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.

In probability, our first approach is to define the event and then we approach to find out the probability of an event or we approach to find out the chance of the occurrence of an event.

**Sample space **This is defined with respect to a random experiment and denotes the set representing all the possible outcomes of the random experiment. For example: – Sample space when a coin is tossed is S = (Head, Tail). Sample space when a dice is thrown is S = (1, 2, 3, 4, 5, 6). Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

**Event **The set representing the desired outcome of a random experiment is called the event. Consider the experiment of tossing a coin two times. An associated sample space is S = {HH, TT, TH, HT}. Suppose, getting two heads is the desired outcome. We find that (HH) is the only element corresponds to the desired outcome.

Therefore, event E = {HH}

Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Suppose, we want to get a number less than 4. We find that (1, 2, 3) are the only element of S corresponds to our requirement. Therefore, event E = {1, 2, 3}.

Event (E) is always a subset of sample space (S).

When the sets A and B are two events associated with a sample space, then event ‘either A or B or both’ is represented by A ∪ B.

Event ‘A and B’ is represented by A ∩ B.

**Mutually exclusive events **A set of events is mutually exclusive when the occurrence of any one of them means that the other event cannot occur. Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Consider events A ‘an odd number appears’ and B ‘an even number appears’

A = {1, 3, 5} B = {2, 4, 6}

Occurrence of event A means that the other event B cannot occur.

A ∩ B = f (null)

Therefore, A and B are mutually exclusive events.

**Equally likely events **If two events have the same probability they are called equally likely events. In a toss of a coin, the chance of getting head is equal to chance of getting tail. Therefore, getting head and getting tail are equally likely events.

**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. Let us define the following events

A: ‘a number less than 3 appears’

B: ‘3 appears’

C: ‘a number more than three appears’

A = {1, 2}

B = {3}

C = {4, 5, 6}

A U B U C = {1, 2} U {3} U {4, 5, 6} = {1, 2, 3, 4, 5, 6} = S

Therefore, A, B and C are called exhaustive set of events.

**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.)

**Formula for finding probability**

Let A be an event. Probability of event A is denoted by P(A).

Probability of an event A = P(A) =

**(Number of outcomes favorable to event)/ (Total number of outcomes possible)**

Probability of an event ‘not A’ is denoted by P(A’).

**P(A’) = 1 – P(A)**

**Some other Formulae:**

Let A and B be two events associated with a random experiment. Then,

**P (either A or B) = P(A) + P(B) – P (A and B)**

** or**

**P (A U B) = P(A) + P(B) – P (A ∩ B)**

**P (A U B)’ = P (A’ ∩ B’)**

Let A, B and C be three events associated with a random experiment. Then,

**P (A U B U C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ B) – P (A ∩ B) + P (A ∩ B ∩ C)**

**Problem 1) In a throw of two dice, find the probability of getting one prime and one composite number?**

**Sol. **Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Let A denote the event of getting a prime and a composite number

A = (1, 2) (1, 4) (1, 6)

(2, 1) (2, 3) (2, 5)

(3, 2) (3, 4) (3, 6)

(4, 1) (4, 3) (4, 5)

(5, 2) (5, 4) (5, 6)

(6, 1) (6, 3) (6, 5)

No. of outcomes favorable to A = 18

Total no. of outcomes = 36

Probability = (18)/ (36) = 1/ 2

**Problem 2) The letters of the word LUCKNOW are arranged among themselves. Find the probability of always having NOW in the word?**

**Sol. ** Number of arrangements of word LUCKNOW = 7!

Number of arrangements of word LUCKNOW having NOW = 5!

Probability = (Total number of outcomes favorable to event)/ (Total number of outcomes)

= (5!)/ (7!)

