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Tree Diagram is a visual way of using a tree-like structure to represent a hierarchy. A tree structure typically consists of a node, leaf node and root node. A Root node is a member without any superior parent. The several nodes are attached from the root nodes that are joined together by using line connections called branches or links which demonstrates the members’ relationship. The Leaf nodes or the end nodes are the members without any child or children nodes. We use a tree diagram for several reasons like In mathematics and computer science, to convey family relations or to show classification in taxonomy.
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Key Words: Tree, Diagram, Probability, Leaf, Root, Node, Branch, Statistics, Events, Leaf Node, Taxonomy
What is a Tree Diagram?
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The tree diagram is used in Statistics and Probability in Mathematics. It helps us to calculate the total amount of possible results of a particular event when the results are given in an organised way. A tree diagram’s branch represents the outcome of an event. It can be thought of as a simple way of showing the events’ sequences by recording all the possible results in a clear and uncomplicated manner.
Typically, a tree diagram begins with an item or a node that branches into more than or equal to two, then each node will branch into more than or equal to two, and so on. Then the final diagram looks like a tree with multiple branches and a trunk.
Tree Diagram in Probability Theory
A tree diagram basically consists of a root node, node and leaf node.
- Root Node: A root node can be defined as a member that has no superior parent.
- Node: The nodes are are the family members of the root node and are connected from the root nodes that are linked together with the help of line connections called links or branches that helps to display the relationship between the members.
- Leaf Node: Leaf nodes, also called end nodes or child nodes, are the members which do not contain any children or child nodes.
Uses of Tree Diagram
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A tree diagram is used for the following purposes:
- In Mathematics and Computer Science
- To depict family relations
- To show classification in taxonomy
- In business organizations
- In the Science of classification
Read More: Theoretical Probability
Tree Diagram in Probability
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A tree diagram could be utilised by probability theory to show a probability space. These diagrams show a sequence of different events (like a set of a coin tossed) or probabilities with conditions (for example, drawing cards from a deck without substituting the cards in it).
Every node on the tree diagram shows an incident and is linked with the probability of that occurring. The root node denotes the particular incident and thus it has a probability of 1. All sets of sibling connections or nodes appoint a self-dependent and complete distribution of the parent incident.
Occasionally we experience difficulties while figuring probabilities and it's tough to decide what to do. Tree diagrams will end this difficulty. Think of a case in probability to draw a tree diagram for tossing a coin. There are two branches, head and tail. The probability of an incident is mentioned on the branch and the resultant is mentioned at the end of the branch.
Read More: Sum of Probabilities
Calculation of Overall Probability in Tree Diagram
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Probability means the liability of the incidents. The probability value is a numerical value and it constantly lies between 0 and 1. As we know that the probability of an impossible event is zero and the probability of the sure event is 1. The formula to check out the probability is written as
P(E) = Number of favourable outcomes / Total number of outcomes
Also, there are several probability formulas in probability theory, each depending upon the type of incident.
While tossing a coin, the probability of getting heads is 0.5 or 1/2 and that of getting tails is 0.5 or 1/2.
Steps to Calculate Overall Probability
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Here are the steps to calculate overall probability:
- Firstly we have to mention the probability value on the branches.
- Then we have to multiply the probability value along the branches.
- Finally, we have to add the probability value that is obtained after the multiplication process.
Upon the addition of all the probability values obtained, the resultant should be equal to 1.
Read More: Some Applications of Trigonometry
Example of Tree Diagram
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Let’s Consider the experimentation of tossing a coin. However, toss it again but if it shows a tail, again throw a die, If the coin shows the head. Calculate what will be the conditional probability of the incident that the die shows a number higher than 4 given that there is at least one tail.
The conclusions of the experimentation are shown in a diagrammatic form called the tree diagram, where branches and nodes of the tree illustrate the incident or a happening.
The sample space of the experimentation may be shown as S = { (H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) } where (H, H) denotes that both the tosses result into the head and (T, i) express the first toss result into a tail and the number appears on the die for i =. Therefore, the chances assigned to the 8 basic events (H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6) are 1/4, 1/4, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12 respectively.
