## What you need to know

Probability is the study of how likely things are to happen. We express probabilities either as fractions, decimals, or percentages, and they always fall between 0 and 1, where 0 represents total impossibility and 1 represents total certainty. For example, the probability of the result of coin flip being ‘heads’ is 0.5, 50%, or \dfrac{1}{2}. If we have an **exhaustive** list of outcomes (i.e. all possible outcomes are accounted for), then the sum of their probabilities must be 1. Continuing with the coin example, the only two possible outcomes are heads or tails, both of which have probability 0.5, and 0.5+0.5=1, as expected.

There are two types of probability you will see:

**Theoretical probability**– this is the kind of probability that we have prior understanding of. For example, we know that the chance of rolling a 6 on a fair dice is \dfrac{1}{6}.**Relative frequency**– this is the kind of probability that we determine from a survey or experiment. We calculate the relative frequency for a specific outcome as such:

\text{relative frequency }=\dfrac{\text{number of times outcome happens}}{\text{total number of trials}}

For example, if you conducted a survey asking people what genres of music they enjoyed, then the relative frequency of people who liked rock music would be

\text{relative frequency }=\dfrac{\text{number who liked rock music}}{\text{total number asked}}

We can use a probability value to determine **expected frequency**, which is the number of times we would expect an outcome to happen if were to do an experiment. It is calculated as such:

\text{expected frequency }=\text{ probability }\times\text{number of trials}

For example, if we roll a dice 60 times, the expected frequency of sixes is \dfrac{1}{6}\times60= 10. In practice, it probably won’t end up being exactly ten times, i.e. our relative frequency won’t be \dfrac{1}{6}. However, if you do more trials – 600 rolls, 6,000 times – the relative frequency will get closer to \dfrac{1}{6}. In general, we say that the **relative frequency will tend toward the theoretical probability** as the number of trials increases. We say an experiment with **more trials **is** more accurate**.

**Example: **Stella and Gary are throwing darts at a target. By calculating relative frequencies, work out who is more likely to get a hit. Also, state which one of their relative frequencies gives a more accurate statement of their chance of hitting the target and explain why.

To calculate the relative frequencies, we will divide each person’s hits by their total attempts. Gary:

\dfrac{18}{18+42}=0.3

Then, Stella:

\dfrac{45}{45+75}=0.375

Therefore, Stella is more likely to get a hit. Furthermore, Stella made 45+75=120 attempts in total, compared to Gary’s 18+42=60, therefore Stella’s relative frequency is a more accurate representation of her chance of hitting the target than Gary’s is for him.

Often, we are interested in the probability of more than one event occurring. In this case, the events might be **independent **– the outcome of one does not affect the other – or **dependent **– the outcome of one does affect the other. In either case, we can use **tree diagrams** to solve these problems.

**Example: **Penny will play 2 games of badminton against Monica. Penny’s chances of winning each game are outlined in the tree diagram below, where P denotes a Penny win and M denotes a Monica win. Work out the probability of Penny winning exactly one match.

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On a tree diagram, the first match is described by the left set of branches, and the second match is described by the right set of branches . As you can see, both outcomes of the first match then have their own pair of branches describing the outcomes of the second match – this way, all 4 possible outcomes of the 2 matches are included.

In this case, the 2 matches are **dependent** – if Penny wins the first match, then the chance of her winning the second is 0.8, but if she loses the first match, her chance of winning the second is 0.5.

To use a tree diagram, we have to introduce the **and/or rule** of probability. To understand this rule, imagine we have two events, A and B. To find the probability of A **and **B occurring, we **multiply **their probabilities, but to find the probability of A **or **B occurring, we **add **their probabilities. On a tree diagram, this means we multiply along the branches, and then add the resulting probabilities at the end.

So, we are looking for the probability that Penny wins exactly one match. This means she could either win the first and lose the second or lose the first and win the second. By multiplying along the P line for the first match and the M line for the second, we get

\text{P(P wins 1st, M wins 2nd)}=0.7\times0.2=0.14.

Similarly, by multiplying along the M line for the first match and the P line for the second, we get

\text{P(M wins 1st, P wins 2nd)}=0.3\times0.5=0.15.

Either the first **or **second of these situations results in exactly one win for Penny, so to find the probability of one win for Penny, we add them: P(\text{Exactly 1 P win) }=0.14+0.15=0.29.

If you are on the **higher course**, you also need to understand **conditional probability** – the chance of one thing happening given that you know something else has already happened. We use P(A|B) to denote “probability of A happening given B has happened”. In the case of a tree diagram, conditional probability works nicely – if we were to ask, “What is the probability that Monica wins the second match given that Penny has already won the first?”, all we need to do is look to the branch that represents the event of Monica winning the 2nd match after a Penny win. So, the probability of Monica winning the second given that Penny wins the first is 0.2. In other words,

\text{P(Monica wins 2nd | Penny wins 1st)} = 0.2

### Example Questions

1) A bag contains beads that are either red, blue, or green. Work out the probability of a) picking a blue bead out of the bag, and b) picking a green bead out of the bag.

The key to answering this question is recognising that these are **exhaustive** probabilities (the bead you choose has to be either red, blue, or green), so they must add up to 1. Therefore,

0.25+5x+4x=1

Subtract 0.25 from both sides to get

9x=0.75

So, we get that x=0.75\div9=\dfrac{1}{12}. Now we have x, we know that the **probability of getting a blue bead is**

5x=\dfrac{5}{12},

and the **probability of getting a green bead is**

4x=\dfrac{4}{12}.

2) Hannah and Mike want to know how many cars are black in their neighbourhood. To do so, they count the cars that pass by their window at home and note down if they are black or not. Hannah spends 30 minutes looking at cars in the morning and Mike spends 1 hour looking at cars in the evening.

- a) Based on her data, Hannah claims that 40% of the cars in their neighbourhood are black. Comment on the accuracy of her claim.
- b) Using all the data collected, calculate the relative frequency of a car being black to 3dp.
- c) Given that there are 5,000 people in their neighbourhood, use your answer to part b) to work out the expected frequency of black cars in Hannah and Mike’s neighbourhood.

- a) If we calculate the relative frequency of a car being black using Hannah’s data, we get

\dfrac{32}{32+48}=0.4=40\%.

So, according to her data, her statement is correct. However, this might not be accurate compared to the relative frequency according to Mike’s data (since he collected more) or according to all the data which we’ll see in part b).

- b) We’ll calculate the relative frequency using all the data by combining Hannah’s and Mike’s data. In other words, we will divide the total number of black cars seen by the total number of cars seen. Doing so, we get:

\dfrac{32+41}{32+41+111+48} = 0.315\text{ (3dp)}.

- c) To find the expected frequency of black cars in their neighbourhood, we will multiply the probability of a car being black (i.e., the relative frequency calculated in part b) by the total number of cars. We get the expected frequency of black cars in the neighbourhood to be

0.315\times5,000=1,575\text{ black cars}.

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