Chapter 7: Probability

Life's a Game, Let's Find the Odds 🎲

Introduction

1. Introduction

Ever wondered about the chances of your favorite artist dropping a surprise album? Or the probability of snagging that legendary skin in your next game loot box? That's probability in action! It's not just some abstract math concept; it's the logic behind everything from your Spotify shuffle to predicting if your crush will snap you back. In this chapter, we're going to decode the math of chance. We'll start with the basics of single events (like rolling a dice for a game). Then, we'll level up to see what happens when events are combined, using Tree Diagrams to map out sequences of events, like picking two different snacks from the cupboard. We'll also use Venn Diagrams to sort out overlapping situations, like figuring out how many of your friends are on both Insta and Snapchat. Finally, we'll tackle the boss level: Conditional Probability, which is all about how one event can change the odds for another. Let's get started and turn uncertainty into something you can actually calculate!

2. Introduction to Single Event Probability

\frac{1}{2}

A horizontal number line from 0 to 1. 0 is labelled 'Impossible'. 0.25 is 'Unlikely'. 0.5 is 'Even Chance'. 0.75 is 'Likely'. 1 is 'Certain'.

P(\text{Event}) = \frac{\text{Number of ways your event can happen}}{\text{Total number of possible outcomes}}

Worked example

Worked Example: Calculating Probability from a Table

Let's Crack This Playlist Problem 🎧

You're putting your music library on shuffle for a road trip. The genres in your playlist are shown in the table below. | Genre | Number of Songs | |---|---| | Pop | 60 | | Hip-Hop | 45 | | Indie | 35 | | K-Pop | 10 | A song is chosen at random. a) Find the probability that it is a Hip-Hop song. b) Find the probability that it is not a Hip-Hop song.

3. Combined Events: Using Tree Diagrams

Alright, let's talk about tree diagrams. Think of them like the branching storylines in a game like The Witcher or Detroit: Become Human . Every choice you make opens up new paths, and each path has a certain chance of happening. That's exactly what a tree diagram does for probability - it helps you visualize and calculate the chances of a sequence of events. First, you start with your initial event - let's say, picking a song from a shuffled playlist. You draw branches for each possible outcome (e.g., picking a rap song or a pop song). On each branch, you write the probability. Super important: the probabilities branching from a single point must always add up to 1 . Why? Because it's 100% certain that one of those outcomes will happen. Now for the second event, like picking another song. You draw new branches coming off the end of your first ones. This is where it gets interesting. If the events are independent (like if you can pick the same song twice), the probabilities for the second set of branches are the same as the first. But if they're dependent - the classic 'without replacement' scenario - things change. Imagine you're grabbing snacks from a bag for you and a friend. Once you take one, it's gone! This means there's one less item in the bag, so the total number of outcomes changes, affecting the probabilities for the second pick.

A detailed tree diagram for a 'without replacement' problem. It shows the first pick from a bag of 10 sweets (6 Red, 4 Blue). The first branches are P(Red) = 6/10 and P(Blue) = 4/10. The second set of branches shows updated probabilities: after picking Red, the new branches are P(Red) = 5/9 and P(Blue) = 4/9. After picking Blue, the new branches are P(Red) = 6/9 and P(Blue) = 3/9.

\times

Worked example

Worked Example: Dependent Events without Replacement

The Grab Bag Problem 🛍️

A bag contains 7 green counters and 4 blue counters. You pick one counter at random, do not replace it, and then pick a second counter. (a) What is the probability that you pick two green counters? (b) What is the probability that you pick counters of different colours?

1
$First, let's set up the tree diagram. We start with 11 counters in total (7 Green, 4 Blue). The first branches will show the probabilities for the first pick.$
$P(\text{First is Green}) = \frac{7}{11} \quad \text{and} \quad P(\text{First is Blue}) = \frac{4}{11}$
2
$Now, let's figure out the probabilities for the second pick. These are dependent events because we don't replace the first counter. After one is picked, there are only 10 counters left. If we picked a green first, there are now 6 greens left. If we picked a blue first, there are still 7 greens left.$
$P(\text{Second is Green | First was Green}) = \frac{6}{10} P(\text{Second is Blue | First was Green}) = \frac{4}{10} P(\text{Second is Green | First was Blue}) = \frac{7}{10} P(\text{Second is Blue | First was Blue}) = \frac{3}{10}$
3
$For part (a), we need the probability of picking two green counters (Green and then Green). We use the 'AND' rule, which means we multiply the probabilities along the G-G branch.$
$P(G \text{ and } G) = P(G_1) \times P(G_2 | G_1) = \frac{7}{11} \times \frac{6}{10} = \frac{42}{110} = \frac{21}{55}$
4
$For part (b), 'counters of different colours' means we could either pick Green then Blue (G-B) OR Blue then Green (B-G). We need to calculate the probability for each of these paths by multiplying along the branches.$
$P(G \text{ and } B) = \frac{7}{11} \times \frac{4}{10} = \frac{28}{110} P(B \text{ and } G) = \frac{4}{11} \times \frac{7}{10} = \frac{28}{110}$
5
$Finally, since either of these outcomes satisfies the condition, we use the 'OR' rule and add their probabilities together to get the final answer.$
$P(\text{different colours}) = P(G \text{ and } B) + P(B \text{ and } G) = \frac{28}{110} + \frac{28}{110} = \frac{56}{110} = \frac{28}{55}$

Answer

P(\text{different colours}) = P(G \text{ and } B) + P(B \text{ and } G) = \frac{28}{110} + \frac{28}{110} = \frac{56}{110} = \frac{28}{55}

4. Probability of Combined Events using Venn Diagrams

\mathcal{E}

$A standard Venn diagram with a rectangle labelled with the universal set symbol . Inside are two overlapping circles labelled A and B. The overlapping section is shaded and labelled 'A \cap B' (A and B). The combined area of both circles is outlined and labelled 'A \cup B' (A or B). The area outside the circles but inside the box is labelled '(A \cup B)'' (Neither A nor B).$

A \cap B

Worked example

Worked Example: Calculating Probabilities from a Venn Diagram

The Part-Time Job & Driving Hustle 🚗💼

In a survey of 100 students, 45 have a part-time job (J), 32 have their driver's license (D), and 15 have both. A student is chosen at random from the group. a) Draw a Venn diagram to represent this information. b) Find the probability that the student has a job or a license. c) Find the probability that the student has a license but no job.

1
$J \cap D$
$\text{Intersection } (J \cap D) = 15$
2
$45 - 15$
$\text{Job only} = 45 - 15 = 30 \text{License only} = 32 - 15 = 17$
3
$Next, let's figure out how many students have neither. We add up everyone inside the circles and subtract that from the total of 100 students. This number goes outside the circles but inside the box.$
$\text{Inside circles} = 30 (J \text{ only}) + 15 (Both) + 17 (D \text{ only}) = 62 \text{Neither} = 100 - 62 = 38$
4
$P(J \cup D)$
$P(J \cup D) = \frac{\text{Number in J or D}}{\text{Total}} = \frac{30 + 15 + 17}{100} = \frac{62}{100} = \frac{31}{50}$
5
$Finally, for part (c), we need the probability of having a license but no job. This corresponds to the 'License only' section of the Venn diagram, which we found to be 17.$
$P(D \cap J') = \frac{\text{Number in D only}}{\text{Total}} = \frac{17}{100}$