# How Algebra can be used to learn Probability?

The probability of any event is defined as the chance of occurrence of the events to the total possible outcomes. If there are ‘n’ exhaustive, mutually exclusive, and equally likely outcomes of a random experiment. Out of which, ‘m’ are favorable to the occurrence of an event E

**Formula of Probability**

The most basic and general formula to calculate the probability is

P(n) = (favorable events)/(total events

= m/n

**Probability Definition of an Event**

- Probability of event
**A or B**= P(A ∪ B) = P(A) + P(B) – P(A ∩ B). - Probability of event
**not A**= P(A’) = 1 – P(A) - Probability of event
**A but not B**= P(A ∩ B’) = P(A) – P(A ∩ B) - Probability of event not
**A not B**= P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – Probability of event A or B.

**Explanation of Probability**

Imagine you and your friends are having fun playing. You have to throw dice in this game, and getting six is lucky. The likelihood of winning increases with the number of sixes you obtain. How do you figure up your odds of winning? Does everybody have the same chance of winning? Here, scoring a six is considered a win.

Do all events happen the same for everyone? We are now familiar with the definition of probability. We can provide solutions to probability formulas. We will learn about events, the different kinds of occurrences, and a brief definition of algebra and probability in this part.

**Example: **Say two coins are tossed, and you must find the probability of the following events.

- At least one tail turns up
- No heads turn up
- At the most one tails turns up

So to find the probability of an event A in a finite space S.

P(A)= (number of sample points in A)/(total number of sample points in S)

= n(a)/ n(s)Here,

S = total smaple space = { HH, HT, TH, TT }

=> n(s) = 4

**At least one tail turns up**

Here let A be event when at least one head turns up,

So A = {HT, TH, TT};

n(A) = 3;

P (A) = n(A)/n(S)

= 3/4

**No heads turn up**

Here A is the event where no heads turn up,

so A= {TT};

n(A) = 1;

P (A) = n(A)/n(S)

= 1/4

**At the most one tails turns up**

Here A is the event when at the most one tails turns up,

so A = {HH, TH, HT};

n(A) = 3;

P (A) = n(A)/n(S)

= 3/4

**Terms related to Probability**

**Random Experiment**: A random experiment is one in which all the possible results are known in advance but none of them can be predicted with certainty.**Outcome**: The result of a random experiment is called an outcome.**Sample Space**: The set of all the possible outcomes of a random experiment is called a Sample Space, and it is denoted by ‘S’.**Event:**A subset of the sample space is called an Event.

**Events and Its Algebra**

Any subset of a sample space is an event. In other words, a combination of outcomes of a random experiment is an event. It is denoted by capital letters. In a random experiment of throwing a die, an event can be of getting any of the numbers from 1 to 6 on its uppermost face. We can calculate the probability of any of the possible events. For example, the probability of an event of getting 5 in a single throw of a die is 1⁄6.

**Types of Events**

There are various types of events that are used in probability which are explained below

**Simple Event**

Any event is Simple if it corresponds to a single possible outcome of the experiment. In other words, if an event has only one sample point of a sample space it is a simple event. In a random experiment of throwing a die, the sample space

S = {1, 2, 3, 4, 5, 6}. The event, E, of getting 5 on the uppermost face is a simple event.

**Compound Event**

Any event is Compound if it corresponds to more than a single possible outcome of the experiment. In other words, if any event has more than one sample point of a sample space it is a compound event. In a random experiment of throwing a die, the sample space

S = {1, 2, 3, 4, 5, 6}. The event E of getting a multiple of 2 is a compound event as E = {2, 4, 6}.

Based on the Set theory, we can perform some algebra on events. Some of them are the union or the intersection of events. Let us study them in detail.

**Complimentary Event**

For any event E, the complementary event E’ shows not E. Every outcome which is not in E can be assumed to be in E’. In simple terms, if E denotes the glass is half empty, the event E’ shows that the glass is half-filled.

