Probability of an event

After looking at Example 2, we can now ask: What is Hassan's chance of success? Or: What is the probability of Hassan's success?

All six outcomes are equally likely. Of these, only two  – 1 and 4 – are favourable to the event 'the result is a perfect square'. So we say the probability of success is \(\dfrac{2}{6}=\dfrac{1}{3}\) .

So we can define the probability of an event.

For an experiment in which all the outcomes are equally likely:

We use the letter 'P' instead of 'probability of'. In the above example, we write the result as:

P (perfect square) = \(\dfrac{1}{3}\)

Example 3

A bag contains 10 marbles numbered 1 to 10. One marble is removed.

1. List the sample space.
2. Write down the outcomes favourable to the event 'a marble with an odd number is taken out'.
3. What is the probability of getting an odd number?

Solution

1. Sample space: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
2. Favourable outcomes (odd numbers): 1, 3, 5, 7, 9
3. \begin{align}P(\text{getting an odd number})&=\dfrac{5}{10}\\\\ &=\dfrac{1}{2}\end{align}

Representations of probabilities

We have defined the probability of an event occurring as a fraction. At times, it is more appropriate to represent these probabilities as either decimals or percentages.

For example, when one is talking about the weather, one may say 'There is a 50% chance that it will rain tomorrow'. This is the same as saying that there is a probability of \(\dfrac{1}{2}\) that it will rain tomorrow, or a 0.5 chance of rain tomorrow.

When using probabilities, we can express them as fractions, decimals or percentages.

Example 4

I toss a die. What is the probability of getting a 2 or a 3?

Solution

1. Sample space: 1, 2, 3, 4, 5, 6
2. Favourable outcomes (2 or 3)
3. \begin{align}P(\text{two or three})&=\dfrac{2}{6}\\\\ &=\dfrac{1}{3}\\\\ &=0.\dot{3}\end{align}

Probability of an event not happening

Sometimes we are interested in an event not happening. For example, the weather bureau predicts there is a \(\dfrac{1}{5}\) chance that it will rain tomorrow. However, it is sports day tomorrow so we would like to know the probability of it not raining.

As the sum of the probabilities is 1,

\begin{align}P(\text{not raining})&=1\ –P(\text{raining})\\\\ &=1\ –\dfrac{1}{5}\\\\ &=\dfrac{4}{5}\end{align}

In this case, the probabilities would most likely be expressed as percentages – that is, a 20% chance of rain and an 80% chance of it not raining.