**Use of conjunction AND **If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A and B occur is equal to

P(A) x P(B)

**Use of conjunction OR **If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A or B occur is equal to

P(A) + P(B)

**Problem 3) If we have the probability of A winning a race is 1/3 and that of B winning the race is 1/ 2, then the probability that either A or B win a race is given by?**

**Sol. ** Probability that either A or B wins a race is given by P(A) + P(B)

= 1/3 + 1/ 2 = 5/6

**Problem 4) One card is drawn from a well-shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be**

**a) a diamond**

**b) not an ace**

**c) a black card**

**d) not a diamond**

**e) not a black card**

**Sol. ** Let A be the event ‘the card drawn is a diamond’ = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = P (A) = 13/52 = 1/ 4

**b)** Let B be the event ‘the card drawn is an ace’

P(B) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 4/ 52

P(B)’ = 1 – 4/52 = 48/52

**c)** Let C be the event ‘the card drawn is a black card’

P(C) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 26/52 = 1/2

**d)** Let D be the event ‘the card drawn is a diamond’

P(D) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 13/52 = 1/ 4

P(D)’ = 1 – 1/4 = 3/4

**e)** Let E be the event ‘the card drawn is a black card’

P(E) = (Number of outcomes favorable to event)/ (Total number of outcomes possible)

= 26/52

P(E)’ = 1 – 1/2 = 1/2

**Problem 5) Find the probability that a year chosen at random will have 53 Sundays provided that year was not leap year?**

**Sol. ** A Year has 365 days. Year has 52 complete weeks and one day. One day maybe a Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday.

Probability that this day will be a Sunday = 1/7

Therefore, probability that a year chosen at random will have 53 Sundays = 1/7

**Problem 6) If two dice are thrown, what is the probability that the sum of the numbers is not less than 10?**

**Sol. ** Let A be an event ‘the sum of numbers is equal to 10’

Let B be an event ‘the sum of numbers is equal to 11’

Let C be an event ‘the sum of numbers is equal to 12’

Probability that the sum of numbers is not less than 10 = P(A) + P(B) + P(C)

Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

A = {(4, 6), (6, 4), (5, 5)}

B = {(5, 6), (6, 5)}

C = {(6, 6)}

P(A) = 3/36

P(B) = 2/36

P(C) = 1/36

Probability that the sum of numbers is not less than 10 = 1/36 + 2/36 + 3/36 = 6/36 = 1/6

**Problem 7) 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?**

**Sol. ** Let A be an event ‘Selection of the first bag’

Let B be an event ‘Selection of second bag’

One bag is selected at random out of two bags. Therefore, P(A) = P(B) = 1/2

Let C be an event ‘Black ball drawn from the first bag’

P(C) = 6/14

Let D be an event ‘Black ball drawn from the first bag’

P(D) = 7/14

Probability of this ball being black = P(A). P(C) + P(B). P(D) = (1/2). (6/14) + (1/2). (7/14)

= (1/2). (13/14) = 13/28

**Problem 8) The probability that Ajeet will solve a problem is 1/5. What is probability that**

**a) he does not solve a single problem out of ten problems**

**b) he solves at least one problem out of ten problems**

**Sol. ** Let A be an event ‘Ajeet will solve a problem’

P(A) = 1/5

P(A)’ = 4/5

He solves no problems i.e. he does not solve the first problem and he does not solve the second problem……….. and he does not solve the tenth problem.

Probability that he does not solve a single problem out of ten problems = (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) = (1/5)^{10}

**b)** Let B be an event ‘Ajeet does not solve a single problem out of ten problems’

P(B) = (1/5)^{10}

P(B)’ = 1 – (1/5)^{10}

**Problem 9) In a four-game series between Radha and Anand, the probability that Anand wins a particular game is 2/5 and that of Radha 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 series is drawn (2-2)?**

**Sol. ** Let R be an event ‘Radha will win a particular match’

Let A be an event ‘Anand will win a particular match’

P(A) = 2/5 and P(R) = 3/5

The event definition for the series to end in a draw can be described as

(**R** AND **R** AND **A** AND **A**) OR (**R** AND **A** AND **R** AND **A**) OR (**R** AND **A** AND **A** AND **R**) OR (**A** AND **A** AND **R** AND **R**) OR (**A** AND **R** AND **R** AND **A**) OR (**A** AND **R** AND **A** AND **R**)

= (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}

= (2/5)^{2}(3/5)^{2 }x (6)

= (2/5)^{2}(3/5)^{2 }x (^{4}C_{2})

Where ^{4}C_{2} gives us the number of ways in which Radha can win two games and Anand can win two games.

