Probability
Probability is often tested in the Quantitative Aptitude section of various exams like- SBI PO, IBPS PO & SO, SEBI Grade A, SSC CGL and many others. It is a scoring yet tricky chapter.
Therefore, it’s essential that you are familiar with the short tricks and tips to solve these problems quickly and accurately.
Definition of Probability
Probability deals with the uncertainty of the occurrence of an event using numbers. The chance that a particular event will or will not occur is expressed on a scale ranging from 0-1. Another way to represent probability is by way of percentage. For instance, 0% denotes an impossible event and 100% denotes a certain event.
Types of Probability
It is based on actual experiments and adequate recordings of the happening of events. A series of experiments are conducted to determine the occurrence of any event. Experiments with no results are called random experiments, where the outcome of such experiments is uncertain. To determine their likelihood, random experiments are repeated multiple times. The experiment is repeated a fixed number of times and each repetition is known as a trial.
The formula of experimental probability is-
P(E)= Number of times an event occurs/Total number of trials.
Let’s understand with an example-
You ask 3 friends A, B and C to toss a fair coin 15 times each in a row and the outcome of this experiment is given below-
Coins Tossed by | Number of Heads | Number of Tails |
---|---|---|
A |
6 |
9 |
B |
7 |
8 |
C |
8 |
7 |
Calculate the probability of occurrence of heads and tails.
Answer:
The experiential probability for the occurrence of of heads and tails can be calculated as:
Coins Tossed by | Number of Heads | Number of Tails | Experimental Probability of Occurrence of heads | Experimental Probability of Occurrence of tails |
---|---|---|---|---|
A |
6 |
9 |
6/15 = 0.4 |
9/15 = 0.6 |
B |
7 |
8 |
7/15 = 0.47 |
8/15 = 0.53 |
C |
8 |
7 |
8/15 = 0.53 |
7/15 = 0.47 |
It can be observed that if the number of tosses of the coin increases, the probability of occurrence of heads or tails increases to 0.5.
Geometric Probability
Geometric probability is the calculation of the likelihood that one will hit a particular area of a figure. It is calculated by dividing the desired area by the total area. In the case of Geometrical probability, there are infinite outcomes.
Let’s understand the basics of Probability in detail.
Outcome and Event
Outcome is the result of a random experiment.
For instance, when we roll a dice, getting four is an outcome.
On the other hand, Event is a set of outcomes of an experiment. It is denoted by E.
An event in probability is the subset of the corresponding sample space.
Sample space is the entire possible set of outcomes of a random experiment. It is denoted by S.
For instance, when we roll a dice, the probability of getting a number less than 5 is an event.
Please note- An Event can have a single outcome
Probability of Occurrence of an Event
Probability of occurrence of any event is the number of favourable outcomes divided by the total number of outcomes.
The formula is as follows:
P(E) = Number of Favourable Outcomes/ Total Number of Outcomes
Let’s understand with an example-
Q) Find the probability of rolling a 5 on a fair dice
Answer:
To find the probability of getting 5 while rolling a dice, an experiment is not needed. We know that there are 6 possible outcomes when rolling a dice. They are 1, 2, 3, 4, 5, 6.
Therefore, the probability is,
Probability of Event P(E) = No. of. Favourable outcomes/ No. of. Possible outcomes.
P(E) = 1/6.
Hence, the probability of getting 5 while rolling a fair dice is 1/6.
Types of Events in Probability
Following are the important probability events-
Where the probability of occurrence of an event is 0, such an event is called an impossible event.
Let us understand with an example of impossible event-
Q) A bag contains only orange flavoured candies. Radha takes out one candy without looking into the bag. What is the probability that she takes out a lemon flavoured candy?
Answer:
Let us take the number of candies in the bag to be 100
Number of lemon flavoured candies= 0 (since the bag contains only orange flavoured candies)
Thus, the probability that she takes out lemon flavored candies= P= Number of lemon flavoured candies/ Total number of candies= 0/100= 0
Therefore, the probability that Radha takes out a lemon flavoured candy is 0 which proves that the probability of an impossible event is 0.
If the probability of occurrence of an event is1, it will be a sure event.
Let’s look at an example of a sure event-
Q) What is the probability that a number obtained after throwing a dice is less than 7?
