Probability is one of the most important mathematical concepts that has 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. The students who didn't have Mathematics as their core subject after Class X/XII th classes have been seeing CAT exam as a test of Class XII standard. This leads to students taking a negative approach while tackling/preparing for the Quantitative aptitude section. EduRev suggests Students that the Math asked in MBA entrance is mainly logical while studying the chapter.
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. For those who are experienced in the topic, you advise you to use this chapter both for revising the basic concepts and practice the 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.
While you move into the application & problems in probability, one question that strikes in every question of probability is that whenever one is asked the question, what is the probability? the immediate question that arises/should arise in one’s mind is the probability of what?
The answer to this question is the probability of an EVENT. The EVENT is the cornerstone or the bottomline of probability. Hence, the first aspect of solving a question on probability is to define the event.
Question: 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?
Solution: 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.
Can you try this question?
We have the event of getting any number greater than 1, you have following possibilities for such an event - 2. 3. 4. 5 & 6.
The individual probabilities of each of these are 1/6, 1/6, 1/6,1/6 and 1/6 respectively. Hence, the required probability is 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6.
We have the event of getting any number greater than 1, you have following possibilities for such an event - 2. 3. 4. 5 & 6.
Usually you would encounter two major types of problems in Probability with the the use of conjunctions AND and OR:
In general, event definition means breaking up the event to the most basic building blocks, which are commonly through the two English conjunctions— AND and OR.
‘AND’ as a binding conjunctive.
Whenever you see AND as the natural conjunction joining two separate parts of the event definition, you can replace the AND by the multiplication (X) sign. Thus, if E AND R have to occur, and if the probability of their occurrence are P(E) and P(R) respectively, then the probability that E AND R occur is obtained by connecting P(E) AND P(R). Replacing the AND by multiplication sign you get the required probability as:
Req Probability = P(E) X P(R)
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?
Solution: Event here is that E hits the target AND R hits the target, both hitting the target. Hence the resulting probability is going to be
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
EduRev 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.)
Question: In a throw of two dice, find the probability of getting one prime and one composite number.
Solution: The probability of getting a prime number - 2,3 & 5 when a dice is thrown is 3/6 = 1/2.
Similarly, in a throw of a dice, there are only 2 possibilities of getting composite numbers viz : 4 or 6 and this gives a probability of 1/3 for getting a composite number.
The event is— getting one prime and one composite number.
This can be got as:
The first number is prime and the second is composite OR the first number is composite and the second is prime.
= (1/2) X (1/3) + (1/3) X (1/2) = 1/3
EduRev Tip: ‘1’ is not a prime number.
‘OR’ as a binding conjunctive.
Whenever you use OR as the natural conjunction joining two separate parts of the event definition, you replace the OR by the addition (+) sign.
Thus, if E OR R have to occur, and if the probability of their occurrence are P(E) and P(R) respectively, then the probability that E OR R occur is obtained by connecting P(E) AND P(R). Replacing the OR by addition sign you get the required probability as:
Req Probability = P(E) + P(R)
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?
Solution: As you can see whether E OR R hits the target, the outcome is counted even if one of them hits the target. Hence the resulting probability is going to be
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
Unlike above example where the event will happen only with both E & R events occurring , here if any one of E, R events can occur.
EduRev Tip: Look for the language of question, conjunction ‘OR’ is used to connect the events, which means that you need to add them up.
Combination of ‘AND’ and ‘OR’
You would encounter where in once the event is defined, the probability of each sub event within the broad event is calculated and all the sub events are connected by Multiplication (for AND) or by Addition (for OR) to get the final solution.
If two dice are thrown, what is the chance that the sum of the numbers is not less than 10.
Solution: Event Definition- The sum of the numbers is not less than 10 if it is either 10 OR 11 OR 12.
Which can be done by
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)
that is, 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 1/6⅙ = 6/36 = 1/6
Hence for the event of sum less than 10, P(E’) = 1 - P(E) = 1 - 1/6 = 5/6.
can see that no matter how many broad & sub events the problem of probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.
