1.10 Probability
Introduction:
The above picture summarises chances of some thing happening from
'impossible' to 'possible'.
As we know ‘probability’ means chance of some thing
happening. It could be any of the following:
1.
The
chances of rains on any particular day
2.
The
chances of
3.
A
cricketer scoring century in a particular test match
4.
5.
Number
of times head comes up when 1 rupee coin is tossed up 100 times
6.
Number
of times number 2 appears when a dice is
drawn 500 times
The outcome of any of the above event is pure
chances. It may happen or it may not happen. Do you notice the difference
between first 4 statements and the last
2 statements? In case of first 4 cases, the result depends upon external
factors such as place, time, strength of opposition team..
where as result of 5 and 6 will almost be same irrespective of place and
time of the event. There are cases when
outcome of certain events can be forecast with some certainty. One such event is
tossing of coin. When we toss the coin there can only be 2 outcomes, either head
appears or tail appears. When coin is tossed up several times obviously, the
chances of head coming up can vary from 0% to 100% and tail coming up can vary
from 100% to 0%. However when
coin is tossed up in large number of times(in thousands) head coming up may be 50% and tail coming up
could be 50%.
In such cases, the probability is 1/2. Note that when head
appears tail can not appear at the same time. Similarly when tail appears, head
can not appear at the same time. This we say that events are exclusive. That
is to say, there can not be any other possibility other than head appearing or
tail appearing. In such case, we note that the sum of their probability is
1/2+1/2= 1. Note that probability is expressed as a
ratio and it is always ≤ 1. why
does dice have dots numbers from 1 to 6?.
Note that dice is a cube and it has 6 faces. Hence their
faces are represented by 1 to 6 dots representing numbers from 1 to 6. When a
dice is cast what are the chances of a number appearing from 1 to 6? It is one
out of 6 and hence the probability is 1/6.
What is the sum of probabilities of number appearing between
1 and 6?
It is = Probability of number 1 appearing+ Probability of
number 2 appearing + Probability of number 3 appearing + Probability of number
4 appearing + Probability of number 5 appearing + Probability of number appearing between 1 and 6
= 1/6+1/6+1/6+1/6+1/6+1/6= 1
(Note that, when one number appears when a die is cast,
other number can not appear at the same time. Since probability of each number
appearing is excusive of others, sum of their probabilities is equal to 1.
Probability of number 1 not appearing
= Probability of number 2 appearing + Probability of
number 3 appearing + Probability of number 4 appearing + Probability of number
5 appearing + Probability of number 6 appearing = 1/6+1/6+1/6+1/6+1/6= 5/6
This also can be calculated as
Probability of number 1 not appearing = 1  Probability of
number 1 appearing = 11/6= 5/6
Let us learn some of the definitions of terms used
No 
Term 
Meaning 
Examples 
1 
Experiment 
Test or
a procedure or operation which produces a result. In this lesson all
experiments will be random. 

2 
Trial 
Performing
an experiment 
tossing
a coin/throwing a dice 
3 
outcome 
Result
of trial 
head or tail/ appearance of number from 1 to 6. 
4 
Sample
space 
Set of
all possible outcomes 
S = {H,T}. S = {1,2,3,4,5,6} 

Event 
Every subset of
Sample Space 
Getting
the head=A= {H}, Getting
the number 4=B={4} Getting
even numbers= C = {2,4,6} 
5 
Elementary
event 
Each of
the outcome in sample space. That is each of the element in sample space.
(Every subset of sample space having only one element) 
Getting
the head=A= {H}, Getting
the number 4=B={4} 
6 
Compound
event 
Obtained
by combining one or more elementary events ( Subsets of sample space having
one or more element) 
Getting
all three heads A= {HHH} Getting
at least 2 tails B = {HTT, TTH,
THT, TTT} 
7 
Favourable
event 
An
elementary element that occurs 

