How to Prepare for Quantitative Aptitude for CAT (2014)

Block VI: Counting

Chapter 18. PROBABILITY

CONCEPT AND IMPORTANCE OF PROBABILITY

Probability is one of the most important mathematical concepts that we use/come across in our day-to-day life. Particularly important in business and economic situations, probability is also used by us in our personal lives. For a lot of students who are not in touch with Mathematics after their Xth/XIIth classes, this chapter, along with permutations and combinations, is seen as an indication that XIIth standard Mathematics appear in the MBA entrance exams. This leads to students taking negative approach while tacking/preparing for the Mathematics section. Students are advised to remember that the Math asked in MBA entrance is mainly logical while studying the chapter.

As I set out to explain the basics of this chapter, I intend to improve your concepts of probability to such an extent that you feel in total control of this topic.

For those who are reasonably strong, my advice would be to use this chapter both for revisiting the basic concepts as well as for extensive practice.

Probability means the chance of the occurrence of an event. In layman terms, we can say that it is the likelihood that something—that is defined as the event—will or will not occur. Thus probabilities can be estimated for each of the following events in our personal lives:

(a)the probability that an individual student of B.Com will clear the CAT,

(b)The chance that a candidate chosen at random will clear an interview,

(c)The chance that you will win a game of flush in cards if you have a trio of twos in a game where four people are playing,

(d)The likelihood of India’s winning the football World Cup in 2014.

(e)The probability that a bulb will fuse in it’s first day of operation, and so on.

The knowledge of these estimations helps individuals decide on the course of action they will take in their day-to-day life. For instance, your estimation/ judgement of the probability of your chances of winning the card game in Event c above will influence your decision about the amount of money you will be ready to invest in the stakes for the particular game. The application of probability to personal life helps in improving our decision making.

However, the use of probability is much more varied and has far reaching influence on the world of economics and business. Some instances of these are:

(a)the estimation of the probability of the success of a business project,

(b)the estimation of the probability of the success of an advertising campaign in boosting the profits of a company,

(c)the estimation of the probability of the death of a 25-year old man in the next 10 years and that of the death of a 55-year old man in the next 10 years leading to the calculation of the premiums for life insurance,

(d)the estimation of the probability of the increase in the market price of the share of a company, and so on.

UNDERLYING FACTORS FOR REAL-LIFE ESTIMATION OF PROBABILITY

The factors underlying an event often affect the probability of that event’s occurrence. For instance, if we estimate the probability of India winning the 2015 World Cup as 0.14 based on certain expectations of outcomes, then this probability will definitely improve if we know that Sachin Tendulkar will score 800 runs in that particular World Cup.

As we now move towards the mathematical aspects of the chapter, one underlying factor that recurs 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 the EVENT.

The EVENT is the cornerstone or the bottomline of probability. Hence, the first objective while trying to solve any question in probability is to define the event.

The event whose probability is to be found out is described in the question and the task of the student in trying to solve the problem is to define it.

In general, the student can either define the event narrowly or broadly. Narrow definitions of events are the building blocks of any probability problem and whenever there is a doubt about a problem, the student is advised to get into the narrowest form of the event definition.

The difference between the narrow and broad definition of event can be explained through an example:

Example: What is the probability of getting a number greater than 2, in a throw of a normal unbiased dice having 6 faces?

The broad definition of the event here is getting a number greater than 2 and this probability is given by 4/6. However, this event can also be broken down into its more basic definitions as:

The event is defined as getting 3 or 4 or 5 or 6. The individual probabilities of each of these are 1/6,1/6,1/6 and 1/6 respectively.

Hence, the required probability is 1/6 + 1/6 + 1/6 + 1/6 = 4/6 = 2/3.

Although in this example it seems highly trivial, the narrow event-definition approach is very effective in solving difficult problems on probability.

In general, event definition means breaking up the event to the most basic building blocks, which have to be connected together through the two English conjunctions—AND and OR.

The Use of the Conjunction AND (Tool No. 9)

Refer Back to school section. Whenever we use AND as the natural conjunction joining two separate parts of the event definition, we replace the AND by the multiplication sign.

Thus, if A AND B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A AND B occur is got by connecting P(A) AND P(B). Replacing the AND by multiplication sign we get the required probability as:

P(A) × P(B)

Example: If we have the probability of A hitting a target as 1/3 and that of B hitting the target as 1/2, then the probability that both hit the target if one shot is taken by both of them is got by

Event Definition: A hits the target AND B hits the target.

Æ P(A) × P(B) = 1/3 × 1/2 = 1/6

(Note that since we use the conjunction AND in the definition of the event here, we multiply the individual probabilities that are connected through the conjunction AND.)

The Use of the Conjunction OR (Tool No. 10)

Refer Back to school section. Whenever we use OR as the natural conjunction joining two separate parts of the event definition, we replace the OR by the addition sign.

Thus, if A OR B have to occur, and if the probability of their occurrence are P(A) and P(B) respectively, then the probability that A OR B occur is got by connecting

P(A) OR P(B). Replacing the OR by addition sign, we get the required probability as

P(A) + P(B)

Example: If we have the probability of A winning a race as 1/3 and that of B winning the race as 1/2, then the probability that either A or B win a race is got by

Event Definition: A wins OR B wins.

