Probability and Expected Outcomes for the ESAT

Defining Sample Spaces and Events

In the study of probability, the sample space is the complete set of all possible outcomes of an experiment. For instance, if an experiment consists of tossing a coin twice and recording the results, the sample space is written as $\{HH, HT, TH, TT\}$.

An event is a specific outcome or a subset of the sample space that we are interested in. In the previous coin tossing example, we might define an event as 'obtaining exactly one head'. This event corresponds to the subset $\{HT, TH\}$ within the sample space.

Fairness and Equally Likely Outcomes

When we describe an object as fair, we mean that every possible outcome has an equal chance of occurring. A fair die has an equally likely chance of landing on any of its six faces, and a fair spinner has an equally likely chance of stopping on any of its sections. When events are equally likely, the probability of any single outcome is 1 divided by the total number of possible outcomes.

Calculating Expected Outcomes

We can use the theoretical probability of an event to predict the results of multiple future experiments. This is done by multiplying the probability by the total number of trials. The formula is expressed as:

$\text{Expected Outcome} = P(\text{Event}) \times n$

where $n$ is the number of times the experiment is repeated. For example, to find the expected number of 6s when rolling a fair six-sided die 30 times, we calculate $\frac{1}{6} \times 30 = 5$.

Understanding Experimental Variability

It is important to understand that if an experiment is repeated, the outcome may be different every time. Calculating an expected outcome does not guarantee that the result will occur exactly as predicted. If the expected number of 6s in 30 rolls is 5, it remains entirely possible to roll a 6 fewer than 5 times or more than 5 times. The expected outcome is a theoretical average rather than a certain result.

Worked Example: Die Rolling Trials

Suppose a fair six-sided die is rolled 90 times. We can determine the expected frequency for various outcomes based on the fact that each face has a probability of $\frac{1}{6}$.

a) How many times would it be expected to land on 5?

The probability of landing on 5 is $P(5) = \frac{1}{6}$. Expected number of 5s in 90 rolls = $\frac{1}{6} \times 90 = 15$.

b) How many times would it be expected to land on an even number?

The even outcomes on a die are 2, 4, and 6. Therefore: $P(\text{even number}) = P(2) + P(4) + P(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$. Expected number of even numbers in 90 rolls = $\frac{1}{2} \times 90 = 45$.

c) How many times would it be expected to land on a square number?

The square numbers between 1 and 6 are 1 and 4. Therefore: $P(\text{square number}) = P(1) + P(4) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$. Expected number of square numbers in 90 rolls = $\frac{1}{3} \times 90 = 30$.

Worked Example: Four Sided Spinner

A fair four-sided spinner has four sections: red, yellow, green, and blue.

If the spinner is spun 100 times, how many times would 'yellow' be expected?

Since the spinner is fair, the probability of yellow is $\frac{1}{4}$. Expected number of 'yellow' in 100 spins = $P(\text{yellow}) \times 100 = \frac{1}{4} \times 100 = 25$.

Explain why the actual number of 'yellow' in 100 spins might not be exactly 25.

The actual number of times the spinner lands on yellow will not necessarily match the theoretical expectation because of randomness. Any total from 0 to 100 is possible, although totals closer to 25 are much more likely to occur.