Table of contents 
Questions Covered on Topics: 
Introduction 
What is Probability? 
Probability With the Use of Conjunctions AND and OR 
Important Consideration during defining Events 
Binomial Probability Distribution 
Solved Previous Year Questions 
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Probability is one of the most important mathematical concepts that have a real presence & impact in our daily lives. There is a consistent presence of probability questions in CAT & other B school entrances at a min of 2 questions every year. This makes this chapter extremely important for an aspirant.
In this chapter, you intend to improve your concepts of probability to such an extent that you feel in total control of this topic through easy methods of explanations & practice examples.
Probability is simply how likely something is to happen. Whenever you’re not certain about the outcome of an event, you can talk about the probabilities of certain outcomes—how likely they are.
Examples:
(i) The best example for understanding probability is flipping a coin
(ii) Probability (Chance) of rain on a given day.
(iii) Dice outcomes
Tip:
a. Probability always lies between 0 and 1b. Probability of a sure event is 1
c. Probability is 0 for an impossible event
Example 1: Imagine an instance where you are throwing a dice, what is the probability of getting a number greater than 3, in a throw of a normal unbiased dice having 6 faces?
Here the event is throwing the dice at first place & condition of getting a number more than 3.
⇒ The event is defined as getting 4 or 5 or 6.
The individual probabilities of each of these are 1/6,1/6 and 1/6 respectively.
Hence, the required probability is 1/6 + 1/6 + 1/6 = 3/6 = 1/2.
In general, event definition means breaking up the event into the most basic building blocks, which are commonly through the two English conjunctions— AND and OR.
Example 2: Let’s try solving an example, If you have the probability of E hitting a target as 1/2 and that of R hitting the target as 1/5, then the probability that both hit the target if one shot is taken by both of them is?
The event here is that E hits the target AND R hits the target, both hitting the target. Hence the resulting probability = P(E) AND P(R) = P(E) x P(R)
Using the values given of P(E) & P(R),
P(E) x P(R) = 1/2 x 1/5 = 1/10
Tip: When both events are connected by ‘AND’, please remember to use multiplication of events.
(Note: Remember the above explanation, where you used Multiplication when both events are connected by conjunctive AND. That is the reason why you multiplied them.)
Example 3: If you have the probability of E hitting a target as 1/2 and that of R hitting the target as 1/5, then the probability that both hit the target if one shot is taken by either of them is?
As you can see whether E OR R hits the target, the outcome is counted even if one of them hits the target.
Hence, resulting probability = P(E) OR P(R) = P(E) + P(R)
Using the values given of P(E) & P(R),
P(E) + P(R) = 1/2 + 1/5 = 7/10
Tip: Look for the language of the question, conjunction ‘OR’ is used to connect the events, which means that you need to add them up.
Unlike in example, 2 where the event will happen only with both E & R events occurring, here if any one of E, R events can occur.
Example 4: If two dice are thrown, what is the chance that the sum of the numbers is not less than 10?
Event Definition: The sum of the numbers is not less than 10 i.e. either 10 OR 11 OR 12.
⇒ P(E) = (6 AND 4) OR (4 AND 6) OR (5 AND 5) OR (6 AND 5) OR (5 AND 6) OR (6 AND 6)
= [(1/6) x (1/6)] + [(1/6) x (1/6)] + [(1/6) x (1/6)] + [(1/6) x (1/6)] + [(1/6) x (1/6)] + [(1/6) x (1/6)] = 6/36 = 1/6
Hence for the event of sum less than 10, P(E’) = 1  P(E) = 1  1/6 = 5/6.
EduRev Tip: You can see that no matter how many broad subevents the problem of probability have, it can always be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.
While attempting the questions, you need to take into consideration certain activities/events. Questions in CAT exams usually have direct or indirect mentions of such events.
Event: Event refers to any outcome of an experiment that is independent of any other outcome and cannot occur simultaneously with another event.
Experiment: An action that can deliver some welldefined outcomes, is an experiment.
(i) Impossible event and certain event
(ii) Random experiment
Random Experiment gives one or more results under identical conditions.
E.g. a coin is tossed.
Example 5: A coin is tossed and a single 6sided dice is rolled. Find the probability of getting a head on the coin and a 3 on the dice.
P(head) = 1/2, P(3) = 1/6
P (head and 3) = (1/2) x (1/6) = 1/12
(iii) Sample space
(iv) Mutually Exclusive Events
(v) Equally likely events
(vi) Exhaustive set of events
Example 6: If a twodigit no. is selected at random. What is the probability of getting a number, which has both the digits the same?
The total 2 digit numbers are 10 to 99 ⇒ 90
Required numbers are 11, 22,      99 ⇒ 9
Thus, probability = 9/90 = 1/10
(vii) Independent events
(viii) Conditional probability
Sum Rule
 If E and F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by: P(E or F) or P (E ∪ F) = P(E) + P(F)
If the event is not mutually exclusive, thenthe probability that either event E or event F will occur in a single trial is given by: P(E ∪ F) = P(E) + P(F) – P(E and F together)
P (neither E nor F) = 1 – P(E or F)
Multiplication Rule
 When two events, A and B, are independent, the probability of both occurring is: P(A and B) = P(A ∩ B) = P(A) × P(B)
Example 7: From a bag containing 8 green and 5 red balls, three are drawn one after the other. Find the probability of all three balls being green if
(a) the balls drawn are replaced before the next ball is picked
(b) the balls drawn are not replaced.