Æ P(A) + P(B) = 1/3 + 1/2 = 5/6

(Note that since we use the conjunction OR in the definition of the event here, we add the individual probabilities that are connected through the conjunction OR.)

Combination of AND and OR

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 if it is either 10 OR 11 OR 12.

Which can be done by

(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 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 + 1/6 × 1/6 = 6/36 = 1/6

The bottomline is that no matter how complicated the problem on probability is, it can be broken up into its narrower parts, which can be connected by ANDs and ORs to get the event definition.

Once the event is defined, the probability of each narrow event within the broad event is calculated and all the narrow events are connected by Multiplication (for AND) or by Addition (for OR) to get the final solution.

Example: In a four game match between Kasporov and Anand, the probability that Anand wins a particular game is 2/5 and that of Kasporov 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).

For the match to be drawn, 2 games have to be won by each of the players. If ‘A’ represents the event that Anand won a game and K represents the event that Kasporov 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 narrow event definition.]

A&K&K) OR (A&K&A&K) OR (A&K&K&A)

OR (K&K&A&A) OR (K&A&K&A) OR (K&A&A&K)

This further translates into

(2/5)²(3/5)² + (2/5)²(3/5)² + (2/5)²(3/5)² + (2/5)²(3/5)²

+ (2/5)²(3/5)² + (2/5)²(3/5)²

= (36/625) × 6 = 216/625

After a little bit of practice, you can also think about this directly as:

⁴C₂ × (2/5)² × ²C₂ × (3/5)² = 6 × 1 × 36/625 = 216/625

Where, 4C₂ gives us the number of ways in which Anand can win two games and 2C₂ gives us the number of ways in which Kasporov can win the remaining 2 games (obviously, only one).

BASIC FACTS ABOUT PROBABILITY

For every event that can be defined, there is a corresponding non-event, which is the opposite of the event. The relationship between the event and the non-event is that they are mutually exclusive, that is, if the event occurs then the non-event does not occur and vice versa.

The event is denoted by E; the number of ways in which the event can occur is defined as n(E) and the probability of the occurrence of the event is P(E).

The non-event is denoted by E¢; the number of ways in which the non-event can occur is defined as n(E¢) while the probability of the occurrence of the event is P(E¢).

The following relationships hold true with respect to the event and the non-event.

n(E) + n(E¢) = sample space representing all the possible events that can occur related to the activity.

P(E) + P(E¢) = 1

This means that if the event does not occur, then the non-event occurs.

Æ P(E) = 1 – P(E¢)

This is often very useful for the calculation of probabilities of events where it is easier to describe and count the non-event rather than the event.

Illustration

The probability that you get a total more than 3 in a throw of 2 dice.

Here, the event definition will be a long and tedious task, which will involve long counting. Hence, it would be more convenient to define the non-event and count the same.

Therefore, here the non-event will be defined as

A total not more than 3 Æ 2 or 3 Æ (1&1) OR (1&2) OR (2&1) = 1/36 + 1/36 + 1/36 = 3/36 = 1/12.

However, a word of caution especially for students not comfortable at mathematics: Take care while defining the non-event. Beware of a trap like Æ event definition: Total > 10 in two throws of a dice does not translate into a non-event of < 10 but instead into the non-event of < 10.

SOME IMPORTANT CONSIDERATIONS WHILE DEFINING EVENT

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 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).]

Event The set representing the desired outcome of a random experiment is called the event. Note that the event is a subset of the sample space.

Non-event The outcome that is opposite the desired outcome is the non-event. Note that if the event occurs, the non-event does not occur and vice versa.

Impossible Event An event that can never occur is an impossible event. The probability of an impossible event is 0. e.g. (Probability of the occurrence of 7 when a dice with 6 faces numbered 1–6 is thrown).

Mutually Exclusive Events A set of events is mutually exclusive when the occurrence of any one of them means that the other events cannot occur. (If head appears on a coin, tail will not appear and vice versa.)

Equally Likely Events If two events have the same probability 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 exhaustive set of events. (e.g. In a throw of a dice the number is less than three or more than or equal to three.)

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 dices we get a total of 7 or more, given that in the first throw of the dices the number 5 had occurred.)

The Concept of Odds For and Odds Against

Sometimes, probability is also viewed in terms of odds for and odds against an event.

Odds in favour of an event E is defined as:

Odds against an event is defined as:

Expectation: The expectation of an individual is defined as

Probability of winning × Reward of winning

Illustration: A man holds 20 out of the 500 tickets to a lottery. If the reward for the winning ticket is ` 1000, find the expectation of the man.

Solution: Expectation = Probability of winning × Reward of winning = × 1000 = ` 40.

ANOTHER APPROACH TO LOOK AT THE PROBABILITY PROBLEMS

The probability of an event is defined as

This means that the probability of any event can be got by counting the numerator and the denominator independently.

Hence, from this approach, the concentration shifts to counting the numerator and the denominator.

Thus for the example used above, the probability of a number > 2 appearing on a dice is:

=

The counting is done through any of

(a)The physical counting as illustrated above,

(b)The use of the concept of permutations,

(c)The use of the concept of combinations,

(d)The use of the MNP rule.

