without replacement from a standard deck. Let R1 = Red on 1st draw, R2 = Red on 2nd draw.

• \(P(R_1) = \frac{26}{52} = \frac{1}{2}\)
• If R1 happened, there are now 51 cards left, 25 of which are red.
• \(P(R_2 | R_1) = \frac{25}{51}\) (The probability of R2, GIVEN R1 occurred)
• \(P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) = \frac{26}{52} \times \frac{25}{51} = \frac{25}{102}\)

Visualising using Tree Diagrams

Conditional probability problems are often easily solved using tree diagrams, especially when dealing with sequences of events.

1. Start with the branches representing the first event and label them with unconditional probabilities.
2. The subsequent branches are labeled with conditional probabilities (the 'given that' probabilities).
3. To find the probability of a sequence (e.g., Path 1 AND Path 2), multiply along the branches.

Don't worry if this seems tricky at first—tree diagrams simplify complex chains of events by visually separating the sequential probabilities.

Working with Conditional Probability (Step-by-Step Example)

A survey found that 60% of students own a laptop (L), 30% own a tablet (T), and 10% own both. What is the probability a student owns a laptop, given that they own a tablet? (\(P(L|T)\))

• State knowns: \(P(L) = 0.6\), \(P(T) = 0.3\), \(P(L \cap T) = 0.1\).
• Identify target: \(P(L|T)\).
• Apply formula: \[P(L|T) = \frac{P(L \cap T)}{P(T)}\]
• Calculate: \[P(L|T) = \frac{0.1}{0.3} = \frac{1}{3}\]

Interpretation: Out of the 30% of students who own a tablet, 10% also own a laptop. So, the probability among tablet owners of also owning a laptop is 1/3.

5. Exclusive vs. Independent: A Critical Distinction

Students frequently confuse these two terms because "exclusive" and "independent" sound similar in everyday speech, but they are mathematically very different.

If two events A and B are mutually exclusive:

• They share no outcomes: \(P(A \cap B) = 0\).
• They are highly DEPENDENT: if A happens, B definitely CANNOT happen (so \(P(B|A) = 0\)).
• Rule for OR: \(P(A \cup B) = P(A) + P(B)\).

If two events A and B are independent:

• They can happen together (overlap): \(P(A \cap B) = P(A) \times P(B)\).
• They are NOT exclusive (unless \(P(A)=0\) or \(P(B)=0\)).
• Rule for AND: \(P(A \cap B) = P(A) \times P(B)\).
• Crucially: The conditional probability is the same as the unconditional probability: \(P(A|B) = P(A)\).

Key Takeaway: Exclusive events deal with addition (cannot overlap). Independent events deal with multiplication (no influence on each other).