Let us take an example of throwing a die. The sample space, S = {1, 2, 3, 4, 5, 6}. E shows the event of getting even number i.e., E = {2, 4, 6}. The event E’ shows the outcome of not an even number or getting an odd number. E’ = {1, 3, 5}.

**Event A or B**

Event A or B shows the sample points of a random experiment which are either in A or B or both.

Event A or B = A ∪ B

Suppose event A = {1, 3, 4, 7} and B = {2, 3, 5. 6}.

A ∪ B = {1, 2, 3, 4, 5, 6, 7}.

**Event A and B**

Let A and B be two events. The event A and B show the sample points of a random experiment which are common to both A and B. It is similar to the intersection of two sets A and B. Event A and B = A ∩ B.

Let us take an example of the random experiment of throwing a die. A is the event of getting an even number. B is the event of getting a multiple of 3. A ∩ B shows the sample point which is common to both A and B.

Here, A = {2, 4, 6} and B = {3, 6} and A ∩ B = {6}.

**Event A but not B**

Event A but not B shows the sample points which are in A but not in B.

Event A but not B = A ∩ B’ = A – A ∩ B.

This event shows the unique sample points of A other than that in B.

Suppose event A= {1, 3, 4, 5, 6, 7} and B’ = {2, 3, 5. 6}

then, A ∩ B’ = {1, 4, 7}.

**Exhaustive Events**

The total number of possible outcomes of a random experiment is an exhaustive event. The event of getting an odd number and an event of getting an even number in a throw of a die together forms an exhaustive event.

**Favorable Events**

The numbers of outcomes that show the happening of the event in a random experiment are favorable events. They show the number of cases favorable to an event. In a random experiment of tossing two coins, the number of the favorable event for getting two tails together is 1.

**Mutually Exclusive Events**

Events are mutually exclusive if the happening of any one of the events excludes the chance of happening of the other in the same trail. Also, we can say that no two or more events can happen simultaneously. The events of head and tail in tossing a coin are mutually exclusive. Only one of them can happen.

**Equally Likely Events**

An event in which all the outcome has an equal chance to occur. In throwing a die, all six faces are equally likely to come.

**Independent Events**

Two events are independent if there is no effect on the happening and the non-happening of one by the others. In throwing a die, the result of getting 2 in a first throw does not affect the result of the second and other throws.

**Impossible Events**

The events which are impossible to happen come in the category of impossible events. The empty set Φ is an impossible set. Consider an example of a pack of 52 playing cards, the event of getting a card of the number 12 is impossible.

### Sample Problems

**Problem 1: Find the probability for a randomly chosen month to have its 15th day on a Monday,**

**Solution:**

So let us solve this step by step,

The probability of choosing any month from the given 12 months is 1/12

There are 7 possible days the 15th of a month can fall on. So the probability it falls on a Monday is 1/7.

Thus the probability of a randomly chosen month having its 15th day on a Monday is

1/12 × 1/7 = 1/84

**Problem 2: Suppose there are 4 red, 6 blue, and 2 green balls in a bag. Two balls are drawn at random from the bag. Find the probability of that the two balls drawn are red.**

**Solution:**

Total number of balls in the bag = 4 + 6 + 2 = 12.

Two balls are drawn at random.

The total number of ways in which any two balls are drawn =

^{12}C_{2}= 66.The number of favorable cases of getting two red balls =

^{4}C_{2}= 6.Therefore, the required probability = 6⁄66 = 1/11.

** Problem 3: A, B, and C are three mutually exclusive and exhaustive events of a random experiment. If P(A) = 4⁄5 P(C) and P(B) = 3⁄5 P(C). Find P(C).**

**Solution: **

Since A, B and C are mutually exclusive and exhaustive events,

P(A) + P(B) + P(C) = 1.

⇒ 4⁄5 P(C) + 3⁄5 P(C) + P(C) = 12⁄5 P(C) = 1 or,P(C) = 5⁄12.

