Hong Kong
Stage 4 - Stage 5

Relative frequencies of And/Or events

Lesson

Consider the following Venn Diagram that shows the number of people that like the colours red, green, both or neither.

From the diagram we can see that

P(Green)$=$=$\frac{10+18}{60}=\frac{28}{60}$10+1860=2860$=$=$\frac{7}{15}=47%$715=47%

P(Red)$=$=$\frac{10+14}{60}=\frac{24}{60}$10+1460=2460$=$=$\frac{2}{5}=40%$25=40%

And

What about the probability of Green AND Red?   Well this is just the overlap of the circles.

P(Green and Red)$=$=$\frac{10}{60}=\frac{1}{6}$1060=16$=$=$17%$17%

There is a special symbol in probability (and set theory) that we use to denote the word AND it is$\cap$.

So we would write P(Green and Red) like this: P(Green$\cap$Red)

Or

What about the probability of Green OR Red?  Well this relates to anything inside the two circles. We can add up the values of the like green only, like red only or like both.

P(Green OR Red)$=$=$\frac{18+10+14}{60}=\frac{42}{60}$18+10+1460=4260$=$=$\frac{7}{10}=70%$710=70%

We can also see from this that this is P(Green)$+$+P(Red)$-$P(Green$\cap$Red)$=$=$47+40-17%$47+4017%$=$=$70%$70%

By looking at the diagram can you see why?

Well each time we calculated the P(Green) or P(Red) we added in the 'like both'.  This means that if we were to just add the Probabilities of P(Green) and P(Red) [$40+47%$40+47%], we would have counted the 'like both' section twice. So when calculating the OR possibility, we have to subtract the overlap.

You may have also noticed the the P(Green or Red)$=$=$1$1$-$P(like neither) as they are complementary events, so P(Green or Red)$=$=$100-30=70%$10030=70%

There is a special symbol in probability (and set theory) that we use to denote the word OR it is$\cup$.

So we would write P(Green or Red) like this: P(Green$\cup$Red)

Now we have another handy little relationship that we can use when solving probability problems.

Probability Relationships

P(A)$+$+P(A')$=1$=1

P(A$\cup$B)$=$=P(A)$+$+P(B)$-$P(A$\cap$B)

Worked example

A standard die is rolled. What is the probability that the number is even or less than $5$5?

Think:  Firstly we need to define the events, so

Event A: Number is even (i.e the numbers $2,4,6$2,4,6)

Event B: Number is less than $5$5 (i.e the numbers $1,2,3,4$1,2,3,4)

We can also easily identify the probabilities of both

P(A)$=$=$\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{3}{6}$total favourable outcomes total possible outcomes =36

P(B)$=$=$\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}=\frac{4}{6}$total favourable outcomes total possible outcomes =46

and the probability of both, P(A$\cap$B)$=$=$\frac{\text{total favourable outcomes }}{\text{total possible outcomes }}$total favourable outcomes total possible outcomes

Total favourable will be $\left\{2,4\right\}${2,4}, so two:

P(A$\cap$B)$=$=$\frac{2}{6}$26

We can now use the relationship P(A$\cup$B)$=$=P(A)$+$+P(B)$-$P(A$\cap$B)

Do: P(A$\cup$B)$=$=P(A)$+$+P(B)$-$P(A$\cap$B)

P(A$\cup$B)$=$=$\frac{3}{6}+\frac{4}{6}-\frac{2}{6}$36+4626

P(A$\cup$B)$=$=$\frac{5}{6}$56

Practice questions

QUESTION 1

A standard six-sided die is rolled and then a coin is tossed.

1. Create a probability tree for this situation.

2. What is the probability of an even number and a head?

3. What is the probability of an even number, a head, or both?

QUESTION 2

A coin is tossed, then the spinner shown is spun.

The blue segment is twice as big as the yellow one.

1. Create a probability tree that represents all possible outcomes.

2. What is the probability of throwing a heads and spinning a yellow?

3. What is the probability of throwing a heads, or spinning a yellow , or both?

QUESTION 3

A florist collected a sample of her flowers and divided them into the appropriate categories.

1. What is the probability that a flower is not red but has thorns?

2. What is the probability that a flower is not red and does not have thorns?