[Refer to the chapter on Permutations and Combinations to understand b, c and d above.]

WORKED-OUT PROBLEMS

Problem 18.1 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 when a dices is thrown is 3/6 = 1/2. (This occurs when we get 2, 3 or 5 out of a possibility of getting 1, 2, 3, 4, 5 or 6.)

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.

Now, let us look at defining the event. 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) × (1/3) + (1/3) × (1/2) = 1/3

Problem 18.2 Find the probability that a leap year chosen at random will have 53 Sundays.

Solution A leap year has 366 days. 52 complete weeks will have 364 days. The 365th day can be a Sunday (Probability = 1/7) OR the 366th day can be a Sunday (Probability = 1/7). Answer = 1/7 + 1/7 = 2/7.

Alternatively, you can think of this as: The favourable events will occur when we have Saturday and Sunday or Sunday and Monday as the 365th and 366th days respectively. (i.e. 2 possibilities of the event occurring). Besides, the total number of ways that can happen are Sunday and Monday OR Monday and Tuesday … OR Friday and Saturday OR Saturday and Sunday.

Problem 18.3 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 we get:

(1/2) × (6/14) + (1/2) × (7/11) = (3/14) + (7/22) = (66 + 98)/(308) = 164/308 = 41/77

Problem 18.4 The letters of the word LUCKNOW are arranged among themselves. Find the probability of always having NOW in the word.

Solution The required probability will be given by the equation

= No. of words having NOW/Total no. of words

= 5!/7! = 1/42 [See the chapter of Permutations and

Combinations to understand the logic behind these values.]

Problem 18.5 A person has 3 children with at least one boy. Find the probability of having at least 2 boys among the children.

Solution The event is occurring under the following situations:

(a)Second is a boy and third is a girl OR

(b)Second is a girl and third is a boy OR

(c)Second is a boy and third is a boy

This will be represented by: (1/2) × (1/2) + (1/2) × (1/2) + (1/2) × (1/2) = 3/4

Problem 18.6 Out of 13 applicants for a job, there are 5 women and 8 men. Two persons are to be selected for the job. The probability that at least one of the selected persons will be a woman is:

Solution The required probability will be given by

First is a woman and Second is a man OR

First is a man and Second is a woman OR

First is a woman and Second is a woman

i.e. (5/13) × (8/12) + (8/13) × (5/12) + (5/13) × (4/12) = 100/156 = 25/39

Alternatively, we can define the non-event as: There are two men and no women. Then, probability of the non-event is

(8/13) × (7/12) = 56/156

Hence, P(E) = (1– 56/156) = 100/156 = 25/39

[Note: This is a case of probability calculation where repetition is not allowed.]

Problem 18.7 The probability that A can solve the problem is 2/3 and B can solve it is 3/4. If both of them attempt the problem, then what is the probability that the problem gets solved.

Solution The event is defined as:

A solves the problem AND B does not solve the problem

OR

A doesn’t solve the problem AND B solves the problem

OR

A solves the problem AND B solves the problem.

Numerically, this is equivalent to:

(2/3) × (1/4) + (1/3) × (3/4) + (2/3) × (3/4)

= (2/12) + (3/12) + (6/12) = 11/12

Problem 18.8 Six positive numbers are taken at random and are multiplied together. Then what is the probability that the product ends in an odd digit other than 5.

Solution The event will occurs when all the numbers selected are ending in 1, 3, 7 or 9.

If we take numbers between 1 to 10 (both inclusive), we will have a positive occurrence if each of the six numbers selected are either 1, 3, 7 or 9.

The probability of any number selected being either of these 4 is 4/10 (4 positive events out of 10 possibilities)

[Note: If we try to take numbers between 1 to 20, we will have a probability of 8/20 = 4/10. Hence, we can extrapolate up to infinity and say that the probability of any number selected ending in 1, 3, 7 or 9 so as to fulfill the requirement is 4/10.]

Hence, answer = (0.4)⁶

Problem 18.9 The probability that Arjit will solve a problem is 1/5. What is the probability that he solves at least one problem out of ten problems?

Solution The non-event is defined as:

He solves no problems i.e. he doesn’t solve the first problem and he doesn’t solve the second problem … and he doesn’t solve the tenth problem.

Probability of non-event = (4/5)¹⁰

Hence, probability of the event is 1-(4/5)¹⁰

Problem 18.10 A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly 2 of them will be defective?

Solution The probability that exactly two balls are defective and exactly two are not defective will be given by (4C₂) × (8/25) × (7/24) × (17/23) × (16/22)

Problem 18.11 Out of 40 consecutive integers, two are chosen at random. Find the probability that their sum is odd.

Solution Forty consecutive integers will have 20 odd and 20 even integers. The sum of 2 chosen integers will be odd, only if

(a)First is even and Second is odd OR

(b)First is odd and Second is even

Mathematically, the probability will be given by:

P(First is even) × P(Second is odd) + P(First is odd) × P(second is even)

= (20/40) × (20/39) + (20/40) × (20/39)

= (2 × 20²/40 × 39) = 20/39

Problem 18.12 An integer is chosen at random from the first 100 integers. What is the probability that this number will not be divisible by 5 or 8?