Answer:
P(E)= P (getting a number less than 7)= 6/6= 1
Any event having only one outcome of the experiment is called a simple event.
For example- If we toss a coin ‘n’ number of times, we will get only two possible outcomes: Heads or Tails. Thus, for an individual toss, it has only one outcome i.e. Heads or Tails.
The sum of the probabilities of all simple events of an experiment is one.
For example- Let us take the coin toss example again . P(Heads) + P(Tails)= ½ + ½ = 2/2 = 1.
On the contrary, if any event consists of more than one single point of the sample space, it is called a compound event.
For example-
If S= {56, 78, 96, 54, 89}
E1= {56, 54}
E2= {78,56, 89}
Then, E1 and E2 represent compound events.
If the occurrence of any event is completely unaffected by the occurrence of any other event, such events are known as independent events. The events which are affected by other events are called dependent events.
Here, the occurrence of one event excludes the occurrence of another event. Such events are mutually exclusive events. In other words, two events do not have a common point.
For example-
S= {1, 2, 3, 4, 5, 6} and E1 and E2 are two events such that E1 consists of numbers less than 3 and E2 consists of numbers greater than 4.
Then, E1= {1,2} and E2= {5,6}
Which means, E1 and E2 are mutually exclusive.
If all the events consume the entire sample space, the set of events is called exhaustive.
For any event E1 there exists another event E1’ which represents the remaining elements of the sample space S.
E1= S - E1’
For example-
If a dice is rolled, then the sample space S is given as S= {1, 2, 3, 4, 5, 6}.
If event E1 represents all the outcomes which is greater than 4, then E1= {5,6} and E1’ = {1, 2, 3, 4}
Thus, E1’ is the complement of the event E1.
Likewise, the complement of E1, E2, E3……….En will be represented as E1’, E2’, E3’...........En’
If two events E1 and E2 are associated with Or means that it can be either E1 or E2 or both. The union symbol (U) is used to represent Or in probability.
Therefore, the event E1 U E2 denotes E1 Or E2.
If there are mutually exhaustive events E1, E2, E3……….En associated with sample space S then,
E1 U E2 U E3 U……….En= S
If two events E1 and E2 are connected with And, this means that there is an intersection of elements which is common to both the events. The intersection symbol (n) is used to denote And in probability.
Thus, the events E1 n E2 denotes E1 and E2
It denotes the difference between both the events. E1 but not E2 represents all the outcomes which are present in E1 but not in E2. So, the event E1 but not E2 is denoted as E1, E2= E1 - E2
Practice Questions
Q1) It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992 . What is the probability that the
2 students have the same birthday?
Answer:
It is provided to us that, probability of 2 students not having the same birthday is 0.992 .
So,
P(2 students having the same birthday)+P(2 students not having the same birthday)=1
= P(2 students having the same birthday)+0.992=1
Simplifying further,
= P(2 students having the same birthday)=0.008.
Q2) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i)red and (ii) not red?
Answer:
The bag has 3 red balls and 5 black balls
(i) So, the probability of getting a red ball= number of red balls/total number of balls
P(E)= 3/ 3+5= ⅜
(ii) the probability of getting a red ball= 1- number of red balls/total number of balls
So, P(E)= 1- ⅜= ⅝
Q3) A dice is thrown once. Find the probability of getting (i) a prime number;
Answer:
There are 6 results that can be obtained from a dice.
There are 3 prime numbers, 2,3,5 among those results.
Thus, the probability of getting a prime number= no of prime numbers in a dice/total numbers on dice= 3/6= ½
(ii) a number lying between 2 and 6?
Answer:
There are 3 numbers between 2 and 6, 3,4,5.
Thus, the probability of getting a number between 2 and 6= no. of numbers between 2 and 6 in a dice/total numbers on a dice= 3/6= ½ no
Q4) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a king of red color?
Answer:
We know there are 52 cards in the deck.
There are 2 kings of red colour in the deck.
Thus, P(E)= total number of red kings/total number of cards= 2/52= 1/26
Q5) 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer:
There are a total (132+12)=144 number of pens in the lot.
And also there are 132 good pens in the given collection.
Thus, the probability of getting a good pen= number of good pens/total number of pens= 132/144= 11/12
To sum up, probability is an easy yet tricky chapter. Therefore, this chapter requires a lot of practice to solve questions with minimal error.