S1 Probability: Fundamental Concepts and Rules

Rolling the Dice on Probability! 🎲

Introduction

1. Introduction

Alright, let's dive into the world of probability! It might seem like it's all about luck and chance, but it's actually a super powerful branch of maths for making sense of uncertainty. Think about it - from weather forecasts to game strategies, probability is everywhere. In this chapter, we're going to build your skills from the ground up. We'll start with the basics of calculating probabilities, then move on to the essential Addition and Multiplication Rules for combining events. Next, we'll tackle Conditional Probability, which is all about how one event can change the likelihood of another. Finally, we'll clear up the difference between Independent and Mutually Exclusive Events, two key concepts that will make solving complex problems a whole lot easier. Ready to become a pro at predicting outcomes? Let's get started! 💪

P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}

Your success hinges on one question you must ask yourself every time:'Does the order of selection matter?' Get that right, and you've unlocked the problem. Mastering this isn't just about exams; it's fundamental to fields like data science, cryptography, and risk analysis in finance. So, let's get this locked down.

Worked example

Assembling the Dream Committee 🏆

A committee of 4 people is to be chosen at random from a group of 7 men and 5 women. Find the probability that the committee consists of exactly 2 men and 2 women.

1
$n(S)$
$n(S) = \text{Total ways to choose 4 from 12} = {}_{12}C_4 = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495$
2
$n(A)$
$n(A) = (\text{Ways to choose 2 men from 7}) \times (\text{Ways to choose 2 women from 5})$
3
$Now, let's calculate each part of that expression. Again, we're using combinations because the order of selection doesn't matter.$
$\text{Ways to choose 2 men} = {}_{7}C_2 = \frac{7!}{2!5!} = 21 <br> \text{Ways to choose 2 women} = {}_{5}C_2 = \frac{5!}{2!3!} = 10$
4
$n(A)$
$n(A) = 21 \times 10 = 210$
5
$Finally, we calculate the probability by dividing the number of favourable outcomes by the total number of outcomes. Always remember to simplify your fraction if possible!$
$P(\text{2 men, 2 women}) = \frac{n(A)}{n(S)} = \frac{210}{495} = \frac{14}{33}$

Answer

P(\text{2 men, 2 women}) = \frac{n(A)}{n(S)} = \frac{210}{495} = \frac{14}{33}

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cap B)

Worked example

The University Offer Dilemma 🎓

P(A) = 0.7

P(A|B)

P(A|B)

Worked example

Let's Diagnose This Probability Problem 🩺

0.2\%

1
$First things first, let's define our events and lay out the probabilities. This helps us structure the problem. Let 'D' be the event that a person has the disease, and 'P' be the event that they test positive. A tree diagram is perfect for visualizing this.$
$P(D) = 0.002$
2
$P(D|P)$
$P(D|P)$
3
$P(A|B) = \frac{P(A \cap B)}{P(B)}$
$P(D|P) = \frac{P(D \cap P)}{P(P)}$
4
$P(D \cap P)$
$P(D \cap P) = P(D) \times P(P|D) = 0.002 \times 0.99 = 0.00198$
5
$P(P)$
$P(P) = P(D \cap P) + P(D' \cap P) = (0.002 \times 0.99) + (0.998 \times 0.05) = 0.00198 + 0.0499 = 0.05188$
6
$Finally, we plug our numerator and denominator back into the formula to get our answer. Notice how even with a positive test, the probability of having the disease is still surprisingly low! This is a famous result in statistics and shows why understanding conditional probability is so critical in real-world applications.$
$P(D|P) = \frac{0.00198}{0.05188} \approx 0.03816$

Answer

P(D|P) = \frac{0.00198}{0.05188} \approx 0.03816

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

A

Worked example

The University Offer Dilemma 🎓

P(M)

1
$First, let's clearly define our events and list the probabilities given in the problem. This helps us organise our thoughts before we jump into the calculation.$
$M$
2
$Now, we state the mathematical condition that must be met for two events to be independent. This is the core formula we need to test.$
$M$
3
$P(M) \times P(E)$
$P(M) \times P(E) = 0.3 \times 0.8 = 0.24$
4
$P(M \cap E)$
$P(M \cap E) = 0.24$
5
$We must write a clear concluding statement that answers the original question. It's crucial to state why you've reached your conclusion, referencing the test you performed.$
$P(M \cap E) = P(M) \times P(E)$

Answer

P(M \cap E) = P(M) \times P(E)

Unlock the full chapter: practice questions, flashcards, mock papers and notes, free.

Continue revising