TREE DIAGRAMS

In this lesson we will look at more examples of probabilty problems We will use tree diagrams to help solve the problems. We will see that tree diagrams can be used to represent the set of all possible outcomes involving one or more experiments.

Example:

Julia spins 2 spinners; one of which is labelled 1, 2 and 3, and the other is labelled 4, 5 and 6.

Solution:

a) Tree diagram for the experiment.

n(S ) = 9
Probability that the spinners stop at (3,4) =

c) The probability that the spinners do not stop at “3” and “4”

Probability that the spinners do not stop at (3,4) =

d) The probability that the first spinner does not stop at “1”

Probability that the first spinner stop at “1” =
Probability that the first spinner does not stop at “1” =

Example:

Box A contains 3 cards numbered 1, 2 and 3. Box B contains 2 cards numbered 1 and 2. One card is removed at random from each box.

a) Draw a tree diagram to list all the possible outcomes.
b) Find the probability that:
(i) the sum of the numbers is 4
(ii) the sum of the two numbers is even.
(iii) the product of the two numbers is at least 5.
(iv) the sum is equal to the product.

Solution:

a) A tree diagram of all possible outcomes.

b) The probability that:

(i) the sum of the numbers is 4.

Let S be the sample space and A be the event that the sum is 4.

n(S) = 6; n(A) = 2

P(A) =

(ii) the sum of the two numbers is even.

Let B be the event that the sum is even.

n(B) = 3

P(B) =

(iii) the product of the two numbers is at least 5.

Let C be the event that the product of the two numbers is at least 5.

n(C) = 1

P(C) =

(iv) the sum is equal to the product.

Let D be the event that the sum of the two numbers is equal to the product.

n(D) = 1

P(D) =

Example:

A bag contains 4 cards numbered 2, 4, 6, 9. A second bag contains 3 cards numbered 2, 3, 6. One card is drawn at random from each bag.

a) Draw a tree diagram for the experiment.
b) With the help of the tree diagram, calculate the probability that the two numbers obtained:
(i) have different values.
(ii) are both even.
(iii) are both prime.
(iv) have a sum greater than 5.
(v) have a product greater than 16.

Solution:

a) A tree diagram of all possible outcomes.

b) The probability that the two numbers obtained:

(i) have different values.

Let S be the sample space and A be the event that the two values are different

n(S) = 12 ; n(A) = 10

P(A) =

(ii) are both even.

Let B be the event that both values are even.

n(B) = 6

P(B) =

(iii) are both prime.

Let C be the event that both values are prime.

n(C) = 2

P(C) =

(iv) have a sum greater than 5.

Let D be the event that the sum of both values is greater than 5.

n(D) = 10

P(D) =

(v) have a product greater than 16.

Let E be the event that the product of both values is greater than 16.

n(E) = 6

P(E) =

The following videos gives more examples of solving probability problems using tree diagrams.

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