**Problem 4: What is the probability of drawing a black card or a ten in a deck of cards?**

**Solution:**

There are 4 tens in a deck of cards P(10) = 4/52

There are 26 black cards P(black) = 26/52

There are 2 black tens P(black and 10) = 2/52

P(black or ten) = 4/52+26/52−2/52

=30/52−2/52

=28/52

=7/13

**Problem 5: Three events A, B, and C are mutually exclusive, exhaustive, and equally likely. What is the probability of the complementary event of A?**

**Solution:**

Since A, B and C are mutually exclusive, we have

P(AUBUC) = P(A) + P(B) + P(C) —— (1)

Since they are exhaustive,

P(AUBUC) = 1 —— (2)

Since they are equally likely events,

P(A) = P(B) = P(C) = K, Say —— (3)

Combining equations (1), (2) and (3), we have

1 = K + K + K

1 = 3K

1/3 = K

Thus P(A) = P(B) = P(C) = 1/3

Therefore, P(A’) = 1 – P(A)

P(A’) = 1 – 1/3

P(A’) = 2/3.

**Problem 6: The odds in favor of an event are 4:3. and The odds against another independent event are 2:3. What is the probability that at least one of the events will occur?**

**Solution:**

Assume that the given events are A and B

Then by the problem, probability of occurrence of A = P(A) = 4/(4+3)=4/7

And probability of occurrence of B = P(B) = 3/(2+3)=3/5

Therefore, the probability of occurrence of at least one of the events A and B

= P(A⋃B) = P(A) + P(B) – P(A⋂B)

= P(A) + P(B) – P(A).P(B)= 4/7+3/5−(4/7.3/5)

=(20+21−12)/35

=29/35

**Problem 7: Given that the events A and B are such that P(A) = 1/2, P (A ∪ B) = 3/5, and P(B) = p. Find p if they are**

**(i) mutually exclusive**

**(ii) independent**

**Solution: **

Given, P(A) = 1/2 ,

P (A ∪ B) = 3/5

and P(B) = p.

(i) For Mutually Exclusive:Given that, sets A and B are mutually exclusive.

Thus, they do not have any common elements

Therefore, P(A ∩ B) = 0

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Substitute the formulas in the above-given formula, we get

3/5 = (1/2) + p – 0

Simplify the expression, we get

(3/5) – (1/2) = p

(6 − 5)/10 = p

1/10 = p

Therefore, p = 1/10

Hence, the value of p is 1/10, if they are mutually exclusive.

(ii) For Independent events:If the two events A & B are independent,

we can write it as P(A ∩ B) = P(A) P(B)

Substitute the values,= (1/2) × p

= p/2

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Now, substitute the values in the formula,

(3/5) = (1/2)+ p – (p/2)

(3/2)– (1/2)= p – (p/2)

(6 − 5)/10 = p/2

1/10 = p/2

p= 2/10

P = 1/5

Thus, the value of p is 1/5, if they are independent.

**Problem 8: The probability of solving the specific problem independently by persons A and B are 1/2 and 1/3 respectively. In case, if both persons try to solve the problem independently, then calculate the probability that the problem is solved.**

**Solution:**

Given that, the two events say A and B are independent if P(A ∩ B) = P(A). P(B)

From the given data, we can observe that P(A) = 1/2 & P(B) = 1/3

The probability that the problem is solved =( probability that person A solves the problem )or( probability that person B solves the Problem)

This can be written as:

= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

If A and B are independent, then P(A ∩ B) = P(A). P(B)

Now, substitute the values,

= (1/2) × (1/3)

P(A ∩ B) = 1/6

Now, the probability of problem solved is written as

P(Problem is solved) = P(A) + P(B) – P(A ∩ B)

= (1/2) + (1/3) – (1/6)

= (3/6) + (2/6) – (1/6)

= 4/6

= 2/3

Hence, the probability of the problem being solved is 2/3.

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