# Probability Basics & Listing Outcomes Worksheets, Question sand Revision

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## What you need to know

Probability

Probability is the study of how likely things are to happen.

All probabilities must fall between 0 and 1, where 0 means ‘impossible’ and 1 means ‘certain’. If the probability of an outcome is closer to 0 then it’s unlikely to happen, but if it’s closer to 1 then it’s likely to happen.

We can express probabilities as fractions, decimals or percentages (as shown on the diagram above).

NOTATION: To say “the probability of X happening” we write P(X). This notation will be used throughout the rest of this page.

If you consider all possible outcomes of an event – this is known as exhausting all options – then their probabilities must add up to 1. For example, the chance of flipping and coin and landing on heads is one half, and the chance of landing on tails is the same: one half. These are the only two possibilities (we’ve exhausted all options), and indeed we get that

$\text{P(heads)}+\text{P(tails)}=\dfrac{1}{2}+\dfrac{1}{2}=1$

Example: Sarah and Charlie are playing a game. There are 3 possible outcomes for the game: Sarah wins, Charlie wins, or it’s a draw. The chance of Sarah winning is 0.6. A draw is half as likely as Sarah winning. What is the probability that Charlie wins?

To find the probability of a draw, we should half the probability that Sarah wins:

$\text{P(draw) }=0.6\div 2=0.3$

Exhausting all options, we know that P(Sarah wins), P(draw), and P(Charlie wins) must all add up to 1. Written as an equation, we get

\begin{aligned} 0.6+0.3+\text{P(Charlie wins)}&=1 \\ 0.9+\text{P(Charlie wins)}&=1\end{aligned}

Then, subtracting 0.9 from both sides of the equation, we get

$\text{P(Charlie wins)}=1-0.9=0.1$

The And/Or Rule

Suppose we’re considering two outcomes that might happen: A and B. Then, the and/or rule of  probability is

$\text{P(both A and B)}=\text{P(A)}\times\text{P(B)}$

$\text{P(Either A or B)}=\text{P(A)}+\text{P(B)}$

The rule can be summarised as: if it’s and, we times, but if it’s or, we add.

Example: Fae and Sunny play badminton together. It’s impossible to draw in badminton. The chance of Fae winning any match against Sunny is $\frac{1}{6}$. Calculating the probability that Sunny wins 2 matches of badminton against Fae.

Firstly, the only two possible outcomes are Sunny wins or Fae wins. Therefore, their probabilities must add up to 1. In other words, if we subtract P(Fae wins) from 1, we get

$\text{P(Sunny wins)}=1-\dfrac{1}{6}=\dfrac{5}{6}$

Now, how do we think about Sunny winning two matches? We have to split it up into her winning one, and then winning another. The key word there is ‘and’ – we’re finding the probability of 1 win and then another win, which means we need to multiply the chance of her winning 1 game by the chance of her winning another game. Doing this, we get

$\text{P(Sunny wins 2 games)}=\dfrac{5}{6}\times\dfrac{5}{6}=\dfrac{25}{36}$

Listing Outcomes

Listing outcomes is exactly what it sounds like – given a scenario, list every possible outcome. Usually, to save writing time, we’ll use single letters to denote outcomes, e.g. when flipping a coin, we’ll say H = heads and T = tails. Let’s see how this works out in the following example.

Example: Megan flips a coin 3 times and records the results of each 3 flips, in order. List all possible outcomes.

As mentioned, we’re going to use H = heads and T = tails. This means that to represent the outcome of getting “tails, heads, tails”, we’ll write THT.

Furthermore, she’s recording the order of the flips too, meanings THT is different to TTH.

Now, the main aspect of listing outcomes is making sure you don’t miss any out. To try to avoid this, we’re going to approach the problem methodically. We’re going to start by considering all scenarios where the first throw is heads. These are all going to be of the form H _  _. Doing so, we get

HHH, HHT, HTT, HTH

Take a moment to make sure that’s all of them. If you’re still slightly unsure, be even more methodical about your approach. Start by considering all options where the first two throws are heads, i.e. all the ones of the form HH_ and so on.

Now this is done we can move on to the next bit: considering all scenarios where the first throw is tails. They will all be of the form T _ _. Doing this, we get

TTT, THT, THH, TTH

Given that heads and tails are the only possible outcomes of the first throw, and that we’ve covered both of them, we must have covered everything. And indeed, we have! Therefore, the complete list of outcomes is

HHH, HHT, HTT, HTH, TTT, THT, THH, TTH

## KS3 Maths Revision Cards

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### Example Questions

Since A, B, and C are the only possible outcomes, their probabilities must add up to 1. Therefore, if we take both P(A) and P(B) away from 1, we’ll get P(C). In order to do this, we should express both probabilities in the same form. Here, we’re going to choose to turn P(A) into a fraction.

‘Percent’ means ‘parts out of 100’, so

$55\%=\dfrac{55}{100}=\dfrac{11}{20}$

Now we get

$\text{P(A)}+\text{P(B)}=\dfrac{2}{5}+\dfrac{11}{20}=\dfrac{19}{20}$

Then, subtracting this from 1 we get the answer to be

$\text{P(C)}=1-\dfrac{19}{20}=\dfrac{1}{20}$

Answers in the form of 0.05 or 5% are also acceptable.

#### Is this a topic you struggle with? Get help now.

a) If the probability of a horror movie is the same as for a sci-fi movie, let’s call that probability $x$. In other words,

$\text{P(Horror movie)}=\text{P(Sci-fi movie)}=x$

Then, since he only has 3 choices of genre and probabilities must add up to 1, we get

$0.56+x+x=1$

Firstly, $x+x$ becomes $2x$. Then, subtracting 0.56 from both sides we get

$2x=0.44$

Lastly, dividing by two gives us our answer:

$\text{P(Sci-fi movie)}=0.44\div 2=0.22$

b) This is an or scenario, which means we simply need to add the probability of Jimmy picking a sci-fi movie to that of him picking a romantic comedy.

$\text{P(Sci-fi or romantic comedy)}+\text{P(Sci-fi)}+\text{P(Romantic comedy)}=0.56+0.22=0.78$

#### Is this a topic you struggle with? Get help now.

Let’s start by assuming Macy chooses black trousers. Then she has 3 choices for the colour of her jumper, and so the 3 possible outcomes are

BO, BY, BW

Suppose instead she chooses navy trousers. Then the 3 possible outcomes are

NO, NY, NW

Lastly, suppose she chooses purple trousers. Then the remaining possible outcomes are

PO, PY, PW

BY was already given in the question, so the full list of other possible outcomes is

BO, BW, NO, NY, NW, PO, PY, PW

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