Let F be the incident that there is at least one tail and E be the incident the die shows a number higher than 4.
Again,
\(F = \{ (H, T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) \}\)
\(E = \{ (T,5), (T,6) \} \ and\ E \cap F = \{ (T,5), (T,6) \} \)
Now \(P(F) = P\{(H,T)\} + P \{(T,1)\} + P \{(T,2)\} + P \{(T,3)\} + P \{(T,4)\} + P\{(T,5)\} + P\{(T,6)\}\)
\(= 1/4 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12\)
\(= 3/4\)
and \(P(E \cap F) = P \{(T,5)\} + P \{(T,6)\} = 1/12 + 1/12 = 1/6\)
Therefore, \(P(E|F) = P(E \cap F)/P(F) = (1/6)/(3/4) = 2/9\)
Read More: Conditional Probability Formula
Things To Remember
- A Tree Diagram is a visual way of using a tree-like structure to represent a hierarchy.
- A tree structure typically consists of a node, leaf node and root node.
- A Root node is a member without any superior parent.
- The several nodes are attached from the root nodes that are joined together by using line connections called branches or links which demonstrates the members’ relationship.
- The Leaf nodes or the end nodes are the members without any child or children nodes.
- The tree diagram is used for calculating statistics and probability in mathematics.
- It helps us to calculate the total amount of possible results of a particular event when the results are given in an organised way.
- A tree diagram’s branch represents the outcome of an event. It can be thought of as a simple way of showing the events’ sequences by recording all the possible results in a clear and uncomplicated manner.
Sample Questions
Ques. A bag contains 3 black and 5 white balls. Sketch out the probability tree diagram for two draws for the situation. (3 Marks)
Ans. Given:
No. of black balls = 3
No. of white balls = 5
Total Number of balls = 8
Thus, the probability of getting black balls = 3/8
Probability (Getting white balls) = 5/8
Hence, the tree diagram for two draws of balls with possible outcomes and probability will be drawn as follows.
Ques. What is the Sample Space? (3 Marks)
Ans. A sample space is the set of all possible resultants in experimentation like choosing a ball, drawing a card, etc). It's generally expressed by the letterS. Sample space can be written using the set notation {}.
The total of all the chances of the different resultants within a sample space is 1.
Ques. A bag contains 10 balls. 3 are blue, and 7 are red. Now, a ball is drawn at random from the bag and NOT replaced in the bag. Draw a tree diagram to represent the probabilities of drawing two consecutive balls of the same colour. (3 Marks)
Ans. Given below is the solution for the above problem:
Notice that the probabilities of drawing a Red or Blue ball are non-identical in the new draw as compared to the foremost draw. For example, in the foremost draw, we've. 3 blue and 7 red balls, so the probability of drawing a blue ball is 3/10.
For the new draw, if we think that a Blue ball was drawn in the foremost draw, again there would be 2 Blue and 7 Red balls left, and thus the probability of drawing another Blue ball is 2/9 as displayed in the top branch of the new draw. We calculate all the new draw probabilities using a matching argument and display them on top of their branches. Eventually, the probability of drawing two balls of identical colour is found by adding the probabilities corresponding to (B, B) and (R, R) resultants, i.e, P (Two balls of the same colour) =P (R, R) P (B, B) = 7/15 + 1/15 = 8/15
Ques. Suppose a factory is producing light bulbs. The probability that any light bulb is defective is p = 0.01. Now, a tester has been testing light bulbs randomly. Calculate what will be the probability of the following events: (5 Marks)
1) Getting 2 defective light bulbs in 3 tests
2) Getting no defective light bulbs in 3 tests
3) First defective light bulb is discovered at the third attempt.
4) First defective light bulb is discovered within the first two attempts
Ans. Let us assume that D represents a ”defective light bulb” and D’ represents a ”not defective light bulb”.
The probability of a defective light bulb is given to be P(D)=0.01.