(a) When the balls drawn are replaced, you can see that the number of balls available for drawing out will be the same for every draw.
⇒ The probability of a green ball appearing in the first draw and a green ball appearing in the second draw as well as one appearing in the third draw is equal to each other.
Hence, Required probability = (8/13) x (8/13) x (8/13) = (8/13)^{3}
(b) When the balls are not replaced, the probability of drawing any color of ball for every fresh draw changes.
Hence, Required probability = 8/13 X 7/12 X 6/11
Tip: If n dice are thrown then total possible outcomes = 6^{n} & if n coins are tossed then possible total outcomes = 2^{n}
The Binomial Probability distribution is an experiment that possesses the following properties:
The Binomial Probability distribution of exactly x successes from n number of trials is given by the below formula:
Where,
Example 8: A dice is tossed 5 times. What is the probability that 5 shows up exactly thrice?
Here the 'random experiment' consists of tossing a die 5 times and observing the number '5' as success,
⇒ p = Probability of getting '5' with single dice =
Since the value of p is constant for each dice and the trials are independent, using formula for Binomial probability law, the probability of r successes is given by:
Example 9: What is the chance of drawing an ace from a deck of cards?
There are 52 cards in a deck of cards, the total number of possible outcomes of drawing a card is 52.
Since there are 4 aces in a pack of cards, the total number of favorable outcomes is 4. The chance or probability of drawing an ace from the given pack = 4/52 = 1/13
Example 10: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:
(i) an ace,
(ii) red,
(iii) either red or king,
(iv) red and a king
Out of 52 cards, one card can be drawn in ^{52}C_{1} ways.
Therefore exhaustive number of cases = ^{52}C_{1} = 52
(i) There are four aces in a pack of 52 cards, out of which one can be drawn in ^{4}C_{1}
Therefore Favourable number of cases = ^{4}C_{1} = 4
So, required probability = 4/52 = 1/13
(ii) There are 26 red cards, out of which one red card can be drawn in ^{26}C_{1} ways. Therefore, Favourable number of cases = ^{26}C_{1} = 26
So, required probability = 26/52 = 1/2
(iii) There are 26 red cards including 2 red kings and there are 2 more kings. Therefore, there are 28 cards, one can be drawn in ^{28}C_{1} ways.
Therefore favourable number of cases = ^{28}C_{1} = 28
So, required probability = 28/52 = 7/13
(iv) There are 2 cards, which are red and king.
Therefore, favourable number of cases = ^{2}C_{1} = 2.
So, required probability = 2/52 = 1/26.
Example 11: Harshad and Amit throw with two dice. If Harshad throws 10, what is Amit’s chance of throwing a higher number?
Total number of outcomes of throwing two dice is 6 × 6 = 36.
Amit must throw either 11 or 12.
This can be done in 3 ways namely (6 + 5, 5 + 6, 6 + 6)
Thus Amit’s chance of throwing a higher number is 3/36 or 1/12
Q.1. In a throw of two dice, find the probability of getting one prime and one composite number.
(a) 1/2
(b) 1/3
(c) 2/3
(d) 3/4
Ans: B
Sol: The probability of getting a prime number (2,3 or 5) when a dice is thrown = 3/6 = 1/2.
The probability of getting a composite number (4 or 6) when a dice is thrown = 3/6 = 1/2.
The event is— getting one prime and one composite number.
This can be got as first number prime and second composite OR first number composite and second prime = [(1/2) x (1/3)] + [(1/3) x (1/2)] = 1/3
EduRev Tip: ‘1’ is not a prime number.
Q.2. There are two bags containing white and black balls. In the first bag, there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black.
(a) 40/77
(b) 41/77
(c) 37/77
(d) 43/77
Ans: B
Sol: The event definition: 1st bag and blackball OR 2nd Bag and Black Ball.
The chances of picking up either the 1st OR the 2nd Bag are 1/2 each.
The chance of picking up a black ball from the first bag is 6/14
The chance of picking up a black ball from the second bag is 7/11.Thus, using these values and the ANDs and ORs, Probability = [(1/2) x (6/14)] + [(1/2) x (7/11)] = (3/14) + (7/22) = (66 + 98)/(308) = 164/308 = 41/77
Q.3. An unbiased dice is thrown. What is the probability of getting
(i) an even number;
(ii) a multiple of 3;
(iii) an even number or a multiple of 3;
(iv) an even number & a multiple of 3?
Choose the most appropriate option.
(c) (1/2), (1/6), (2/3), (1/3)
(d) (1/6), (1/3), (2/3), (1/2)
Ans: B
In a single throw of an unbiased dice, you can get any one of the outcomes 1, 2, 3, 4, 5, 6. So, the exhaustive number of cases = 6.
(i) An even number is obtained if you obtain any one of 2, 4, 6 as an outcome.
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