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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.

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 Use of the Conjunction AND**

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)

**Q. 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**

A hits the target AND B hits the target

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**

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

**A wins OR B wins. so 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.

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

**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.

**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 P(E) / P(E’)

Odds against an event is defined as: P(E’) / P(E)

Expectation: The expectation of an individual is defined as Probability of winning × Reward of winning

**Q. 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.**

Expectation = Probability of winning × Reward of winning = 20500∗1000=40

Q.

In a throw of two dice, find the probability of getting one prime and one composite number

1.1/6

2.11/51

3.1/2

4.1/3

Explanation :

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

Q.

Find the probability that a leap year chosen at random will have 53 Sunday

1.2/7

2.12/17

3.2/17

4.1/4

Explanation :

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.

Q.

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

1.41/77

2.2/3

3.21/77

4.36/77

Explanation :

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

Q.

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

1.1/42

2.13/42

3.12/42

4.11/42

Explanation :

The required probability will be given by the equation

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

= 5!/7! = 1/42

Q.

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

1.3/4

2.3/114

3.13/14

4.3/14

Explanation :

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

Q.

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

1.5/9

2.25/139

3.2/39

4.25/39

Explanation :

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

Q.

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.

1.11/12

2.7/12

3.4/12

4.1/12

Explanation :

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

Q.

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.

1.(0.4)4

2.(0.4)3

3.(0.4)2

4.(0.4)6

Explanation :

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

Q.

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?

1.(4/5)10

2.1-(4/5)8

3.1-(4/5)7

4.1-(4/5)10

Explanation :

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

Q.

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?

1.(4C2) × (8/25) × (7/24) × (17/23) × (16/22)

2.(4C2) × (8/25) × (7/24) × (17/23)

3.(4C2) × (8/25) × (17/23) × (16/22)

4.(4C2) × (7/24) × (17/23) × (16/22)

Explanation :

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)

Q.

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

1.20/39

2.2/9

3.20/9

4.2/39

Explanation :

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 × 20

2

/40 × 39) = 20/39

Q.

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?

1.17/100

2.7/100

3.7/10

4.47/100

Explanation :

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

Q.

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

1.313/1313

2.713/1313

3.913/1313

4.813/1313

Explanation :

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 = \( \frac{8}{13} * \frac{8}{13} * \frac{8}{13} = \frac{8^3}{13^3}\)