Finding the Probability of a Compound Event | Grade 7 Math | LevelKite

A compound event is an event that combines the results of two or more actions, such as flipping a coin and rolling a number cube, or drawing two marbles from a bag.

You already know how to look at a sample space (the set of all possible results, called outcomes) and pick out the outcomes that make up an event. In this lesson, you count those outcomes and turn the counts into a probability: a number from $0$ to $1$ that tells you how likely the event is.

Start with a small case. You flip a fair coin twice. The coin can land heads (H) or tails (T), and fair means both sides have the same chance on every flip.

Write each result as an ordered outcome: the first letter is the first flip and the second letter is the second flip. The sample space is HH, HT, TH, TT, with each ordered outcome listed once. HT and TH are different outcomes, because the head comes on a different flip.

Each flip is fair, and one flip does not change what the other flip can show. So these $4$ ordered outcomes are equally likely, which means each one has the same chance.

Now take the event "exactly one head". The outcomes that make it happen are called the favourable outcomes: here, HT and TH. That is $2$ of the $4$ equally likely outcomes, so the probability is $\frac{2}{4} = \frac{1}{2}$ . Half of the equally likely outcomes make the event happen.

Key concept

Probability of a compound event

When a sample space lists every ordered outcome exactly once and all of those outcomes are equally likely, the probability of a compound event is the fraction of the outcomes that are favourable: favourable over total. Write $P(\text{event})$ for "the probability of the event".

Both conditions matter. If a list leaves out or merges ordered outcomes, or its outcomes are not equally likely, the fraction can be wrong.

Watch for drawing without replacement (you do not put the first object back). The first draw changes what is available later. You count the later draw out of fewer objects, so the denominator for that draw is smaller. Build the sample space from what is really left.

P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

Sample space for one flip of a fair coin and one roll of a fair number cube, written (coin, cube)

Number cube shows	Coin shows H	Coin shows T
1	(H, 1)	(T, 1)
2	(H, 2)	(T, 2)
3	(H, 3)	(T, 3)
4	(H, 4)	(T, 4)
5	(H, 5)	(T, 5)
6	(H, 6)	(T, 6)

Worked example

A coin and a number cube

You flip a fair coin once: it lands heads (H) or tails (T), and each side is equally likely. You also roll a fair number cube once: it shows $1$ , $2$ , $3$ , $4$ , $5$ or $6$ , and each face is equally likely. What is the probability that the coin shows heads and the cube shows a number less than $5$ ? Use the sample space in the table above.

Count the whole sample space. Each ordered outcome is written (coin, cube). The table above has 6 rows with 2 outcomes in each row, and every possible pair appears exactly once.
$6 \times 2 = 12$
Check that the 12 outcomes are equally likely. The coin is fair, the cube is fair, and the coin does not change what the cube can show. So no pair has a better chance than another, and you can use favourable over total.
Find the favourable outcomes: heads together with a number less than 5. The numbers less than 5 are 1, 2, 3 and 4, because 5 itself is not less than 5. That gives 4 favourable outcomes.
$\text{(H, 1), (H, 2), (H, 3), (H, 4)}$
Write favourable over total, which works here because the 12 ordered outcomes are each listed once and are equally likely. Then divide the numerator and the denominator by 4. Dividing both by the same non-zero number keeps the value of the fraction. The result is between 0 and 1, as every probability is.
$P(\text{heads and less than 5}) = \dfrac{4}{12} = \dfrac{4 \div 4}{12 \div 4} = \dfrac{1}{3}$

Answer

The probability of heads and a number less than $5$ is $\frac{1}{3}$ , because $4$ of the $12$ equally likely ordered outcomes are favourable.

Many compound events come from drawing twice from the same set of objects. Two phrases tell you how the drawing works:

With replacement: you put the first object back before the second draw, so the second draw has the same choices as the first.
Without replacement: you keep the first object out, so the second draw has one fewer choice.

The next two examples use the same bag. It holds $4$ marbles of the same size: $2$ red and $2$ blue. You draw without looking, and each marble in the bag has the same chance of being picked. To list ordered outcomes, name the marbles R1, R2, B1 and B2. The two red marbles are different objects, so (R1, R2) and (R2, R1) are different ordered outcomes.