Solution For a number from 1 to 100 not be divisible by 5 or 8, we need to remove all the numbers that are divisible by 5 or 8.

Thus, we remove 5, 8, 10, 15, 16, 20, 24, 25, 30, 32, 35, 40, 45, 48, 50, 55, 56, 60, 64, 65, 70, 72, 75, 80, 85, 88, 90, 95, 96, and 100.

i.e. 30 numbers from the 100 are removed.

Hence, answer is 70/100 = 7/10 (required probability)

Alternatively, we could have counted the numbers as number of numbers divisible by 5 + number of numbers divisible by 8 – number of numbers divisible by both 5 or 8.

= 20 + 12 – 2 = 30

Problem 18.13 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.

Solution

(a)When the balls drawn are replaced, we 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 = = (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 =

LEVEL OF DIFFICULTY (I)

Directions for Questions 25 to 27: Read the following passage and answer the questions based on it.

The Bangalore office of Infosys has 1200 executives. Of these, 880 subscribe to the Time magazine and 650 subscribe to the Economist. Each executive may subscribe to either the Time or the Economist or both. If an executive is picked at random, answer questions 25–27.

LEVEL OF DIFFICULTY (II)

LEVEL OF DIFFICULTY (III)

Level of Difficulty (I)

Solutions and Shortcuts

Level of Difficulty (I)

1.Out of a total of 6 occurrences, 3 is one possibility = 1/6.

2.5 or 6 out of a sample space of

1, 2, 3, 4, 5 or 6 = 2/6 = 1/3

3.Event definition is:

(1 and 1) or (1 and 2) or (1 and 3) or (1 and 4) or (1 and 5) or (1 and 6) or (2 and 1) or (3 and 1) or (4 and 1) or (5 and 1) or (6 and 1)

Total 11 out of 36 possibilities = 11/36

4.Event definition: First is blue and second is blue = 7/12 × 6/11 = 7/22.

5.Knave and queen or Queen and Knave Æ

4/52 × 4/51 + 4/52 × 4/51 = 8/663

6.(i) 13/52 = 1/4

(ii)4 kings and 4 queens out of 52 cards.

Thus, 8/52 = 2/13.

(iii)13 spades + 3 kings Æ 19/52.

7.(i) Event definition is: T and T and H or T and H and T or H and T and T = 3 × 1/8 = 3/8.

(ii)Same as above = 3/8.

8.(i) Probability of 3 heads = 1/8

Also, Probability of 3 tails = 1/8.

Required probability = 1– (1/8 + 1/8) = 6/8 = 3/4.

(ii)H and H and H = 1/8.

9.At least one tail is the non-event for all heads.

Thus, P (at least 1 tail) = 1 – P(all heads)

= 1 – 1/8 = 7/8.

10.(i) Positive outcomes are 2(1 way), 4(3 ways), 6(5 ways) 8(5 ways), 10 (3 ways), 12 (1way).

Thus, 18/36 = 1/2

(ii)Positive outcomes are: 4, 8 and 12

4 (3 ways), 8(5 ways) and 12(1 way)

Gives us 9/36 = 1/4.

(iii)Positive outcomes are 2(1 way), 3(2 ways), 5(4 ways), 7(6 ways). Total of 13 positive outcomes out of 36.

Thus, 13/36.

11.(i)First is white and second is white 7/10 × 6/9 = 7/15.

(ii)White and Green or Green and White

7/10 × 39 + 3/10 × 7/9

42/90 = 7/15.

12.

From the figure it is evident that 80 students passed at least 1 exam. Thus, 20 failed both and the required probability is 20/100 = 1/5.

13.3 or 4 or 5 or 6 = 4/6 = 2/3

14.(i)Since 2 is the only prime number out of the three numbers, the answer would be 1/3

(ii)Since all the numbers are even, it is sure that the number drawn out is an even number. Hence, the required probability is 1.

(iii)Since there are no odd numbers amongst 2, 4 and 8, the required probability is 0.

15.(i)The event would be Head and Head Æ ½ × ½ = ¼

(ii)The event would be Head and Tail OR Tail and Head Æ = ½ × ½ + ½ × ½ = 1/8

(iii)The event would be Tail and Tail Æ ½ × ½ = ¼

16.(i)With a six on the first dice, there are 6 possibilities of outcomes that can appear on the other dice (viz. 6 & 1, 6 & 2, 6 & 3, 6 & 4, 6 & 5 and 6 & 6). At the same time with 6 on the second dice there are 5 more possibilities for outcomes on the first dice: (1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6)

Also, the total outcomes are 36. Hence, the required probability is 11/36.

(ii)Out of 36 outcomes, 5 can come in the following ways – 1 + 4; 2 + 3; 3 + 2 or 4 + 1 Æ 4/36 = 1/9.

17.Odds against an event =

In this case, the event is: All black, i.e., First is black and second is black and third is black.

P(E) = 5/9 × 4/8 × 3/7 = 60/504 = 5/42.

Odds against the event = 37/5.

18.Event definition is: 15 or 16 or 17 or 18.

