A given event's Probability measures how likely it is to occur. It is used to describe and analyze uncertainty in everyday situations and games of chance.
P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
The addition rule of probability is used to find the probability that at least one of two events occurs. For any two events A and B, it is given by:
P(A or B) = P(A) + P(B) − P(A and B)
Probability with Venn Diagrams
Mutually Exclusive Events
Two events, A and B, are said to be mutually exclusive if they cannot occur simultaneously during a single trial.
Example: In a coin toss, the events "Getting Heads" and "Getting Tails" are mutually exclusive because both cannot happen at the same time.
Venn Diagram: Mutually exclusive events are represented as non-overlapping circles in a Venn diagram since there is no intersection between the two events.
Addition Rule: Since P(A ∩ B) = 0 (no overlap), the formula simplifies to: P(A ∪ B) = P(A) + P(B)
Here:
- P(Getting Heads) = 1/2,
- P(Getting Tails) = 1/2,
- P(Getting Heads or Tails) = 1/2 + 1/2 = 1
When the probabilities of all possible events in a sample space are added, their sum is equal to 1.
Non-Mutually Exclusive Events
Two events, A and B, are said to be non-mutually exclusive if they can occur simultaneously during a single trial.
Example: Rolling a die, let A represent rolling an odd number ({1, 3, 5}) and B represent rolling a 3 ({3}). In this case, the number 3 belongs to both events, meaning A and B overlap.
Venn Diagram: Non-mutually exclusive events are represented as overlapping circles in a Venn diagram, with the shared outcomes placed in the intersection.
- The two circles representing Event A ({1, 3, 5}) and Event B ({3}) intersect, showing an overlap at 3.
Addition Rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Here:
- P(A): Probability of rolling an odd number = 3/6 = 1/2,
- P(B): Probability of rolling a 3 = 1/6,
- P(A ∩ B): Probability of rolling a number that is both odd and 3 = 1/6.
Substitute into the formula: P(A ∪ B) = 1/2 + 1/6 − 1/6 = 1/2.
Adding Probabilities
In probability theory, when you add the probabilities of all possible outcomes, the sum always equals 1. This is a result of the fact that one of the possible outcomes must occur.
Example: If a trial has three possible outcomes, A, B, and C.
P(E1) + P(E2) + ... + P(En)= 1
Elementary Events
Sometimes we have only one outcome in which we are interested. Let's say 8 teams are participating in the cricket World Cup. We are interested in finding the probability of winning the World Cup for India. We are not interested in finding out the probability for every other team. So we will formulate the problem in the following way,
Let's say event A denotes India winning the World Cup. So, another event B denotes India not winning the World Cup.
P(A) + P(B) = 1
P(A) = 1 - P(B)
Such events are called elementary events.
Addition Rules For Probability
For any two events A and B, based on the fact whether both the events are Mutually Exclusive or not, Two different Rules are described,
Rule 1: When the events are Mutually Exclusive
When the events are mutually exclusive, the probability of the events occurring is the sum of both events.
P(A∪ B) = P(A) + P(B)
Rule 2: When the events are not mutually exclusive
There is always some overlapping between two non-mutually exclusive events, Therefore, the Probability of the events will become,
P(A∪ B) = P(A) + P(B) - P(A∩ B)
Related Articles :
Sample Examples
Question 1: Let's say a die was rolled. Answer the following questions:
- What is the probability of getting a number greater than 4?
- What is the probability of getting an even number?
Solution:
When a die is rolled, there are six possible outcomes: 1, 2, 3, 4, 5 and 6
1. Probability of getting a number greater than 4:
Number of favorable outcomes = 2
Total number of outcomes = 6P(Number Greater than 4) = 2/6 = 1/3
2. Probability of getting an even number:
Number of favorable outcomes = 3
Total number of outcomes = 6P(Getting a Even Number) = 3/6 = 1/2
Question 2: Let's say a card was drawn from a well-shuffled deck of cards. Find the probability of getting a Queen on one draw.