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.
Solution: The event definition here is: 1st bag and black ball OR 2nd Bag and Black Ball. The chances of picking up either the 1st OR the 2nd Bag are 1/2 each.
Besides, the chance of picking up a black ball from the first bag is 6/14 and the chance of picking up a black ball from the second bag is 7/11. Thus, using these values and the ANDs and ORs you get: (1/2) X (6/14) + (1/2) X (7/11) = (3/14) + (7/22) = (66 + 98)/ (308) = 164/308 = 41/77
In a four game match between Monica and Chandler, the probability that Chandler wins a particular game is 2/5 and that of Monica winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the match is drawn (Score is 2–2).
Solution: For the match to be drawn, 2 games have to be won by each of the players. If ‘A’ represents the event that Monica won a game and K represents the event that Chandler won a game, the event definition for the match to end in a draw can be described as: [The student is advised to look at the use of sub event definition.] (M & M & C& C) OR (M & C & C & K) OR (M & C & C & M) OR (C & C & M & M) OR (C & M & C & M) OR (C & M & M & C) This further translates into (2/5)X(2/5)X(3/5) X (3/5) + (2/5)X(2/5)X(3/5) X (3/5) + (2/5)X(2/5)X(3/5) X (3/5) + (2/5)X(2/5)X(3/5) X (3/5) + (2/5)X(2/5)X(3/5) X (3/5) = (36/625) X 6 = 216/625
EduRev Tip: In the above question you tried to identify the sub-events & try to add them based on the event definition.
Important Consideration during defining Events:
Brief Intro, While attempting the questions, you need to take into considerations certain activities/events. Questions in CAT exams usually have direct or indirect mention of such events.
Impossible event and certain event
An impossible event has no chance of occurring. If event A is impossible, then P(A) = 0 A certain event is certain to occur. If event A is certain, then It is usually represented by P(A) & the value of P(A) = 1.
Random experiment An experiment whose outcome has to be among a set of events that are completely known but whose exact outcome is unknown is a random experiment (e.g. Throwing of a dice, tossing of a coin). Most questions on probability in CAT are based on random experiments.
Sample space This is defined in the context of a random experiment and denotes the set representing all the possible outcomes of the random experiment. [e.g. Sample space when a coin is tossed is (Head, Tail). Sample space when a dice is thrown is (1, 2, 3, 4, 5, 6).]
Mutually Exclusive Events
Two or more events are said to be mutually exclusive; these events cannot occur simultaneously. Two or more events are said to be compatible if they can occur simultaneously. Two events (A and B) are mutually exclusive if the intersection of two events is null or they have no common element i.e. A ∩ B = φ .
e.g. In drawing a card from a deck of 52 cards:
A: The event that it is red.
B: The event that it is black.
C: The event that it is a king.
In the above case events A and B are mutually exclusive but the events B and C are not mutually exclusive or disjoint since they may have common outcomes.
Equally likely events: If two events have the same prob- ability or chance of occurrence they are called equally likely events. (In a throw of a dice, the chance of 1 showing on the dice is equal to 2 is equal to 3 is equal to 4 is equal to 5 is equal to 6 appearing on the dice.)
Exhaustive set of events: A set of events that includes all the possibilities of the sample space is said to be an exhaus- tive set of events. (e.g. In a throw of a dice the number is less than three or more than or equal to three.)
Example: If a two digit no. is selected at random. What is the probability of getting a number, which has both the digits same?
Solution: The total 2 digit numbers are 10 to 99 ⇒ 90
Required numbers are 11, 22, - - - - - 99 ⇒ 9
So probability = 9/90 = 1/10
Independent events An event is described as such if the occurrence of an event has no effect on the probability of the occurrence of another event. (If the first child of a couple is a boy, there is no effect on the chances of the second child being a boy.)
Conditional probability It is the probability of the occurrence of an event A given that the event B has already occurred. This is denoted by P(A|B). (E.g. The probability that in two throws of a dice you get a total of 7 or more, given that in the first throw of the dices the number 5 had occurred.)
