Economics Notes for class 11 Chapter Number six Probability question and answer for all pakistani boards.

**Q.6.1) Define a set, null set, sub-set, universal set, and equal set with examples.**

Table of Contents

**Answer:****Set:**

A set is any well-defined collection or list of distinct objects.These objects are called the elements or members of the set**Examples: **A group of students, the books in a library, people living on the earth, first ten natural numbers etc.

Sets are usually denoted by capital letters such as A,B,C,……,X,Y,Z, while the elements of the set are denoted by small letters such as a,b,c x,y,z etc.

Elements are enclosed in braces elements are separated by commas to represent a set e.g.A = {a,b,c}**Null Set:**

A set which have no element is called null set or empty set. It is denoted by { }.**Example: **If we throw a dice and A represents points on the dice which are greater than 6. Then A is a null set.**Sub Set:**Set A is said to be the subset of B if and only if every element of set A is also the element of set B.

**Example:**

A = {2,4,6,8}

B = {1,2,3 ……….10}

Since every element of A is present in B therefore A is the

subset of B i.e. AcB

**Note:**Null set is the subset of every set.

**Equal Sets:**Two sets A and B are said to be equal sets if and only if every elements of set A is also the element of set B and every element of set u is the element of set A. Symbolically, it can be written as

A = B.

**Example:**

A = {1,2,3,4, x, y,z}

B = {x, y,z, 3,4,2,1}

Since every element of set A is the element of set B and every element of set B is the element of set A Therefore

A = B

**Universal Set:**The large or the original set of which all sets, under consideration, are subsets is called the universal set, denoted by S or Q.

**Example:**

A = {2,4,6,8} B = {1,3,5,7}

Than the universal set in this case is S = {1,2,3,4,5,6,7,8}

Read more: Statistics Chapter 5 (Index number)

**Q.6.2) Find AUB,AUC,BUC,A-B,, ifA=( 1,2,3,4,5,6) B=( 1,3,5,7,9), C=( 1,3,5,7)**

**solution:**AuB = {1,2,3,4,5,6} u {1,3,5,7,9}

= {1.2,3,4,5,6,7,9}

AuC = {1,2,3,4,5,6} u {1,3,5,7}

= {1,2,3,4,5,6,7}

BuC = {1,3,5,7,9} u {1,3,5,7}

= {1,3,5,7,9}

A-B = {1,2,3,4,5,6}-{1,3,5,7,9}

= {2,4,6}

**Q.6.3) **

a) Define fundamental rules of counting.

b) How many distinct permutations can be formed from all the letters of each word?

i) INFINITY

ii) UNUSUAL

iii) SOCIOLOGICAL?

**solution:FUNDAMENTAL RULES OF COUNTING**

The solution of many problems in probability may require some knowledge of rules of counting which are given below.

**1) Factorial**

The factorial of n is the product of first n natural numbers. Factorial of n is denoted by n!

**Example:**Finds! 4!

**Solution:**5! = 5x4x3x2x1 = 120

4! = 4x3x2x1 = 24

2. Permutations

A permutation is any ordered subset from a set of ‘n’ distinct objects

**OR**

A permutation is an arrangement of all or some of a set of objects in a finite order.

**Example:**We might want to know how many arrangements are possible for six people to sit around a table, we might ask how many different orders are possible to draw two lottery tickets from a total of 20.

The different arrangements are called permutations.

The number of permutations of r objects selected in a definite order from n distinct objects is denoted by

^{ n}P

_{r}and is given by the following formula.

**3. Combination **

A combination is any subset of r objects, selected without regard to their order, from a set of n distinct objects.

The difference between a permutation and a combination is that for a permutation the order is taken into account while combination is concerned with a distinct set of r items taken from n items in which the order does not matter.

The total number of combination is denoted by nC_{r} and is written as

Read more: Statistics-Chapter 3 (Measures of Location)

**Q.6.8) **

(a) Define and explain the following terms:

i) Mutually exclusive events

ii) Equally likely events

iii) Exhaustive events

iv) Independent events

v) Random Experiment.

b) Find the probability of each of the following.

i) A head appears in tossing a fair coin.

ii) A “ 5” appears in rolling a six-faced cubical die.

iii) An even number appears when a perfect die is rolled.

iv) At least one head appears in three tosses of a fair coin.

v) Greater than four with a six-faced cubical die.

vi) Throwing an ace with a six-faced cubical dice.

**SOLUTION:PART A:MUTUALLY EXCLUSIVE EVENTS:**

In logic and probability theory, two propositions (or events) are mutually exclusive or disjoint if they cannot both be true (occur). A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.