15 can be got as: 5 and 5 and 5 (one way)

Or

6 and 5 and 4 (Six ways)

Or

6 and 6 and 3 (3 ways)

Total 10 ways.

16 can be got as: 6 and 6 and 4 (3 ways)

Or

6 and 5 and 5 (3 ways)

Total 6 ways.

17 has 3 ways and 18 has 1 way of appearing.

Thus, the required probability is: (10 + 6 + 3 + 1)/216 = 20/216 = 5/54.

19.(i)Positive outcomes = 2 (187 or 215)

Total outcomes = 9 × 8 × 7

Required probability = 2/504 = 1/252

(ii)= 2/729.

20.1/6 + 1/10 + 1/8 = 47/120

21.The event definition would be given by:

First is blue and second is blue = 3/11 × 2/10 = 3/55

22.(i)There are six doubles (1, 1; 2 & 2; 3 & 3; 4 & 4; 5 & 5; 6 & 6) out of a total of 36 outcomes Æ 6/36 = 1/6

(ii)Sum greater than 10 means 11 or 12 Æ 3/36 = 1/12

(iii)Sum less than 10 is the non event for the case sum is 10 or 11 or 12. There are 3 ways of getting 10, 2 ways of getting 11 and 1 way of getting a sum of 12 in the throw of two dice. Thus, the required probability would be 1 – 6/36 = 5/6.

23.Expectation = Probability of winning × Reward of winning = (10/5000) × 1 crore = (1 crore/500) = 20000.

24.1/⁴C₂ = 1/6.

25.

(i)880/1200 = 11/15

(ii)650/1200 = 13/24

26.330/1200 = 11/40

27.330/880 = 3/8

28.Black and Black and Black = 4/9 × 3/8 × 2/7

= 23/504 = 1/21.

29.1/⁵P₂ = 1/20.

30.6 × (4/52) × (3/51) × (2/50) = 6/5525.

31.Positive Outcomes are: 5, 7, 10, 14, 15 or 20

Thus, 6/20 = 3/10.

32.972/1972 = 243/ 493.

33.Black and black = (7/16) × 6/15 = 7/40

34.(i) The required probability would be given by:

All are Red OR All are white OR All are Blue

= (6/18) × (5/17) × (4/16) + (4/18) × (3/17) × (2/16) + (8/18) × (7/17) × (6/16)

= 480/(18 × 17 × 16) 5/51

(ii)All blue = (8 × 7 × 6)/(18 × 17 × 16) = 7/102

35.The required value of the union of the two non events (of A and B) would be 1 – 1/4 = ¾

36.P (Both are selected) = P(A) × P(B)

Since P(A) = 0.5, we get

0.3 = 0.5 × 0.6.

The maximum value of P(B) = 0.6.

Thus P(B) = 0.9 is not possible.

37.

We have: 19/40 + 1/8 + 5/24 = 97/120

38.P (Amit) = 1/3

P (Vikas) = 2/7

P (Vivek) = 1/8.

Required Probability = 1/3 + 2/7 + 1/8 = 125/168.

39.P(A) + P(B) + P(C) = 1 Æ 2P(B)/3 + P(B) + P(B)/2 = 1 Æ 13P(B)/6 = 1 Æ P(B) = 6/13. Hence, P(A) = 4/13

40.P(A) = 1 – 0.65 = 0.35.

Hence, P(B) = 0.65 – 0.35 = 0.3

41.(i)The required probability would be given by the event:

White from first bag and white from second bag = (4/6) × (3/8) = ¼

(ii)The required probability would be given by the event:

Black from first bag and black from second bag = (2/6) × (5/8) = 10/48 = 5/24

(iii)This would be the non event for the event [both are white OR both are black).

Thus, the required probability would be:

¼ – 5/24 = 13/24.

42.P (E₁) 3/8

P (E₂) = 7/12.

Event definition is: E₁ occurs and E₂ does not occur or E₁ occurs and E₂ occurs or E₂ occurs and E₁ does not occur.

(3/8) × (5/12) + (3/8) × (7/12) + (5/8) × (7/12) = 71/96.

43.Kamal is selected and Monica is not selected or Kamal is not selected and Monica is selected Æ (1/3) × (4/5) ¨ (2/3) × (1/5) = (6/15) = (2/5).

44.(i)1/5 × 1/7 = 1/35

(ii)(1/5) × (6/7) + (4/5) × (1/7) = 10/35 = 2/7.

(iii)(4/5) × (6/7) = 24/35

(iv)Both selected or 1 selected = 1/35 + 2/7 = 11/35.

Level of Difficulty (II)

1.The possible outcomes are:

(1, 1); (1, 2); (2, 1), (2, 2); (3, 1); (1, 3).

Out of six cases, in two cases there is exactly one ‘2’

Thus, the correct answer is 2/6 = 1/3.

2.Event definition is A hits, B hits and C hits OR any two of the three hits.

3.The event can be defined as:

First bag is selected and red ball is drawn.

