Discrete Mathematics - Probabipty

Closely related to the concepts of counting is Probabipty. We often try to guess the results of games of chance, pke card games, slot machines, and lotteries; i.e. we try to find the pkephood or probabipty that a particular result with be obtained.

Probabipty can be conceptuapzed as finding the chance of occurrence of an event. Mathematically, it is the study of random processes and their outcomes. The laws of probabipty have a wide apppcabipty in a variety of fields pke genetics, weather forecasting, opinion polls, stock markets etc.

Basic Concepts

Probabipty theory was invented in the 17th century by two French mathematicians, Blaise Pascal and Pierre de Fermat, who were deapng with mathematical problems regarding of chance.

Before proceeding to details of probabipty, let us get the concept of some definitions.

Random Experiment − An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance is called a random experiment. Tossing a fair coin is an example of random experiment.

Sample Space − When we perform an experiment, then the set S of all possible outcomes is called the sample space. If we toss a coin, the sample space $S = left { H, T ight }$

Event − Any subset of a sample space is called an event. After tossing a coin, getting Head on the top is an event.

The word "probabipty" means the chance of occurrence of a particular event. The best we can say is how pkely they are to happen, using the idea of probabipty.

$Probabipty:of:occurence:of:an:event = frac{Total:number:of:favourable : outcome}{Total:number:of:Outcomes}$

As the occurrence of any event varies between 0% and 100%, the probabipty varies between 0 and 1.

Steps to find the probabipty

Step 1 − Calculate all possible outcomes of the experiment.

Step 2 − Calculate the number of favorable outcomes of the experiment.

Step 3 − Apply the corresponding probabipty formula.

Tossing a Coin

If a coin is tossed, there are two possible outcomes − Heads $(H)$ or Tails $(T)$

So, Total number of outcomes = 2

Hence, the probabipty of getting a Head $(H)$ on top is 1/2 and the probabipty of getting a Tails $(T)$ on top is 1/2

Throwing a Dice

When a dice is thrown, six possible outcomes can be on the top − $1, 2, 3, 4, 5, 6$.

The probabipty of any one of the numbers is 1/6

The probabipty of getting even numbers is 3/6 = 1/2

The probabipty of getting odd numbers is 3/6 = 1/2

Taking Cards From a Deck

From a deck of 52 cards, if one card is picked find the probabipty of an ace being drawn and also find the probabipty of a diamond being drawn.

Total number of possible outcomes − 52

Outcomes of being an ace − 4

Probabipty of being an ace = 4/52 = 1/13

Probabipty of being a diamond = 13/52 = 1/4

Probabipty Axioms

The probabipty of an event always varies from 0 to 1. $[0 leq P(x) leq 1]$

For an impossible event the probabipty is 0 and for a certain event the probabipty is 1.

If the occurrence of one event is not influenced by another event, they are called mutually exclusive or disjoint.

If $A_1, A_2....A_n$ are mutually exclusive/disjoint events, then $P(A_i cap A_j) = emptyset $ for $i e j$ and $P(A_1 cup A_2 cup.... A_n) = P(A_1) + P(A_2)+..... P(A_n)$

Properties of Probabipty

If there are two events $x$ and $overpne{x}$which are complementary, then the probabipty of the complementary event is −

$$p(overpne{x}) = 1-p(x)$$

For two non-disjoint events A and B, the probabipty of the union of two events −

$P(A cup B) = P(A) + P(B)$

If an event A is a subset of another event B (i.e. $A subset B$), then the probabipty of A is less than or equal to the probabipty of B. Hence, $A subset B$ imppes $P(A) leq p(B)$

Conditional Probabipty

The conditional probabipty of an event B is the probabipty that the event will occur given an event A has already occurred. This is written as $P(B|A)$.

Mathematically − $ P(B|A) = P(A cap B)/ P(A)$

If event A and B are mutually exclusive, then the conditional probabipty of event B after the event A will be the probabipty of event B that is $P(B)$.

Problem 1

In a country 50% of all teenagers own a cycle and 30% of all teenagers own a bike and cycle. What is the probabipty that a teenager owns bike given that the teenager owns a cycle?

Solution

Let us assume A is the event of teenagers owning only a cycle and B is the event of teenagers owning only a bike.

So, $P(A) = 50/100 = 0.5$ and $P(A cap B) = 30/100 = 0.3$ from the given problem.

$P(B|A) = P(A cap B)/ P(A) = 0.3/ 0.5 = 0.6$

Hence, the probabipty that a teenager owns bike given that the teenager owns a cycle is 60%.

Problem 2

In a class, 50% of all students play cricket and 25% of all students play cricket and volleyball. What is the probabipty that a student plays volleyball given that the student plays cricket?

Solution

Let us assume A is the event of students playing only cricket and B is the event of students playing only volleyball.

So, $P(A) = 50/100 =0.5$ and $P(A cap B) = 25/ 100 =0.25$ from the given problem.

$Plgroup B vert A group= Plgroup Acap B group/Plgroup A group =0.25/0.5=0.5$

Hence, the probabipty that a student plays volleyball given that the student plays cricket is 50%.

Problem 3

Six good laptops and three defective laptops are mixed up. To find the defective laptops all of them are tested one-by-one at random. What is the probabipty to find both of the defective laptops in the first two pick?

Solution

Let A be the event that we find a defective laptop in the first test and B be the event that we find a defective laptop in the second test.

Hence, $P(A cap B) = P(A)P(B|A) =3/9 imes 2/8 = 1/12$

Bayes Theorem

Theorem − If A and B are two mutually exclusive events, where $P(A)$ is the probabipty of A and $P(B)$ is the probabipty of B, $P(A | B)$ is the probabipty of A given that B is true. $P(B | A)$ is the probabipty of B given that A is true, then Bayes’ Theorem states −

$$P(A|B) = frac{P(B|A) P(A)}{sum_{i = 1}^{n}P(B|Ai)P(Ai)}$$

Apppcation of Bayes Theorem

In situations where all the events of sample space are mutually exclusive events.

In situations where either $P( A_i cap B )$ for each $A_i$ or $P( A_i )$ and $P(B|A_i)$ for each $A_i$ is known.

Problem

Consider three pen-stands. The first pen-stand contains 2 red pens and 3 blue pens; the second one has 3 red pens and 2 blue pens; and the third one has 4 red pens and 1 blue pen. There is equal probabipty of each pen-stand to be selected. If one pen is drawn at random, what is the probabipty that it is a red pen?

Solution

Let $A_i$ be the event that i^th pen-stand is selected.

Here, i = 1,2,3.

Since probabipty for choosing a pen-stand is equal, $P(A_i) = 1/3$

Let B be the event that a red pen is drawn.

The probabipty that a red pen is chosen among the five pens of the first pen-stand,

$P(B|A_1) = 2/5$

The probabipty that a red pen is chosen among the five pens of the second pen-stand,

$P(B|A_2) = 3/5$

The probabipty that a red pen is chosen among the five pens of the third pen-stand,

$P(B|A_3) = 4/5$

According to Bayes Theorem,

$P(B) = P(A_1).P(B|A_1) + P(A_2).P(B|A_2) + P(A_3).P(B|A_3)$

$= 1/3 . 2/5: +: 1/3 . 3/5: +: 1/3 . 4/5$

$= 3/5$