Probability 2: Combined Events

Full lesson notes

Everything covered in this lesson, in one place - useful for revision or printing.

\( 2 \times 6 = 12 \text{ outcomes} \)

In Probability 1 you worked with ONE thing at a time — one die, one coin, one bag.

This lesson is about what happens when two things happen together, and how to reason honestly about experiments.

• Count the outcomes when two events combine
• Build a two-way table to list them without missing any
• Calculate probability from a combined sample space
• Work out how many times to expect an outcome
• Understand why a coin has no memory
• Decide whether data really shows something is biased

1. Combined Events

\( \text{coin} \;+\; \text{die} \)

A combined event happens when you do two things in the same experiment.

Each outcome now has two parts — one from each thing.

\( \{ HH;\ HT;\ TH;\ TT \} \)

Toss two coins. It is tempting to say there are 3 outcomes: two heads, one head, two tails. That is wrong.

Label the coins in your head — "coin 1" and "coin 2" — and the trap disappears.

\( 2 \times 6 = 12 \qquad 6 \times 6 = 36 \)

To count the outcomes of a combined event, multiply the number of outcomes of each part.

Adding gives \(2+6=8\), which is wrong — every one of the 2 coin results can pair with every one of the 6 die results.

\( n(\text{total}) = n(\text{first}) \times n(\text{second}) \)

Key idea: combine by multiplying. And order matters — HT and TH are two different outcomes, not one.

2. Two-Way Tables

\( \text{list them in ORDER} \)

Once there are 12 or 36 outcomes, writing them down as you think of them will fail — you will miss some or repeat some.

A two-way table forces a complete, systematic list.

\( 2 \text{ rows} \times 6 \text{ columns} = 12 \text{ cells} \)

Put one thing down the side and the other across the top. Every cell is one outcome:

Count the cells: 12. The table and the multiplication agree.

\( P(\text{head and } 6) = \dfrac{1}{12} \)

Now any question is just "how many cells match?"

Only one cell is a head with a six, out of 12 cells, so \( P = \tfrac{1}{12} \).

\( \text{every cell} = \text{one outcome} \)

Key idea: build the table first, count the cells to get the total, then count the cells that match the event. A tree diagram does the same job — use whichever you prefer.

3. Probability of Combined Events

\( P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} \)

Nothing new to learn here — the formula from Probability 1 still works. The only change is that the sample space is bigger.

\( \text{exactly one} \neq \text{at least one} \)

Two coins: \( \{HH;\ HT;\ TH;\ TT\} \). These two questions are not the same.

"At least one" includes two heads. "Exactly one" excludes it. Read the words carefully.

\( P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6} \)

Two dice give 36 outcomes. Which of them add up to 7?

That is 6 outcomes out of 36, so \( P = \tfrac{6}{36} = \tfrac{1}{6} \).
Notice \((1;6)\) and \((6;1)\) are counted separately — same reason as HT and TH.

\( P = \dfrac{\text{matching cells}}{\text{all cells}} \)

Key idea: the formula never changes. Build the sample space, count carefully, and watch for the words "exactly", "at least" and "at most".

4. Expected Number

\( \text{expected number} = P \times \text{number of trials} \)

If you know the probability, you can predict roughly how many times something will happen.

\( 0,2 \times 300 = 60 \)

On a spinner, \( P(\text{red}) = 0,2 \). The spinner is spun 300 times.

The probability can be a fraction, a decimal or a percentage — convert it first, then multiply.

\( \text{expect } 60 \;\neq\; \text{always } 60 \)

"Expected" does not mean guaranteed.

The expected number tells you what is typical, and it becomes more reliable the more trials you do. Getting 58 instead of 60 is not evidence of anything.

\( E = P \times n \)

Key idea: multiply the probability by the number of trials. Write your answer as "about 60 times" — an expected value is a prediction, not a fact.

5. Independent Events

\( P(\text{head}) = \tfrac{1}{2} \text{ every single toss} \)

Two events are independent when the first one does not change the chances of the second.

Every toss starts fresh at \( \tfrac{1}{2} \).

\( HHHHH \;\rightarrow\; P(\text{next head}) = \dfrac{1}{2} \)

A coin lands on heads five times in a row. What is the probability the next toss is a head?

The five heads have already happened. They cannot reach forward and change the sixth toss.
Tails is not "due".

\( \text{"tails is due"} = \text{wrong} \)

Believing that a run of heads makes tails more likely is such a common mistake that it has a name: the gambler's fallacy.

In an exam, the answer is always: still \( \tfrac{1}{2} \), because the tosses are independent.

\( \text{past results} \;\nrightarrow\; \text{next result} \)

Key idea: for independent events, what already happened tells you nothing about what happens next. Say "independent" or "the coin has no memory" to earn the reason mark.

6. Fair or Biased?

\( \text{actual} \quad \text{vs} \quad \text{expected} \)

To judge whether something is fair, compare what actually happened with what you expected.

Both parts of step 3 matter.

\( 13 \text{ heads out of } 20 \)

A coin is tossed 20 times and gives 13 heads. Expected: \( \tfrac{1}{2} \times 20 = 10 \). Is it biased?

13 instead of 10 is an ordinary result for a fair coin. To call a coin biased you would need hundreds of tosses.

\( 210 \text{ sixes out of } 600 \qquad (\text{expected } 100) \)

A die is rolled 600 times and gives 210 sixes. Expected: \( \tfrac{1}{6} \times 600 = 100 \).

Now the conclusion is safe: the die is almost certainly biased towards six. Same reasoning as the coin — the difference is the sample size and the size of the gap.

\( \text{big gap} + \text{many trials} \Rightarrow \text{biased} \)

Key idea: a surprising result from a few trials proves nothing; the same result from many trials is strong evidence. Always quote the expected number in your answer.

\[ n(\text{total}) = n_1 \times n_2 \qquad E = P \times n \]

You now know how to:

• Count combined outcomes by multiplying
• Build and read a two-way table
• Tell "exactly one" apart from "at least one"
• Calculate an expected number, and say why it is only an estimate
• Explain independence and spot the gambler's fallacy
• Judge fairness using both the gap AND the number of trials

Test yourself with the quiz below.