1/2 × 5/12 + ½ × 3/15 = (5/24) + (3/30) = 37/120

4.(a)For the least chance of drawing a red ball the distribution has to be 5 Red + 11 white in one bag and 1 white in the second bag. This gives us

=

(b)For the greatest chance of drawing a red ball the distribution has to be 1 Red in the first bag and 4 red + 12 white balls in the second bag. This gives us

=

5.One head and seven tails would have eight positions where the head can come.

Thus, 8 × (1/2)⁸ = (1/32)

6.A leap year has 366 days – which means 52 completed weeks and 2 more days. The last two days can be (Sunday, Monday) or (Monday, Tuesday) or ……. (Saturday, Sunday).

2 scenarios out of 7 have a 53^rd Sunday.

Thus, 2/7 is the required answer.

7.The count of the event will be given by:

The number of all 2 digit integers – the number of all 2 digit integers divisible by 7

8.Blue and Red or Red and Blue

= (10/25) × (15/24) + (15/25) × (10/24) = (1/2).

9.(i) 1/5 + ½ = 7/10.

(ii)1 – (7/10) = (3/10)

10.A total of six can be made in any of the following ways (1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1)

11.The event definitions are:

(a)Passes the first AND Passes the second AND Passes the third AND Passes the fourth

(b)Fails the first AND Fails the second AND Fails the third AND Fails the fourth

(c)Fails all is the non-event.

12.(i)First is cricket and second is cricket and third is cricket Æ (3/9) × (3/9) × (3/9) = (1/27)

(ii)(2/9) × (3/9) × (4/9) = (8/243)

(iii)All are cricket or all are tennis or all are squash.

(3/9)³ + (2/9)³ + (4/9)³ = (99/729) = (11/81)

13.(i)(3/9) × (2/8) × (1/7) = (1/84)

(ii)(2/9) × (3/8) × (4/7) = (1/21)

(iii)(3/9) × (2/8) × (1/7) + 0 + (4/9) × (3/8) × (2/7)

30/504 = 5/84

14.The event definitions are

(i)The book transferred is by Stephen Covey

(ii)The book transferred is by Stephen Covey AND The book picked up is by Stephen Covey

(iii)The book transferred is by Stephen Covey AND The book picked up is by Vinay Singh.

15.(2/3) × (3/4) + (1/3) × (2/7) = (1/2) + (2/21) = (25/42)

16.(i)(3/50) × (2/49) = (3/1225)

(ii)(3/50) × (47/49) + 47/50 × (3/49) = (141/1225)

(iii)(47/50) × (46/49) = (1081/1225)

17.(i)(3/5) × (3/7) = (9/35)

(ii)(3/5) × 4/7) = (12/35)

(iii)(2/5) × (3/7) = (6/35)

(iv)(3/5) × (4/7) + (2/5) × (3/7) + (3/5) × (3/7) = 27/35

18.They will contradict each other if: A is true and B is false or A is false and B is true.

(3/4) × (1/6) + (1/4) × (5/6) = 1/3.

19.For the counting of the number of events, think of it as a circular arrangement with n – 1 people (by considering the two specified persons as one). This will give you n(E) = (n – 2)! × (2)!

20.The whole numbers selected can only be 1, 3, 7 or 9 and cannot contain 2, 4, 6, 8, 0 or 5.

21.⁴C₂ × (6/36)² × (30/36)² = 6 × (1/36) × (25/36) = 25/216.

22.The required probability will be given by the expression:

23.Girl and Boy and Boy or Boy and Girl and Boy

Or

Boy and Boy and Girl

= (3/4) × (2/4) × (2/3) + (1/4) × (2/4) × 2/3)

+ (1/4) × (2/4) × (1/3) = 18/48 = 3/8.

24.n(E) = 1

n(S) = ³C₁ × 9 × 9 = 243

25.11/17 (if the first one is black, there will be 11 black balls left out of 17)

26.6/17

27.There must have been two sixes in the first five throws. Thus, the answer is given by:

28.

29.There will be ⁶C₃ triangles formed overall. Out of these visualise the number of equilateral triangles.

30.(5/9) × (4/8) + (4/9) × (3/8) = 32/72 = 4/9

As matched shoes means both red or both black.

31.The event definition here is that first is not a heart and second is not a heart and third is a heart = (3/4) × (3/4) × (1/4) = 9/64.

32.(39/52) × (38/51) × (13/50) = 247/1700.

33.The event definition will be: Event X happens and Y doesn’t happen Or Y happens and X does not happen.

34.X happens and Y does not happen or X doesn’t happen and Y happens or X happens and Y happens

P × (1 – P) + (1 – P) × P¢ + P × P¢ = P + P¢ – PP¢

35.Same logic as question 6.

36.(2/4) × (1/3) = 1/6. ( Faulty and faulty)

37.Faulty and not Faulty and Faulty or Not Faulty and Faulty and Faulty = (2/4) × (2/3) × (1/2) + (2/4) × (2/3) × (1/2) = 1/3.

38.When you put the 7 balls with a gap between them in a row, you will have 8 spaces.

39.The appearance of head or tail on a toss is independent of previous occurrences.

Hence, 1/2.

40. = 1/35.

41.P =

=

= 8! × 4!/11! = 24/990 = 4/165.

42. = = 1/1155.