**Examples:**

· Turning left and turning right are Mutually Exclusive (you can’t do both at the same time

**· Tossing a coin:**Heads and Tails are Mutually Exclusive

**· Cards:**Kings and Aces are Mutually Exclusive

**EQUALLY LIKELY EVENTS:**

Equally likely events are events that have the same theoretical probability (or likelihood) of occurring. Example: Each numeral on a die is equally likely to occur when the die is tossed.

Example In the experiment of tossing a coin:

**A:**the event of getting a

**“HEAD”**and

**B:**the event of getting a

**“TAIL”**

Events “A” and “B” are said to be equally likely events

[Both the events have the same chance of occurrence.

**EXHAUSTIVE EVENTS;**

In probability theory and logic, a set of events is jointly or collectively exhaustive if at least one of the events must occur. For example, when rolling a six-sided die, the outcomes 1, 2, 3, 4, 5, and 6 are collectively exhaustive, because they encompass the entire range of possible outcomes.

**For Example:**consider the experiment of a fair die being thrown. Then there are six outcomes and all of them are equally likely to occur. Also the events of getting different numbers taken together are exhaustive as together at least one of them is certain to happen. For getting a 2 or 5, sure will get one of the numbers during the experiment. So events are exhaustive.

**INDEPENDENT EVENTS:**

Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. Some other examples of independent events are: Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die.

**Example:**You toss a coin and it comes up “Heads” three times … what is the chance that the next toss will also be a “Head”?

The chance is simply ½ (or 0.5) just like ANY toss of the coin.

**RANDOM EXPERIMENT:**

A random experiment is an experiment or a process for which the outcome cannot be predicted with certainty.

Here are examples of random experiments. Give the corresponding sample space.

1. Selection of a plastic component and verification of its compliance

2. Lifetime of a computer

3. Number of calls to a communication system during a fixed length interval of time

**Q.6.9) ****a) Determine the probability of a sum of 12 appears in single toss of a pair of unbiased dice.****b) It was found that in a college library 100 students were reading 20 belonged to 3**^{rd }**year, 30 to 2**^{nd}** year, 40 to the 1**^{st}** year and 10 to the postgraduate class. One student is selected at random. Find the probability of his belonging to each class. What is the probability that he is an undergraduate?**

**Q.6.10) Out of 80 members of club 30 drink tea, 20 drink coffee and 10 drink both. One member is selected at random.**

**Find the probability that he****i) drink either tea or coffee.ii) drink neither tea nor coffeeiii) drink only tea.**

**Q.6.11) If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems, and a dictionary, what is the probability that**

a) The dictionary is selected?

b) 2 novels and one book of poems are selected?

**Q.6.12) The probabilities that a service station will pumps gas into 0,1,2,3,4, or 5 or more cars during a certain 30 minutes period are 0.03, 0.18, 0.24, 0.28,0.10 and 0.17.Find the probability that in this 30 minutes period.**

a) More than 2 cars receive gas?

b) At most 4 cars receive gas?

c) 4 or more cars receive gas?

**SOLUTION:**X=Number of cars that recieve gas

P(pumps gas into 0 car)=p[x=0]=0.03

P(pumps gas into 1 car)=p[x=1]=0.18

P(pumps gas into 2 car)=p[x=2]=0.24

P(pumps gas into 3 car)=p[x=3]=0.28

P(pumps gas into 4 car)=p[x=4]=0.10

P(pumps gas into 5 or more cars)=p[x>5]=0.17

**PART A:**

**1)**

If more than two cars recieve gas= P[X > 2]

SO

P(2) +P(3) +P(4) +P(5)

0.24+0.28+0.10+0.17=0.79

**2)**

If atmost four cars recieve gas=P[X<4]

P(3) +P(2) +P(1) +P(0)

0.28+0.24+0.18+0.03=0.73

**Q.6.13) If 200 students, 100 are currently enrolled in mathematics and 80 are currently enrolled in statistics. Their enrollment figures include 20 students who are in fact enrolled in both courses. What is the probability that a randomly chosen student will be enrolled in either mathematics or statistics?**]

**Q.6.14) **

a) Explain the following concepts of probability:

i) A priori probability

ii) A posteriori probability

iii) Axiomatic probability.

b) Box A contain nine cards numbered 1 through 9 ,and the Box B contain five cards numbered 1 through 5.A box is chosen at random and a card drawn. If the number is even, find the probability that the card came from box A.

**SOLUTION:**

**The probability of an event occurring does the number in the sample space divide the number in the event. Again, this is only true when the events are equally likely. A classical probability is the relative frequency of each event in the sample space when each event is equally likely.**

__CLASSICAL OR PRIORI PROBABILITY:__

__POSTERIOR PROBABILITY:__The revised probability of an event occurring after taking into consideration new information. Posterior probability is normally calculated by updating the prior probability by using Baye’s theorem. In statistical terms, the posterior probability is the probability of event A occurring given that event B has occurred.