From basic probability theory, we know that: P(D′)=1−P(D)=1−(0.01)= 0.99
1. Finding 2 defective light bulbs:
P(finding 2 defective light bulbs)=P(D′,D,D)+P(D,D′,D)+P(D,D,D′)
=(0.99×0.01×0.01)+(0.01×0.99×0.01)+(0.01×0.01×0.99).
=0.000099+0.000099+0.000099=0.000297
2. Finding no defective light bulbs:
P(finding no defective light bulbs)=P(D′,D′,D′).
=(0.99×0.99×0.99)=0.9703
3. First defective light bulb is discovered at the third attempt:
P(1st defective light bulb at 3rd attempt)=P(D′,D′,D)
=(0.99×0.99×0.01)=0.009801
4. The first defective light bulb is discovered within the first two attempts:
P(1st defective light bulb at first 2 attempts)=P(D,D,D′)
=(0.01×0.01×0.99)=0.000099
Ques. The letters of the word ‘SUCCESS’ are printed on 7 cards. Justin chooses a card at random, replaces it, then chooses a card again. Calculate what will be the probability using a tree diagram that only one of the cards he chooses has the letter C printed on it. (3 Marks)
Ans. C’ represents Not the letter C.
We observe from the tree diagram that the probability for one of the cards he chooses has ‘C’ printed on it is
P(One of the cards is C)=P(C, C′)+P(C′, C)
=(2/7×5/7)+(5/7×2/7)=20/49
Ques. We roll a single die three times. Calculate what will be the probability of the following events using a tree diagram:(3 Marks)
a) Obtaining an even number in all three attempts.
b) Obtaining at least two even numbers in three attempts.
Ans. Here is the solution for the problem:
P(All even)=P(E,E,E)= 1/216
P(Two evens)=P(E,E,E′)+P(E,E′,E)+P(E′,E,E)= 15/216
Ques. Three fair coins are tossed simultaneously. Using a tree diagram, determine the probability of obtaining:(3 Marks)
a) At least 2 Tails.
b) At most, two Heads.
c) No Tails at all.
Ans. The solution will be as follows:
P(at least two Tails)=P(T,T,H)+P(T,H,T)+P(H,T,T)+P(T,T,T)=1/2
P(at most two Heads)=1–P(H,H,H)= 7/8
P(No tails)=P(H, H, H)= 1/8
Ques. Two cards have been drawn from a deck of 52 cards without the replacement of any card in the deck. What is the probability that: (3 Marks)
a) Both cards are Kings.
b) At least one of the cards is a King
Ans. The solution would be:
P(Both Kings)=P(K,K)= 1/221
P(At least one King)=P(K,K′)+P(K′,K)+P(K,K)= 33/221
Ques. A person has four keys and only one key fits the lock of a door. What is the probability that the locked door can be unlocked in at most three tries? (3 Marks)
Ans. Let U be the event that the door has been unlocked and L be the event that the door has not been unlocked. We illustrate with a tree diagram.
Probability (unlocking the door in the first try) = ¼
Probability (unlocking the door in the second try) = (3/4)(1/3) = 1/4
Probability (unlocking the door in the third try) = (3/4)(2/3)(1/2) = 1/4
Therefore, the probability of unlocking the door in at most three tries = 1/4 + 1/4 + 1/4 = 3/4.
Ques. A circuit consists of three resistors: resistor R1, resistor R2, and resistor R3 joined in a series. In case one of the resistors fails, the circuit will stop working. The probabilities that resistors R1, R2, or R3 will fail are 0.07, 0.10, and 0.08, respectively. Calculate what will be the probability that at least one of the resistors will fail? (3 Marks)
Ans. The probability that at least one of the resistors fails = 1 – none of the resistors fails.
It's relatively easy to find the probability of the incident that none of the resistors fails.
We do not truly need to draw a tree because we can picture the only branch of the tree that assures this result.
The chances that R1, R2, R3 won't fail are 0.93, 0.90, and 0.92 independently. Thus, the probability that none of the resistors fails = (0.93) (0.90) (0.92) = 0.77.
Therefore, the probability that at least one of them will fail = 1 - 0.77 = 0.23
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