## Probability Tree Diagram

The probabilities of all possible outcomes of a series of trials of random experiment can be shown by a diagram called probability tree diagram. Every branch of a tree diagram shows the probability of respective event in a particular trial.

From a tree diagram, the probability of a particular event of a random experiment can be obtained as the product of the probabilities of the event in different trials along the path of the event.

The probability of independent and dependent events can be calculated by using their tree diagram. Let’s study the following examples.

### Worked out examples on probability tree diagram:

Example 1: If a coin is tossed two times, show the probabilities of all events in a tree diagram.

Solution:

Let, H and T represents head and tail of a coin.

n(H) = 1

n(T) = 1

n(S) = 1 + 1 = 2

It is the case of independent events.

Probability tree diagram:

From the tree diagram,

The probability of both head = P(HH) = ½ × ½ = ¼

The probability of first head and second tail = P(HT) = ½ × ½ = ¼

The probability of first tail and second head = P(TH) = ½ × ½ = ¼

The probability of both tail = P(TT) = ½ × ½ = ¼

Example 2: Two children were given birth by a married couple at the interval of four years. Illustrate the probabilities of son or daughter by drawing a tree diagram and calculate the probability of both are girls.

Solution:

Let, S be the event of son and D be the event of daughter.

n(S) = 1

n(D) = 1

Total number of possible outcomes = 1 + 1 = 2

It is the case of independent events.

Probability tree diagram:

From the tree diagram,

The probability of both are girls = P(DD) = P(D) × P(D) = ½ × ½ = ¼

So, the probability of both are girls is ¼.

Example 3: There are 20 boys and 30 girls in a class. If two students are selected at random, show the probability of selecting a boy and a girl by drawing a tree diagram.

Solution:

Let, B and G denote boy and girl student respectively.

n(B) = 20

n(G) = 30

Total number of students, n(S) = 20 + 30 = 50

It is the case of dependent events.

Probability tree diagram:

From the tree diagram,

The probability of selecting a boy and a girl = P(BG or GB)

= P(BG) + P(GB)

= 600/2450 + 600/2450

= 1200/2450

= 24/29

So, the probability of selecting a boy and a girl is 24/29.

Example 4: There are 5 black balls and 4 white balls in a box. Two balls are drawn randomly one after another without replacement. Show the probabilities in a tree diagram. Also find the probability of (i) getting at least one white ball (ii) getting none of white balls.

Solution:

Let, B and W denote the occurrence of black and white ball.

n(B) = 5

n(W) = 4

Total number of balls, n(S) = 5 + 4 = 9

Two balls are drawn randomly one after another without replacement. So, It is the case of dependent events.

Probability tree diagram:

From the tree diagram,

(i)         Probability of at least one white ball = P(BW or WB or WW) = P(BW WB WW)

= P(BW) + P(WB) + P(WW)

= 5/18 + 5/18 + 1/6

= 13/18

So, probability of getting at least one white ball is 13/18.

(ii)        Probability of none of white balls = P(BW WB WW)c

= 1 - P(BW WB WW)

= 1 – 13/18

= 5/18

So, probability of getting none of white balls is 5/18.

Example 5: There is one red, one blue and one white ball in a bag. Two balls are drawn randomly one after another without replacement. Write the sample space of the experiment using a tree diagram and find the probability of getting at least one red ball.

Solution:

Let R, B and W denote the red, blue and white ball.

n(R) = 1

n(B) = 1

n(W) = 1

No. of possible outcomes, n(S) = 1 + 1 + 1 = 3

Two balls are drawn randomly one after another without replacement. So, It is the case of dependent events.

Probability tree diagram:

From the tree diagram,

The sample space, S = {RB, RW, BR, BW, WR, WB}

The probability of getting at least one red ball = P(RB or RW or BR or WR)

= P(RB) + P(RW) + P(BR) + P(WR)

= 1/6 + 1/6 + 1/6 + 1/6

= 4/6

= 2/3

So, the probability of getting at least one red ball is 2/3.

You can comment your questions or problems regarding the probability tree diagram here.