# Aptitude Probability Theory

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## Probability:

Probability means ‘The extent to which something is
probable; the likelihood of something happening or being
the case’.

Probability is quantitative measure of the chance of
occurrence of a particular event.

## Experiment:

An experiment is an operation which can produce well-defined outcomes.

## Random Experiment:

If all the possible outcomes of an experiment are known but the exact output cannot be predicted in advance, that experiment is called a random experiment.

## Examples:

## Tossing of a fair coin:

When we toss a coin, the outcome will be either Head (H) or Tail (T)

## Throwing an unbiased die:

1.Die is a small cube used in games. It has six faces and
each of the six faces shows a different number of dots from
1 to 6. Plural of die is dice.

2.When a die is thrown or rolled, the outcome is the
number that appears on its upper face and it is a random
integer from one to six, each value being equally
likely.

## Drawing a card from a pack of shuffled cards:

1.A pack or deck of playing cards has 52 cards which
are divided into four categories as given below

Spades (♠)Clubs (♣)Hearts (♥)Diamonds (♦)

2.Each of the above mentioned categories has 13 cards, 9
cards numbered from 2 to 10, an Ace, a King, a Queen and
a jack

3.Hearts and Diamonds are red faced cards whereas Spades
and Clubs are black faced cards.

4.Kings, Queens and Jacks are called face cards

5.Taking a ball randomly from a bag containing balls of
different colours

## Sample Space:

Sample Space is the set of all possible outcomes of an experiment. It is denoted by S.

### Examples:

1.When a coin is tossed, S = {H, T} where H = Head
and T = Tail

2.When a dice is thrown, S = {1, 2 , 3, 4, 5, 6}

3.When two coins are tossed, S = {HH, HT, TH, TT} where
H = Head and T = Tail

## Event:

Any subset of a Sample Space is an event. Events are generally denoted by capital letters A, B , C, D etc.

### Examples:

When a coin is tossed, outcome of getting head or tail is an event

When a die is rolled, outcome of getting 1 or 2 or 3 or 4 or 5 or 6 is an event

## Equally Likely Events:

Events are said to be equally likely if there is no preference for a particular event over the other.

### Examples:

When a coin is tossed, Head (H) or Tail is equally likely
to occur.

When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are
equally likely to occur.

## Mutually Exclusive Events:

Two or more than two events are said to be mutually
exclusive if the occurrence of one of the events excludes
the occurrence of the other

This can be better illustrated with the following
examples

1.When a coin is tossed, we get either Head or Tail. Head
and Tail cannot come simultaneously. Hence occurrence of
Head and Tail are mutually exclusive events.

2.When a die is rolled, we get 1 or 2 or 3 or 4 or 5 or 6. All
these faces cannot come simultaneously. Hence occurrences
of particular faces when rolling a die are mutually
exclusive events.

Note : If A and B are mutually exclusive events,
$ A \cap B$ = ϕ where ϕ represents empty set.

3.Consider a die is thrown and A be the event of getting 2
or 4 or 6 and B be the event of getting 4 or 5 or 6.
Then

A = {2, 4, 6} and B = {4, 5, 6}

Here$ A \cap B$ ≠ϕ. Hence A and B are not mutually
exclusive events.

## Independent Events:

Events can be said to be independent if the occurrence
or non-occurrence of one event does not influence the
occurrence or non-occurrence of the other.

### Example :

When a coin is tossed twice, the event of getting Tail(T) in the first toss and the event of getting Tail(T) in the second toss are independent events. This is because the occurrence of getting Tail(T) in any toss does not influence the occurrence of getting Tail(T) in the other toss.

## Simple Events:

In the case of simple events, we take the probability of occurrence of single events.

### Examples:

1.Probability of getting a Head (H) when a coin is tossed 2.Probability of getting 1 when a die is thrown

## Compound Events:

In the case of compound events, we take the probability of joint occurrence of two or more events.

### Examples:

When two coins are tossed, probability of getting a Head (H) in the first toss and getting a Tail (T) in the second toss.