**If E _{1}, E_{2}, E_{3 ……..}E_{n} are mutually exclusive events of an experiment. Then,**

**P (E _{1 }U E_{2 }U E_{3}………E_{n}) = P(E_{1}) + P(E_{2}) + P(E_{3})…………….+ P(E_{n})**

**If E _{1}, E_{2}, E_{3 ……..}E_{n} are mutually exclusive and exhaustive events of an experiment. Then,**

**P(E _{1}) + P(E_{2}) + P(E_{3})…………….+ P(E_{n}) = 1**

**Problem 10) If P(A) = 1/3, P(B) = 1/2, P (A****∩ B****) = 1/4 then find the P(A’ U B’)**

**Sol. ** **P (A U B)’ = P (A’ ∩ B’)**

Replacing A with A’ and B with B’

**P (A’ U B’)’ = P (A ∩ B)**

**P (A ∩ B) = 1/4**

**P (A’ U B’) = 1 – 1/4 = 3/4**

**Problem 11) A and B are two mutually exclusive events of an experiment. If P(A’) = 0.65 and P (A U B) = 0.65 and P(B) = p, find the value of p.**

**Sol. ** P(A’) = 0.65

P(A) = 0.35

P (A U B) = 0.65

A and B are mutually exclusive events.

Therefore, P (A U B) = P(A) + P(B)

0.65 = 0.35 + p

P = 0.30

**Problem 12) A committee of two persons is selected from two men and two women. What is the probability that the committee will have**

**a) no man**

**b) one man**

**c) two men?**

**Sol. ** Total number of persons = 4. Out of these 4 persons, two can be selected in ^{4}C_{2 }ways.

a)There will be two women in the committee.

Out of two women, two can be selected in ^{2}C_{2 }= 1 way

Probability (no man) = ^{2}C_{2}/^{4}C_{2}

b)One man and one woman can be selected in ^{2}C_{1} x ^{2}C_{1}

Probability (One man and one woman) = (^{2}C_{1} x ^{2}C_{1})/ (^{4}C_{2})

c)Out of two men, two can be selected in ^{2}C_{2 }= 1 way

Probability (two-man) = ^{2}C_{2}/^{4}C_{2}

**Problem 13) A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective?**

**Sol. ** 4 bulbs out of 25 bulbs can be drawn in ^{25}C_{4 }ways.

Two bulbs out of 8 defective bulbs can be drawn in ^{8}C_{2 }ways.

Two bulbs out of 17 non-defective bulbs can be drawn in ^{17}C_{2 }ways.

Two defective and two non-defective bulbs out of 25 bulbs can be drawn in (^{8}C_{2}) x (^{17}C_{2})

Probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective = ((^{8}C_{2}) x (^{17}C_{2}))/ (^{25}C_{4})

**Problem 14) From a bag containing 8 green and 5 red balls, three are drawn one after the another. Find the probability of all three balls being green if balls are drawn without replacement?**

**Sol. ** 3 balls out of 13 balls can be drawn in ^{13}C_{3 }ways

3 green balls out of 8 green balls can be drawn in ^{8}C_{3 }ways

probability of all three balls being green if balls are drawn without replacement = (^{8}C_{3})/(^{13}C_{3})

**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)**.

**P(A/B) = (Number of elementary events favorable to A ∩ B)/ (Number of elementary events which are favorable to B)**

**P(A’/B) = 1 – P(A/B)**

**Problem 15) If P(A) = 7/ 13, P(B) = 9/13 and P (A∩ B) = 4/13, evaluate P(A/B).**

**Sol. ** P(A) = 7/13

P(B) = 9/13

P (A ∩ B) = 4/13

P(A/B) = P (A ∩ B)/ P(B) = (4/13)/ (9/13) = 4/9

**Problem 16) Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?**

**Sol. ** Let A be ‘number on the drawn card is even number’

Let B be ‘number on drawn card is more than 3’

P(A/B) = P (A **∩ B**)/ P(B)

= (4/10)/ (7/10) = 4/7