Two draws with replacement, written (first, second)

First draw	Second draw: R1	Second draw: R2	Second draw: B1	Second draw: B2
R1	(R1, R1)	(R1, R2)	(R1, B1)	(R1, B2)
R2	(R2, R1)	(R2, R2)	(R2, B1)	(R2, B2)
B1	(B1, R1)	(B1, R2)	(B1, B1)	(B1, B2)
B2	(B2, R1)	(B2, R2)	(B2, B1)	(B2, B2)

Worked example

Two draws with replacement

The bag holds the marbles R1, R2, B1 and B2. You draw one marble, put it back, mix the bag, and draw again. On each draw, every marble in the bag is equally likely. What is the probability that you draw a red marble both times? Use the table above.

Count the sample space. The first marble goes back, so the second draw has the same 4 choices as the first. Each of the 4 first draws pairs with 4 second draws, and the table above lists every ordered pair exactly once.
$4 \times 4 = 16$
Check equally likely. You put the first marble back, so the first draw does not change what the second draw can be. Every marble has the same chance on each draw, so all 16 ordered pairs are equally likely and you can use favourable over total.
Find the favourable outcomes: red first and red second. A marble can appear twice in a pair here, because you put it back. That gives 4 favourable outcomes.
$\text{(R1, R1), (R1, R2), (R2, R1), (R2, R2)}$
Write favourable over total for these 16 equally likely ordered outcomes. Then divide the numerator and the denominator by 4, which gives an equivalent fraction.
$P(\text{both red}) = \dfrac{4}{16} = \dfrac{4 \div 4}{16 \div 4} = \dfrac{1}{4}$

Answer

With replacement, the probability of drawing a red marble both times is $\frac{1}{4}$ , because $4$ of the $16$ equally likely ordered outcomes are favourable.

Two draws without replacement, written (first, second); the same marble cannot be drawn twice

First draw	Second draw: R1	Second draw: R2	Second draw: B1	Second draw: B2
R1	not possible	(R1, R2)	(R1, B1)	(R1, B2)
R2	(R2, R1)	not possible	(R2, B1)	(R2, B2)
B1	(B1, R1)	(B1, R2)	not possible	(B1, B2)
B2	(B2, R1)	(B2, R2)	(B2, B1)	not possible

Worked example

Two draws without replacement

The same bag holds R1, R2, B1 and B2. This time you draw one marble, keep it out of the bag, and then draw a second marble. On each draw, every marble still in the bag is equally likely. What is the probability that you draw a red marble both times? Use the table above.

Notice what changes. After the first draw, only 3 marbles are left, so the second draw has 3 possible results instead of 4. A pair such as (R1, R1) cannot happen, because R1 is no longer in the bag.
Count the sample space. Each of the 4 first draws pairs with the 3 marbles that are left, and the table above lists every possible ordered pair exactly once.
$4 \times 3 = 12$
Check equally likely. Each of the 4 marbles has the same chance of being first. Whichever marble comes out, each of the 3 marbles left has the same chance of being second. So all 12 ordered pairs are equally likely and you can use favourable over total.
Find the favourable outcomes: red first and red second, using two different marbles. That gives 2 favourable outcomes.
$\text{(R1, R2), (R2, R1)}$
Write favourable over total for these 12 equally likely ordered outcomes. Then divide the numerator and the denominator by 2, which gives an equivalent fraction.
$P(\text{both red}) = \dfrac{2}{12} = \dfrac{2 \div 2}{12 \div 2} = \dfrac{1}{6}$

Answer

Without replacement, the probability of drawing a red marble both times is $\frac{1}{6}$ , because only $2$ of the $12$ equally likely ordered outcomes are favourable.

Takeaway: when a sample space lists every ordered outcome exactly once and all of those outcomes are equally likely, the probability of a compound event is the number of favourable outcomes divided by the total number of outcomes. When you draw without replacement, the later draw has fewer available outcomes, so build the sample space from what is left.

Next, you will choose and design a simulation (an experiment that acts out a situation) and use it to find the probability of a compound event.