43.(1/3) × (1/4) × (4/5) + (1/3) × (3/4) × (1/5) + (2/3)

× (1/4) × (1/5) + (1/3) × (1/4) × (1/5)

= 10/60 = 1/6.

44.⁵C₃ × [ (8/12) × (7/11) × (6/10) × (4/9) × (3/8)] = 14/33.

45.There can be three girls and one boy.

46.P =

= (10! × 2!)/11!

47.Consider the 6 girls to be one person. Then the number of arrangements satisfying the condition is given by n(E) = 7! × 6!

48.⁶C₂ × [ (7/11) × (6/10) × (5/9) × (4/8) × (4/7) × (3/6)] = 5/11.

49.The event definition is Red AND Red AND Not Red OR Red AND Not Red AND Red OR Not Red AND Red AND Red.

50.The numbers having 2 in them are: 2, 12, 22, 32….92 and 21, 23, 24, 25….29. Hence, n(E) = 19.

Level of Difficulty (III)

1.This problem has to be treated as if we are selecting the third card out of the 50 remaining cards.

11 of these are spades.

Hence, 11/50.

2.

The chance that A will win is an infinite GP with a = and r =

Similarly, the chance that B will win is an infinite GP with a = and r = .

3.A total of 6 can be obtained by either (1 + 2 + 3) or by (2 + 2 + 2).

4.A is true and B is true = (3/4) × (7/10) = (21/40).

5.The required probability will be given by:

6.and 7.

We do not have to consider any sum other than 5 or 7 occurring.

A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]

Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]

For 6: n(E) = 4, n(S) = 6 + 4

P = 0.4

For 7: n(E) = 6

n(S) = 6 + 4 P = 0.6

8.The number of ways for a sum of 4 = 3 i.e. [3 + 1, 2 + 2, 1 + 3]

9.Try to find the number of ways in which 0 or 1 bomb hits the bridge if n bombs are thrown.

The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.

10.The required probability will be given by: 1 – [ⁿC₀ + ⁿC₁]for n = 5.

11.The required answer will be given by:

12.n(E) = ⁶C₂ × 16 × 15 × 14 × 13 × 12

n(S) = 16⁶

13.The number of events for the condition that he will sing = 4. [34, 43, 26, 62]

The number of events in the sample = 90.

Probability that he will sing at least once

= 1 – Probability that he will not sing.

14.The required probability would be given by the event definition:

First is white and second is white = 8/22 × 7/21 = 4/33

15.The required probability would be given by the event definition:

First is red and second is red = 5/22 × 4/21 = 10/231

16.The number of ways in which 6 tickets can be selected from 10 tickets is 10⁶.

The number of ways in which the selection can be done so that the condition is satisfied = 7⁶ – 6⁶.

17.In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.

Thus, (7/15) × (8/15) = 56/225.

18.The total number of ways in which the 6 identical balls can be distributed amongst the three boxes such that each box can get 0, 1, 2, 3, 4, 5 or 6 balls is given by the formula: ^{n + r – 1}C_r where n is the number of identical balls and r is the number of boxes. This will give the sample space.

19.Event definition:

Any six of them work AND four leave OR any seven work AND three leave OR any eight work AND two leave OR any nine work AND one leaves OR All ten work.

20.“To make at least 3 trials to draw a heart” implies that he didn’t get a heart in the first two trials.

21.A box containing 4 defectives would get sent to the market if all the 4 articles selected are not defective.

Thus, ¹⁶C₄/²⁰C₄ = 364/969

22.If the box contains only 1 defective, it would be sent back if the defective is one amongst the 4 selected for testing.

To ensure one of the 4 is selected , the no. of ways is ¹⁹C₃, while the total no. of selections of 4 out of 20 is ²⁰C₄.

Thus, ¹⁹C₃/²⁰C₄ = 1/5.

23.For divisibility by 5 we need the units digit to be either 0 or 5.

The units digit in the powers of 7 follow the pattern – 7, 9, 3, 1, 7, 9, 3, 1, 7, 9…….

Hence, divide 1 to 100 into four groups of 25 elements each as follows.

A = 1, 5, 9, ------- Æ 25 elements

B = 2, 6, 10, ------ Æ 25 elements

C = 3, 7, 11, ------ Æ 25 elements

D = 4, 8, 12, ------ Æ 25 elements

Check the combination values of m and n so that 7^m + 7ⁿ is divisible by 5.

24.P(minimum 3) or P (maximum 7)

P(minimum 3) = ⁷C₂/¹⁰C₃ = 21/120

P( max 7) = ⁶C₂/¹⁰C₂ = 15/120

25.For the event to occur, the dice should show values from 2, 3, 4 or 5. This is similar to selection with repetition.

26.Yellow and Red and Blue = (3/6) × (2/6) × (1/6) = (6/216) = 1/36.

27.Two white and one black can be obtained only through the following three sequences:

Ball drawn from A and B are white and the ball drawn from C is Black. or

Ball drawn from A and C are white and the ball drawn from B is Black. or

Ball drawn from B and C are white and the ball drawn from A is Black.

28.At least one means (exactly one + exactly two + exactly three)

At least two means (exactly two + exactly three)

The problem gives the probabilities for passing in at least one, at least two and exactly two.

