Probability of Combined Events

For some probability questions, you may be asked to calculate the probability of more than one event happening.

Example: Using the ‘AND’ Rule

If I flipped a coin and rolled a dice, what is the probability that I got a head and a 6?

In this case, we would need to work out the probability of each event separately, and then multiply the two fractions together.

Solution:

1. Probability of flipping a head = \frac{1}{2}

2. Probability of rolling a 6 on a 6-sided die: \frac{1}{6}

3. Now, multiply these two fractions together:

\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}

Notice that in this question, we were asked for the probability of the head AND the 6. When you are asked for two events together, you must multiply the fractions.

However, sometimes you may be asked for one event or another event. Here, the keyword is ‘OR’ – this means that we would ADD the fractions, not multiply. Here’s an example:

Example: Using the ‘OR’ Rule

If I flipped a coin and rolled a dice, what is the probability that I get a head or a 6?

As we could have either a head or a 6 (don’t need both together), we would need to work out the probability of each event separately, and then add the two fractions together.

Solution:

1. Probability of flipping a head = \frac{1}{2}

2. Probability of rolling a 6 on a 6-sided dice = \frac{1}{6}

3. Now, add these two fractions together:

\frac{1}{2} + \frac{1}{6} = \frac{4}{6}

= \frac{2}{3}

Summary:

Top tip!

When calculating a probability, the answer must NEVER be greater than 1. If you have added two fractions together and found an answer greater than 1, then it is incorrect.

Independent or Dependent Events

Another part of probability to consider is whether the first event affects the outcome of the second event. For example, consider the probability of the bus being late, and the probability of being late to school.

The second event (being late to school) will depend on the outcome of the first event (whether the bus was late). These are known as dependent events- the probability will change depending on the outcome of the first event.

However, if the first event does NOT affect the second event (for example, the probability of doing your homework, and the probability of being late to school), these are known as independent events.

Example: Dependent Events

Let’s revisit Emma’s buttons again. She has 6 pink, 4 green and 5 blue buttons in a box. She selects two buttons out of the box. What is the probability that they’re both blue?

If Emma is taking two buttons consecutively without replacement, then the second draw depends on the outcome of the first draw.

Therefore, these must be dependent events, as the probability of picking a blue button the second time depends on whether she selected a blue button the first time.

Here’s why:

First event: Selecting a blue button out of the original 15 buttons in the box.

Out of the 15 buttons, there are 5 which are blue. Therefore, the probability of the first button being blue is \frac{5}{15} = \frac{1}{3}

Now that she has taken one blue button, this decreases the total number of buttons by 1, and the number of blue buttons by 1:

Second event: Choosing a second blue button

Now, there are a total of 14 buttons, 4 of which are blue. The probability of her picking a second blue button is now \frac{4}{14}

Combining the probabilities:

The question asks us for the probability of picking two blue buttons. This is the same as selecting a blue AND another blue button. Therefore, we must use the ‘AND’ rule- we must multiply the two fractions.

\frac{1}{3} \times \frac{4}{14} = \frac{4}{42}