## Algebra of Events:

Let A and B are two events with sample space S.
Then

1.$A \cup B$ is the event that either A or B or Both occur.
(i.e., at least one of A or B occurs)

2.$A \cap B$ is the event that both A and B occur

3.$\overline{A}$is the event that A does not occur

4.$\overline{A} \cap \overline{B}$ is the event that none
of A and B occurs

## Probability of an Event:

Let E be an event and S be the sample space. Then
probability of the event E can be defined as

P(E) =$\dfrac{n(E)}{n(S)}$

where P(E) = Probability of the event E, n(E) = number of
ways in which the event can occur and n(S) = Total number
of outcomes possible

### Examples:

1.A coin is tossed once. What is the probability of
getting Head?

Total number of outcomes possible when a coin is tossed =
n(S) = 2 (∵ Head or Tail)

E = event of getting Head = {H}. Hence n(E) = 1

$\dfrac{n(E)}{n(S)}$=$\dfrac{1}{2}$

2.Two dice are rolled. What is the probability that the
sum on the top face of both the dice will be greater than
9?

Total number of outcomes possible when a die is rolled = 6
(∵ any one face out of the 6 faces)

Hence, total number of outcomes possible two dice are
rolled, n(S) = 6 × 6 = 36

E = Getting a sum greater than 9 when the two dice are
rolled = {(4, 6), {5, 5}, {5, 6}, {6, 4}, {6, 5}, (6, 6)}

Hence, n(E) = 6

$\dfrac{n(E)}{n(S)}$=$\dfrac{6}{36}$=$\dfrac{1}
{6}$

## Addition Theorem:

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

P(A U B) = P(A) + P(B) – P(A ∩ B)

If A and B are mutually exclusive events, then P(A U B) =
P(A) + P(B) because for mutually exclusive events,
P(A ∩ B) = 0

If A and B are two independents events, then

P(A ∩ B) = P(A).P(B)

### Example :

Two dice are rolled. What is the probability of getting
an odd number in one die and getting an even number in the
other die?

Total number of outcomes possible when a die is rolled, n
(S) = 6 (∵ any one face out of the 6 faces)

Let A be the event of getting the odd number in one die =
{1,3,5}. => n(A)= 3

p(A)=$\dfrac{n(A)}{n(S)}$=$\dfrac{3}{6}$=$\dfrac{1}
{2}$

Let B be the event of getting an even number in the other
die = {2,4, 6}. => n(B)= 3

p(B)=$\dfrac{n(B)}{n(S)}$=$\dfrac{3}{6}$=$\dfrac{1}
{2}$

Required Probability,
P(A ∩ B) = P(A).P(B) = $\dfrac{1}{2} \times\dfrac{1}
{2}$=$\dfrac{1}{4}$

Let A be any event and $\overline{A}$ be its
complementary event (i.e., $\overline{A}$ is the event that
A does not occur). Then

$P(\overline{A})$ = 1 - P(A)

## Odds on an event:

Let E be an event associated with a random experiment.
Let x outcomes are favourable to E and y outcomes are not
favourable to E, then

Odds in favour of E are x:y, i.e., $\dfrac{x}{y}$ and

Odds against E are y:x, i.e., $\dfrac{y}{x}$
P(E) = $\dfrac{x}{x+y}$
$P(\overline{E})$=$\dfrac{y}{x+y}$

### Example :

What are the odds in favour of and against getting a 1
when a die is rolled?

Let E be an event of getting 1 when a die is rolled

Outcomes which are favourable to E, x=1

Outcomes which are not favourable to E, y=5

Odds in favour of getting 1 =$\dfrac{x}{y}=\dfrac{1}{5}$
Odds against getting 1 = $\dfrac{x}{y}$=$\dfrac{y}
{x}$=$4\dfrac{5}{1}$

## Conditional Probability:

Let A and B be two events associated with a random experiment. Then, probability of the occurrence of A given that B has already occurred is called conditional probability and denoted by P(A/B)

### Example :

A bag contains 5 black and 4 blue balls. Two balls are
drawn from the bag one by one without replacement. What
is the probability of drawing a blue ball in the second draw
if a black ball is already drawn in the first draw?

Let A be the event of drawing black ball in the first draw
and B be the event of drawing a blue ball in the second
draw. Then, P(B/A) = Probability of drawing a blue ball in
the second draw given that a black ball is already drawn in
the first draw.

Total Balls = 5 + 4 = 9

Since a black ball is drawn already,

total number of balls left after the first draw = 8

total number of blue balls after the first draw = 4

P(B/A) =$\dfrac{4}{8}$=$\dfrac{1}{2}$

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