Maths

Probability - Higher

If you are studying the higher paper, you will also need to know about tree diagrams, the and/or rule and conditional probabilities.

Tree diagrams allow us to see all the possible outcomes of an event and calculate their probability. Each branch in a tree diagram represents a possible outcome.

If two events are **independent**, the outcome of
one has no effect on the outcome of the other. For example, if we toss two coins, getting heads with the first coin will not affect the probability of getting heads with the
second.

A tree diagram which represent a coin being tossed three times looks like this :

From the tree diagram, we can see that there are eight possible outcomes. To find out the probability of a particular outcome, we need to look at all the available paths (set of branches).

The sum of the probabilities for any set of branches is always 1.

Also note that in a tree diagram to find a probability of an outcome we **multiply along the branches** and **add vertically**.

The probability of three heads is:

P (H H H) = ^{1}/_{2} × ^{1}/_{2}
× ^{1}/_{2}** = **^{1}/_{8}

P (2 Heads and a Tail) = P (H H T) + P (H T H) + P (T H H)

= ^{1}/_{2} × ^{1}/_{2}
× ^{1}/_{2}** + **^{1}/_{2} × ^{1}/_{2} × ^{1}/_{2}** + **^{1}/_{2} × ^{1}/_{2} × ^{1}/_{2}

= ^{1}/_{8}** + **^{1}/_{8}** + **^{1}/_{8}

= ^{3}/_{8}

- Question
- Bag A contains three red marbles and four blue marbles.
- Bag B contains five red marbles and three blue marbles.
- A marble is taken from each bag in turn.

Find the missing probabilities for the tree diagram:

- Answer

- Question
What is the probability of getting a blue bead followed by a red?

- Answer
Multiply the probabilities together:

- P (blue and red) =
^{4}/_{7}×^{5}/_{8} - =
^{20}/_{56} - =
^{5}/_{14}

- P (blue and red) =

- Question
What is the probability of getting a bead of each colour?

- Answer
P (blue and red or red and blue) = P (blue and red) + P (red and blue)

- =
^{4}/_{7}×^{5}/_{8}+^{3}/_{7}×^{3}/_{8} - =
^{20}/_{56}+^{9}/_{56} - =
^{29}/_{56}

- =

When two events are **independent**, the outcome of one has no effect on the outcome of the other.

When two events are said to be **mutually exclusive**
they cannot happen at the same time. For example, if a die is
thrown, the events 'obtaining a 6' and 'obtaining a 1' are mutually
exclusive, as they cannot happen at the same time.

The events 'obtaining a six' and 'obtaining an even number' are
**not** mutually exclusive, because throwing
a six fits into both categories.

When two events, A and B, are **independent**:

**P(A and B) = P(A) x P(B)**

When two events, A and B, are **mutually exclusive**:

**P(A or B) = P(A) + P(B)**

Notice how the word **and** has been replaced by a **multiplication** sign for independent events, and the word **or**
has been replaced by an **addition** sign for mutually exclusive events.
This forms the basis of the **AND/OR** rule.

A bag contains 5 green beads and 4 red beads. A bead is taken from the bag, its colour noted, and it is then replaced. A second bead is then taken from the bag. What is the probability that the two beads are different colours?

We are being asked to find **P**
(1st is green
**and**
2nd is red **or**
1st is red
**and**
2nd is green).

The 1st bead is replaced before the 2nd bead is taken out, so the
first and second beads are **independent**. Therefore,
the word **and**
can be replaced by a multiplication
sign.

The events '1st is green and 2nd is red', or '1st is red and 2nd
is green', are **mutually exclusive**. Therefore, the word **or** can be replaced by an addition sign.

The answer to the question is

^{5}/_{9}×^{4}/_{9}+^{4}/_{9}×^{5}/_{9}- =
^{20}/_{81}+^{20}/_{81} - =
^{40}/_{81}

There are many pitfalls to using the **AND/OR**
rule in this way. Only do so if you are very confident about the method
**and** the question does not require the use of a
tree diagram.

However, remembering the **AND/OR**
rule will make sure
you know when to add and when to multiply when calculating
probabilities from tree diagrams.

Conditional probability is when the probability of one event depends on a previous event - eg, if items are being picked from a bag in turn, but not replaced.

A bag contains 4 red marbles and 5 green marbles. If we draw 2 marbles from the bag, then the colour of the second marble will be **
dependent** on the colour of the first.

To work out the probabilities for the **first marble**:

P (first marble is red) = ^{4}/_{9}

P (first marble is green) = ^{5}/_{9}

Here are two different sets of probabilities. Each set depends on which colour marble was taken out in the first event.

If the first marble out of the bag is red, the probabilities for the **second marble** are:

P (second marble is red, given that the first marble was red) = ^{3}/_{8}P (second marble is green, given that the first marble was red) = ^{5}/_{8}

However, if the first marble is green, there are four red and four green marbles left in the bag.

P (second marble is red, given that the first marble was green) = ^{4}/_{8}

P (second marble is green, given that the first marble was green) = ^{4}/_{8}

These are known as **conditional probabilities**. They are best represented on a tree diagram. After each event, the top and bottom numbers of the probability change. In this example, the bottom number will decrease by one each time an event occurs (one marble is removed).

The easiest way to work out conditional probabilities is to draw a tree diagram with the probabilities on each branch.

To find out the total probability of something which has a couple of possible outcomes, remember to add the probabilities together. Multiply along the branches to find each probability, and add together the outcomes.

Remember that the sum of the probabilities for any set of branches should always be 1.

From the tree diagram, we can see that:

P (both marbles are red) = ^{4}/_{9} × ^{3}/_{8} = ^{12}/_{72} = ^{1}/_{6}

P (one marble of each colour) = ^{4}/_{9} × ^{5}/_{8} + ^{5}/_{9} × ^{4}/_{8} = ^{40}/_{72} = ^{5}/_{9}

P (both marbles are green) = ^{5}/_{9} × ^{4}/_{8} = ^{20}/_{72} = ^{5}/_{18}

- Question
- The probability that Alex arrives home on time is
**0.7**. - If he does arrive home on time, the probability that his dinner is burnt is
**0.1**. - If he does not arrive home on time, the probability that his dinner is burnt is
**0.8**.

What is the probability that Alex arrives home on time and his dinner is not burnt?

- The probability that Alex arrives home on time is

- Answer
The probability that Alex arrives home on time and his dinner is not burnt is:

**0.7 x 0.9 = 0.63**

- Question
What is the probability that Alex's dinner is burnt?

- Answer
0.7 x 0.1 + 0.3 x 0.8 = 0.07 + 0.24 = 0.31

**Now try a **Test Bite