The set of all possible outcomes in a random experiment is called a sample space. It is generally denoted by S.

**Probability**

Few centuries ago,
gambling and gaming were considered to be fashionable and became widely popular
among many men. As the games became more complicated, players were interested
in knowing the chances of winning or losing a game from a given situation. In
1654, Chevalier de Mere, a French nobleman with a taste of gambling, wrote a
letter to one of the prominent mathematician of the time, Blaise Pascal,
seeking his advice about how much dividend he would get for a gambling game
played by paying money. Pascal worked this problem mathematically but thought
of sharing this problem and see how his good friend and mathematician Pierre de
Fermat could solve. Their subsequent correspondences on the issue represented
the birth of Probability Theory as a new branch of mathematics.

A random experiment is an experiment in
which

(i) The set of all possible
outcomes are known (ii) Exact outcome is not known.

**Example : **1. Tossing a coin. 2.
Rolling a die. 3. Selecting a card from a pack of** **52** **cards.

The set of all possible
outcomes in a random experiment is called a sample space. It is generally denoted by
*S*.

**Example : **When we roll a die, the
possible outcomes are the** **face numbers 1,2,3,4,5,6 of the die. Therefore the sample space is
*S *= {1,2,3,4,5,6}

Each element of a sample
space is called a sample point.

Tree diagram allow us to
see visually all possible outcomes of an random experiment. Each branch in a
tree diagram represent a possible outcome.

**Illustration**

(i) When we throw a die,
then from the tree diagram (Fig.8.3), the sample space can be written as *S*
= {1,2,3,4,5,6}

(ii) When we toss two
coins, then from the tree diagram (Fig.8.4), the sample space can be written as
S={*HH,HT,TH,TT*}

**Example 8.18 **Express the sample space
for rolling two dice using tree diagram.** **

** Solution **When we roll two dice,
since each die contain

Hence, the sample space
can be written as

S=
{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

** **In a random experiment,
each possible outcome is called an** **event. Thus,** **an event will be a subset of the sample space.

Getting two heads when
we toss two coins is an event.

**Trial : **Performing an experiment
once is called a** **trial.

**Example : **When we toss a coin
thrice, then each toss of a coin is a trial.

**Note**

**Elementary event: **If an event** ***E***
**consists of only one outcome then it is called an **elementary event**.

In a random experiment,
let S be the sample space and *E* ⊆ *S* . Then if *E* is
an event, the probability of
occurrence of *E* is defined as

This way of defining the
probability is applicable only to finite sample spaces. So in this chapter, we
will be dealing problems only with finite sample spaces.

A bag contains** **5** **blue balls and** **4** **green balls. A ball is
drawn at random** **from the bag. Find the probability that the ball drawn is (i) blue
(ii) not blue.

Total number of possible
outcomes *n*(S) = 5 + 4 = 9

(i) Let *A* be the
event of getting a blue ball.

Number of favourable
outcomes for the event A. Therefore, *n*(A) = 5

Probability that the
ball drawn is blue. Therefore, *P* (*A*) = *n*(A)/* n*(*S*) = 5/9

(ii)* ** *will be the event of not getting a blue ball. So *P* ()
= 1 − *P* ( *A*) = 1 – 5/9 = 4/9

Two dice are rolled.
Find the probability that the sum of outcomes is

(i) equal to 4

(ii) greater than 10

(iii) less than 13

** **When we roll two dice,
the sample space is given by

*S *= { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
};* n *(*S*) = 36

(i) Let *A* be the
event of getting the sum of outcome values equal to 4.

Then *A* =
{(1,3),(2,2),(3,1)}; *n* (*A*) = 3 .

Probability of getting
the sum of outcomes equal to 4 is *P* (*A*) = *n*(*A*)/*n*(*S*) = 3/36 =1/12

(ii) Let B be the event
of getting the sum of outcome values greater than 10.

Then B =
{(5,6),(6,5),(6,6)}; *n* (*B*) = 3

Probability of getting
the sum of outcomes greater than 10 is *P* (*B*) = *n*(*B*)/*n*(*S*) = 3/36 =1/12

(iii) Let C be the event
of getting the sum of outcomes less than 13. Here all the outcomes have the sum
value less than 13. Hence *C* = *S*.

Therefore, *n* (*C*
) = *n* (*S*) = 36

Probability of getting
the total value less than 13 is *P* (*C* ) = *n*(*C*)/*n*(*S*) = 36/36 = 1.

Two coins are tossed
together. What is the probability of getting different** **faces on the coins?

When two coins are
tossed together, the sample space is

S = { HH, HT, TH, TT } ;
n(S) = 4

Let A be the event of
getting different faces on the coins.