Random Experiment: is the one which gives one or more results under identical conditions. E.g. a coin is tossed.
Example A coin is tossed and a single 6-sided dice is rolled. Find the probability of getting ahead on the coin and a 3 on the dice.
Solution: P(head) = 1/2
P(3) = 1/6
P (head and 3) = 1/2 • 1/6 = 1/12
If there are n-elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening of A is denoted by P (A) and is defined as the ratio m/n.
Probability of an event occurring =
Thus, P (A) = m/n
Clearly, 0 ≤ m ≤ n, therefore 0 ≤ m/n ≤ 1, so that 0 ≤ P (A) ≤ 1
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, then
P(E ∪ F) = P(E) + P(F) – P(E and F together)
Note: P (neither E nor F) = 1 – P(E or F).
Since the number of cases in which the event A will not happen is n - m, therefore if denotes not happening of A, then the probability P () of not happening of A is given by
Example From a bag containing 8 green and 5 red balls, three are drawn one after the other. Find the prob-ability 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.
Solution: (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. This means that 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 are equal to each other. Hence answer to the question above will be:
Required probability = 8/13 X 8/13 X 8/13 = (83/133)
(b) When the balls are not replaced, the probability of drawing any color of ball for every fresh draw changes. Hence, the answer here will be: Required probability = 8/13 X 7/12 X 6/11
EduRev Tip: If n dice are thrown then total possible outcomes = 6n & if n coins are tossed then possible total outcomes = 2n
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: A dice is tossed 5 times. What is the probability that 5 shows up exactly thrice?
Solution: Here the 'random experiment' consists of tossing a die 5 times and observing the number '5' as success, then 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:
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.
(a) (1 / 3), (1 / 2), (2 / 3), (1 / 6)
(b) (1 / 2), (1 / 3), (2 / 3), (1 / 6)
(c) (1 / 2), (1 / 6), (2 / 3), (1 / 3)
(d) (1 / 6), (1 / 3), (2 / 3), (1 / 2)
Explanation In a single throw of an unbiased dice, you can get any one of the outcomes 1, 2, 3, 4, 5, 6. So, exhaustive number of cases = 6.
(i) An even number is obtained if you obtain any one of 2, 4, 6 as an outcome. So, favourable number of cases = 3.
Thus, required probability = (3 / 6) = (1 / 2)
(ii) A multiple of 3 is obtained if you obtain any one of 3, 6 as an outcome. So, favourable number of cases = 2.
Thus, required probability = (2 / 6) = (1 / 3)
(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6. So, favourable number of cases = 4.
Thus, required probability = (4 / 6) = (2 / 3)
(iv) An even number and a multiple of 3 is obtained if you get 6 as an outcome. So, favourable number of cases = 1.
Thus, required probability = (1 / 6)
Example: What is the chance of drawing an ace from a deck of cards?
Solution: 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 is the ratio of total number of favorable outcomes to the total number of possible outcomes. Which in this case is: 4/52 = 1/13.
Example: 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,
(iii) either red or king,
(iv) red and a king
Solution: Out of 52 cards, one card can be drawn in 52C1 ways.
Therefore exhaustive number of cases = 52C1 = 52.
(i) There are four aces in a pack of 52 cards, out of which one can be drawn in 4C1.
Therefore Favourable number of cases = 4C1 = 4.
So, required probability = 4/52 = 1/13.
(ii) There are 26 red cards, out of which one red card can be drawn in 26C1 ways. Therefore, Favourable number of cases = 26C1 = 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 28C1 ways.
Therefore favourable number of cases = 28C1 = 28.So, required probability = 28/52 = 7/13.
(iv) There are 2 cards, which are red and king. Therefore, favourable number of cases = 2C1 = 2.
So, required probability = 2/52 = 1/26.
Example: Harshad and Amit throw with two dice. If Harshad throws 10, what is Amit’s chance of throwing a higher number?
Solution: 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
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