Probability Tree Diagram

Probability Tree Diagram

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


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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.


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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 represent the 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 of tossing a coin two times.

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 a 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 of giving birth of two children.

 

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 of selecting two students randomly from a class.


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 of drawing two balls randomly from a box having 2 different coloured balls.

 

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, the 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, the probability of getting none of the 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 of drawing two balls randomly from a box having 3 different colored balls. 


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.


 

Do you have any questions regarding the Probability Tree Diagram?

You can ask your questions or problems here in the comment section below.

1 comment:

  1. A dice is rolled and a coin is tossed draw a tree diagram and find the probability that dice lands on odd number and coin on head

    ReplyDelete