A = { HT,TH } ; n(A) = 2

Probability of getting
different faces on the coins is P(A) =n(A)/n(S) = 2/4 = ½

**Example 8.22**

** **From a well shuffled
pack of**
**52** **cards, one card is drawn
at random.**
**Find the probability of
getting

(i) red card

(ii) heart card

(iii) red king

(iv) face card

(v) number card

*Solution*

*n *(*S*) = 52

(i) Let *A* be the
event of getting a red card.

*n *(*A*)* *=*
*26

Probability of getting a
red card is

*P *(*A*)* *=*
*26/52* *=* *1/2

(ii) Let *B* be the
event of getting a heart card.

*n *(*B*)* *=*
*13

Probability of getting a
heart card is *P*

* P*(*B*)* *=* n*(*B*)/*n*(*S*)*
*=* *13/52* *=* *1/4

(iii) Let *C* be
the event of getting a red king card. A red king card can be either a diamond
king or a heart king.

*n *(*C *)* *=*
*2

Probability of getting a
red king card is

*P*(*C*)* *=* n*(*C*)/*n*(*S*)* *=*
*12/52* *=* *1/26

(iv) Let *D* be the
event of getting a face card. The face cards are Jack (*J*), Queen (*Q*),
and King (*K*).

*n*(*D*)* *=*
*4* *×3* *=* *12

Probability of getting a
face card is

*P *(*D*)* *=*
n*(*D*)/*n*(*S*)* *=*
*12/52* *=* *3/13

(v) Let *E* be the
event of getting a number card. The number cards are 2, 3, 4, 5, 6, 7, 8, 9 and
10.

*n *(*E*)* *=*
*4* *×9* *=* *36

Probability of getting a
number card is

*P *(*E*)* *=*
n*(*E*)/*n*(*S*)* *=*
*36/52* *=* *9/13

What is the probability
that a leap year selected at random will contain** **53** **saturdays. (Hint: 366 =
52 ×7 + 2 )

** **A leap year has

The possible chances for
the remaining two days will be the sample space.

S = {(Sun-Mon, Mon-Tue, Tue-Wed,
Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}

*n *(*S*)* *=* *7

Let *A* be the
event of getting
53^{rd} Saturday.

Then *A *=* *{Fri-Sat,
Sat-Sun};* n *(*A*)* *=* *2

Probability of getting 53 Saturdays
in a leap year is P ( A ) = *n *(*A*)/* n *(*S*) = 2/7

Thinking Corner: What
will be the probability that a non - leap year will have 53 Saturdays?

**Example 8.24 **A die is rolled and a coin is tossed simultaneously. Find the
probability that the die shows an odd number and the coin shows a head.

Sample space

S =
{1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T};

n(S) = 12

Let A be the event of
getting an odd number and a head.

A = {1H, 3H, 5H}; *n*(*A*)
= 3

P (A) = n (A)/* n*(*S*)
= 3/12 = 1/4

A bag contains 6 green
balls, some black and red balls. Number of black balls is as twice as the number of red balls.
Probability of getting a green ball is thrice the probability of getting a red ball. Find (i)
number of black balls (ii) total number of balls.

Number of green balls is
*n*(*G*)
= 6

Let number of red balls
is *n(R)* = *x*

Therefore, number of
black balls is *n(B)* = 2*x*

Total number of balls *n(S)* = 6 + *x* + 2*x* = 6 + 3*x*

It is given that, *P (G)* = 3 × *P(R)*

6/ (6+3*x*) = 3 × *x*/(6 + 3*x)*

3*x* = 6 gives, *x* = 2

(i) Number of black
balls = 2 × 2 = 4

(ii) Total number of
balls = 6 +(3 × 2) = 12

A game of chance
consists of spinning an arrow which is equally likely to** **come to rest pointing to
one of the numbers 1, 2, 3, …12. What is the probability that it will point to

(i) 7

(ii) a prime number

(iii) a composite
number?

Sample space* **S*** **=
{1,2,3,4,5,6,7,8,9,10,11,12};

(i) Let *A* be the
event of resting in 7. *n*(*A*)=1

*P *(*A*)* *=*
n*(*A*)*/n*(*S*)*
*=* *1/12

(ii) Let *B* be the
event that the arrow will come to rest in a prime number.

*B *= {2,3,5,7,11};* n *(*B*)*
*=* *5

*P *(*B*)* *=
*n*(*B*)*/n*(*S*)
=* *5/12

(iii) Let *C* be
the event that arrow will come to rest in a composite number.

*C *= {4,6,8,9,10,12};* n *(*C
*)* *=6

*P *(*C*)* *=
*n*(*C*)*/n*(*S*)
= 6/12 = 1/2

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