Solution:
We know that a deck has 52 cards. So there are total 52 outcomes that are possible if a card is drawn. We also know that there are four queens in the deck. These are our favorable outcomes.
So,
Total number of outcomes = 52
Total number of favorable outcomes = 4P(Getting a Queen) = 4/42 = 1/13
Question 3: A bag contains 3 white balls, 4 black balls, and 2 green balls. A ball is drawn with replacement. Find the probability of getting:
- A White Ball
- A Black Ball
- A Green Ball
Solution:
There are a total of 3 + 4 + 2 = 9 balls.
1. Probability of getting a white ball
Total number of balls = 9,
Favorable outcomes = 3P(Getting a White Ball) =3/9 = 1/3
2. Probability of getting a Black ball
Total number of balls = 9,
Favorable outcomes = 4P(Getting a Black Ball) = 4/9
3. Probability of getting a Black ball
Total number of balls = 9,
Favorable outcomes = 2P(Getting a Green Ball) = 2/9
Question 4: A satellite from space came crashing down on Earth. The figure below denotes the area in which ISRO suspects the satellite crashed. Find the probability that it crashed in the lake.
Solution:
In this we don't know the number of outcomes. This is a continuous case, that is plane can crash anywhere in the area.
So, total area of the region = 10 × 5 = 50 Km2
Total area of the lake = 5 × 3 = 15 Km2Now we can use these areas to calculate the probability.
Total Number of possible outcomes(area in this case) = 50 Km2
Total number of favorable outcomes (area of the lake in this case) = 5 × 3 = 15 Km2P(Satellite crashing in the lake) = 15/50 = 3/10
Thus, the probability of satellite crashing in the lake is 0.3.
Question 5: Let's say we have a well-shuffled deck. We draw two cards and find the probability of getting either a King or a Queen.
Solution:
Let's say drawing a king represents an event A while drawing a queen represents an event B. We are asked for the probability for getting either King or Queen. We will use law of adding probabilities here,
Probability (King or Queen) = Probability (King) + Probability (Queen)
We know that there are 4 Kings and 4 Queens in the deck.
P(King) = 4/52 = 1/13
P(Queen) = 4/52 = 1/13Thus,
Probability (King or Queen) = 1/13 + 1/13 = 2/13
Question 6: We have an urn that contains three black balls, two blue balls, and three white balls. Find the probability of getting one black, one blue, and one white ball if we draw three times with replacement.
Solution:
We have a total of eight balls.
P(getting a black ball) = 3/8
P(getting a blue ball) = 2/8
P(getting a white ball) = 3/8We will find out this probability with law of addition.
So the total probability of getting all three colors = P(Black) + P(Blue) + P(White)
=3/8 × 2/8 × 3/8
= 18/512Notice that the probability sums up to one. This is in accordance with laws of probability.
Question 7: The Union Budget is going to be announced by the government this week. The probability that it will be announced on a day is given,
| Day | Probability |
| Monday | |
| Tuesday | |
| Wednesday | |
| Thursday | |
| Friday |
Find the probability of the budget getting announced between Monday to Wednesday.
Solution:
We need to use the probability addition law,
P(Monday to Wednesday) = P(Monday) + P(Tuesday) + P(Wednesday)
P(Monday) = 1/7
P(Tuesday) = 3/7
P(Wednesday) = 1/7P(Monday to Wednesday) = P(Monday) + P(Tuesday) + P(Wednesday)
= 1/7 + 3/7 + 1/7
= 5/7
Question 8: In a class of 90 students, 50 took Math, 25 took Physics, and 30 took both Math and Physics. Find the number of students who have taken either math or Physics.
Solution:
Since the events of choosing math and physics are non-mutually exclusive, the second rule of addition will be applied here,
P(Math ∪ Physics) = P(Math) + P(Physics) - P(Math ∩ Physics)
P(Math) = 50
P(physics) = 25
P(Math ∩ Physics) = 30
P(Math ∪ Physics) = 50 + 25 – 30
P(Math ∪ Physics) = 45 students.