29.All four are not in the correct envelopes means that at least one of them is in a wrong envelope. A little consideration will show that one letter being placed in a wrong envelope is not possible, since it will have to be interchanged with some other letter.

Since, there is only one way to put all the letters in the correct envelopes, we can say that the event of not all four letters going into the correct envelopes will be given by

5! – 1 = 119

30.n(E) = 44

n(S) = 120

31.A₁ = (112 or 121)

A₂ = (112 or 211)

A₃ = (121 or 211)

Option (b) is correct.

32.

From the venn diagram we get:

(2/3) + (5/9) – x = 4/5 Æ x = (2/3) + (5/9) – (4/5) = 19/45

33.

Also P(A » B) = 1 – P(A¢ « B¢)

(3/7) + (1/2) – x = 13/14 Æ x = 0

Thus, there is no interference between A and B as P(A « B) = x = 0. Hence, A and B are mutually exclusive.

34.The required probability will be given by .

35.The non-event in this case is that the problem is not solved.

36-37.Are similar to Question No. 2 of LOD III.

38.The event definition will be:

The first bag is selected AND a red ball is selected. OR

The second bag is selected AND a red ball is selected.

39.The event definition is:

A girl is selected from the first group and one boy each are selected from the second and third groups. OR A girl is selected from the second group and one boy each are selected from the first and third groups. OR A girl is selected from the third group and one boy each are selected from the first and second groups.

40.The event definition will be:

A solves the problem and B and C do not solve the problem. OR B solves the problem and A and C do not solve the problem. OR C solves the problem and A and B do not solve the problem.

41.The three balls that are taken out can be either 3 black balls or 2 black and 1 red ball or 1 black and 2 red ball or 3 red balls.

Each of these will give their own probabilities of drawing a black ball.

42.3 Blacks and 4^th is Red or 2 Blacks and 1 Red and 4^th is Red or 1 Black and 2 Reds and 4^th is Red

= (5/8) × (4/7) × (3/6) × (3/5) + ³C₁ × ( 5/8) × (4/7)

× (3/6) × (2/5) + ³C₁ × (5/8) × (3/7) × (2/6) × (1/5)

= = 630/1680 = 3/8 = 0.375

43.The event definition would be: Ball transferred is white and white ball drawn

Or

Ball transferred is black and white ball is drawn.

The answer will be given by:

(5/9) × (8/17) + (4/9) × (7/17) = 68/153 = 4/9

44.Solve this on a similar pattern to the example given in the theory of this chapter.

45.Required probability = 1 – probability that no pair is selected

46.We can have a maximum of 5 heads.

For 0 heads Æ P(E) = (1/2¹⁰) × 1

For 1 heads Æ P(E) = (1/2¹⁰) × 1

For 2 heads and for them not to occur consecutively we will need to see the possible distribution of 8 tails and 2 heads.

Since the 2 heads do not need to occur consecutively, this would be given by (All – 2heads together)

Æ (¹⁰C₈ – 9)

P(E) =

Solving in this fashion, we would get 1/2³.

47.This can be got by taking the number of ways in which exactly two people are born on the same day divided by the total number of ways in which 7 people can be born in 7 days of a week. For the first part select two people from 7 in ⁷C₂ ways & select a day from the week on which they have to be born in ⁷C₁ ways & for the remaining 5 people select 5 days out of the remaining six days of the week & then the number of arrangements of these 5 people in 5 days-thus a total of ⁷C₂ × ⁷C₁ × ⁶C₅ × 5! ways. Also, the number of ways in which seven people can be born on 7 days would be given by 7⁷. Hence, the answer is given by: (⁷C₂ × ⁷C₁ × ⁶C₅ 5!/7⁷) = 21 × 7 × 6 ×120/7⁷ = 3 × 6 × 120/7⁵ = 2160/7⁵.

48.This can be got by defining the number of ways in which the player can get a deal of 13 cards if he gets all four kings divided by the number of ways in which the player can get a deal of 13 cards without any constraints from 52 cards:

The requisite value would be given by: ⁴⁸C₉/⁵²C₁₃ = [48!/39! × 9!] × [13! × 39!/52!] = (48! × 13! × 39!)/(39! × 9! × 52!) = 13 × 12 × 11 × 10/52 × 51 × 50 × 49 = 1 × 11/17 × 17 × 5 × 49 = 11/4165

49.P of heads showing on 50 coins = ¹⁰⁰C₅₀ × P⁵⁰(1 – p)⁵⁰

P of heads showing on 51 coins = ¹⁰⁰C₅₁ × P⁵¹(1 – p)⁴⁹

Both are equal

¹⁰⁰C₅₀ × P⁵⁰(1 – P)⁵⁰ = ¹⁰⁰C₅₁ × P⁵¹(1 – P)⁴⁹

or × (1 – P)

= × P

or × (1 – P) = P

or 51 × (1 – P) = 50 P

or 51 – 51 P = 50 P

or 51 = 101 P

\ P =

50.The common side could be horizontal or vertical. Accordingly, the number of ways the event can occur is.

n (E) = 8 × 7 + 8 × 7 = 112

n (S) = 64C₂

= =