**During the XXth century, a Russian mathematician, Andrei Kolmogorov, proposed a definition of probability, which is the one that we keep on using now a days.**

__AXIOMATIC PROBABILITY__

If we do a certain experiment, which has a sample space Ω, we define the probability as a function that associates a certain probability, P (A) with every event A, satisfying the following properties.

The probability of any event A is positive or zero. Namely P (A) ≥0. The probability measures, in a certain way, the difficulty of event A happening: the smaller the probability, the more difficult it is to happen.

The probability of the sure event is 1. Namely P (Ω) =1. Therefore, the probability is always greater than 0 and smaller than 1: probability zero means that there is no possibility for it to happen (it is an impossible event), and probability 1 means that it will always happen (it is a sure event).

The probability of the union of any set of two by two-incompatible events is the sum of the probabilities of the events. That is, if we have, for example, events A,B,C, and these are two by two incompatible, then P(A∪B∪C)=P(A)+P(B)+P(C).

**PART B:**

**Q.6.15) **

a) State and prove addition theorem of probability for mutually exclusive events.

b) The probabilities that three men hit a target are respectively 1/6, 1/4,1/3. Each shoots once at the target.

i) Find the probability that exactly one of them hits the target.

ii) If only one hit the target, what is the probability that it was the first man?

**Addition theorem on probability:**If A and B are any two events then the probability of happening of at least one of the events is defined as P (AUB) = P (A) + P (B) – P (A∩B).

**Proof:**

Since events are nothing but sets, from set theory, we have

P (AUB) = P (A) + P (B) – P (A∩B).

Dividing the above equation by n(S), (where S is the sample space) P (AUB)/ P(S) = P (A)/ P(S) + P (B)/ P(S) – P (A∩B)/ P(S)

Then by the definition of probability,

P (AUB) = P (A) + P (B) – P (A∩B).

**Example:**

If the probability of solving a problem by two students George and James are 1/2 and 1/3 respectively then what is the probability of the problem to be solved.

**Solution:**

Let A and B be the probabilities of solving the problem by George and James respectively. Then P (A) =1/2 and P (B) =1/3. The problem will be solved if it is solved at least by one of them also. So, we need to find P (AUB). By addition theorem on probability,

We have

P (AUB) = P (A) + P (B) – P (A∩B). P (AUB)

P (AUB) = 1/2 +.1/3 – 1/2 * 1/3 = 1/2 +1/3-1/6 = (3+2-1)/6 = 4/6 = 2/3 Note: If A and B are any two mutually exclusive events then P(A∩B)=0.

Then P (AUB) = P (A) +P (B).

**PART B:**

**Q.6.16) ****a) State and prove addition theorem of probability for not mutually exclusive events.**

**b) In a certain college, 25% of the students failed Mathematics, 15% of the students failed Statistics, and 10% of the students failed both Mathematics and Statistics. A student is selected at random.****i) If he failed Statistics, what is the probability that he failed mathematics?ii) If he failed Mathematics, what is the probability that he failed Statistics?iii) What is the probability that he failed mathematics or Statistics**

**solution:Two Non-Disjoint (Non-Mutually Exclusive) Events**For two events “A” and “B” which are not disjoint (or not mutually exclusive), the probability that at least one of the events would occur i.e. the probability of the occurrence of the union of the events is given by

P (A ∪ B) = P (A) + P (B) − P (A ∩ B)

**Three Non-Disjoint (Non-Mutually Exclusive) Events**

For three events “A”, “B” and “C” which are not disjoint (or not mutually exclusive), the probability that at least one of the events would occur i.e. the probability of the occurrence of the union of the events is given by

P (A ∪ B ∪ C) = P (A) + P (B) + P(C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C)

**PART B:**

Probability of the students failed in mathematics= P(M)=25%=0.25

Probability of the students failed in statistics= P(S)=15%=0.15

Probability of the students failed in both P(M/S) =10%=0.10

NOW:

**1) Probability that he failed mathematics given he failed statistics is P(M/S)**

NOW:

P(M/S) =P(MnS) / P(S)

P(M/S) =0.10 / 0.15 = 0.667

**2) Probability that he failed statistics given that he failed mathematics is P(S/M)**

P(S/M) =P(MnS) / P(M)

P(M/S) =0.10 / 0.25 = 0.4

**3) Probability that he fails mathematics or statistics is P(MUS)**

P(MUS)=P(M) + P(S) – P(MnS)

P(MUS)=0.25+0.15-0.10

P(MUS)=0.3

Read more: Statistics Chapter 2 (Collection and presentation of Data) | Class 11 Notes