8 
Certain(sure)
event 
If any
one of the event will occur in the trial (probability
of sure event is 1) 
When a
die is cast the probability of getting a number > 0 and < 7 is 1. (Because when the die is
cast the numbers got are
always between 1 and 6). 
9 
Impossible
event 
An
event which will not occur in any of the trial (Probability
of impossible event is 0) 
when a die is cast the probability of getting number <1
or >6 is 0.(Because this event will not occur at all). 
10 
Complementary
event 
Converse
of elementary event. if element A is denoted by A
then complementary event is denoted by _{}. Note that A
_{} _{} = S and A _{} _{} = {_{} } 
Getting
head A = {H}, converse of this is not getting head that is getting tail.
Hence _{}= {T} Getting
the number 4, B={4}, converse of this is not getting
4, that is getting numbers 1,2,3,5 and 6. Hence_{} = {1,2,3,5,6} 
11 
Mutually
exclusive event 
2 or
more events are said to be mutually exclusive if those events do not occur
simultaneously, that is they do not happen together 
Let S =
{1,2,3,4,5,6}, A = {1,2,3}, B = {4,5} Here A
and B are mutually exclusive, because either A occurs or B occurs. they do not occur at the same time. Also
note that A _{} B = _{} 
Experiment 
Sample
space and Events 
No of
events n(S) 
Event A 
Favourable
outcomes to event A and n(A) 
Probability P(A) = n(A)/n(S) 
Complementary
Event( not A= _{}) 
Probability
of Complementary Event P(_{}) = n(_{})/n(S) 
1 coin
is tossed 
S={ H,T } 
2 
Tail
coming up 
A=(T) n(A)=1 
1/2 
head
comes up= (H) n(_{})=1 
1//2 
2 coins
are tossed 
S={ HH,HT,TH,TT } 
2*2=4 
Tail
should not come up 
A = {(HH)} n(A)=1 
1/4 
Tail
comes up n(_{})=3 
3/4 
3 coins
are tossed 
S={ HHH,HHT,HTH,THH, HTT, THT,TTH,TTT
} 
2*2*2=8 
Same
face coming up 
A= {(HHH),(TTT)} n(A)= 2 
2/8 
Not all
same faces n(_{})=6 
6/8 
1 dice
is cast 
S={1,2,3,4,5,6
} 
6 
Even
number coming up 
A={2,4,6} n(A)= 3 
3/6 
Odd
number coming up n(_{})=3 
3/6 
2 dices
are cast 
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) } 
6*6=36 
Getting the same number 
A=
{(1,1),(2,2),(3,3), (4,4),(5,5),(6,6) n(A)= 6 
6/36 
Different
numbers n(_{})= 30 
30/6 
P(E) = Probability of an event P_{E}
= _{}=_{}
Observe that P(A)+ P(_{}) =1 ( _{} n(A)/n(S) + n(_{})/n(S) ={ n(A)
+ n(_{})}/n(S) = n(S)/ n(S) )
_{} P(A)
=1 P(_{}), P(_{}) =1 P(A)
Probability: It is the ratio of the number of
elementary elements favourable to the event E to the total number of elements
in the sample space
Can probability ratio be 0
or 1?
1. What
is the probability of getting a number 0 or greater than 6 when dice is cast?
In this
case A = {} an empty set, and n(S) = 6
, hence P(A)= n(A)/n(S)= 0 (When a
dice is cast number is one among
1 to 6 only). This we call an impossible event, because such a
event does not happen
2. What
is the probability of getting a number 1 to 6 when dice is cast?
In this
case A = S and
hence n(A) = n(S) = 6 , Thus P(A)=
n(A)/n(S)= 1 (When a dice is cast number
is always one among 1 to 6). This we
call a sure event,
because such a event always happens
Thus 0 ≤ P(A) _{} 1 This is what we
represented in the beginning by a figure:
Problem 1 : A survey of 850 working woman
showed that 158 used own fourwheeler, 416 used twowheeler, and the remaining
used public transport. If one is chosen randomly among these women, what is the
probability of the woman who commute by (1) own four wheeler (ii) Own two wheeler
(iii) public transport and (iv) own vehicles.
Solution:
Number of women travelling by their own vehicles= 158+416=
574
Number of women travelling by public transport= 850574= 276
Probability of woman using own four wheeler = 158/850
Probability of woman using own two wheeler = 416/850
Probability of woman using public transport = 276/850
Probability of woman using own vehicle = 574/850= (158/850+416/850)
Problem 2 : A box contains 12 balls out of which 'x' are red.
(i) If one ball is drawn, what
is the probability of getting red ball?
(ii) If 6 more red balls are put in the box, the
probability of a drawing a red ball will be doubled than earlier. Find 'x'
Solution:
(i)
Since
'x' number of balls are red out of total
of 12 balls, the probability of getting red is x/12
(ii)
When
6 more balls are added, total number of
balls become 18 (=12+6)
Hence the new probability of getting a red ball = (x+6)/18
It is also given that new probability is twice the earlier
probability thus New probability = 2x/12= x/6;
_{} (x+6)/18 = x/6
6x+36=18x
36= 12x
x=3
Problem 3 : A game consists of rolling 2
dices. If the sum is 2,3,4,5,10,11 or 12, player A
wins. If sum is other than these, then player B wins. Of you want to win, would
you be player A or player B?
Solution:
Let A = { combination of dices whose sum is 2,3,4,5,10,11
or 12 }= {
(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)
}
Then B= _{} ={ combination
of dices whose sum is not 2,3,4,5,10,11 or 12 }= { combination of dices whose
sum is 6,7,8, or 9 }
=
{(1,5),(5,1),(2,4),(4,2),(3,3),(1,6)(6,1),(2,5),(5,2),(3,4),(4,3),(2,6),(6,2),(3,5),(5,3),(4,4),(3,6),(6,3),(4,5),(5,4)}
Note that n(A)= 16 and n(_{}) = 20. Thus number of elementary elements= 36( Other wise also note 6*6=36: fundamental principle of
computing)
P(A) = 16/36 = 0.4444 and P(_{})= 20/36= 0.55555
Obviously Any one would like to be B.
Mutually exclusive event
Let S = {A_{1}, A_{2},
A_{3} …. A_{n}} Clearly n(S)= n, P({A_{1}})= 1/n , P({A_{2}})=1/n,
P({A_{3}})=1/n … P({A_{n}})= 1/n
We note that, P({A_{1}})+
P({A_{2}})+ P({A_{3}})+ . . . P({A_{n}})= 1/n+1/n+1/n +
. . . 1/n= n/n= 1
So, sum of the probabilities of
all the elementary events of an experiment is 1.
Let E_{1} and E_{2}
be two mutually exclusive elements, that is they will
not have any common elements. We have observed that E_{1}_{}E_{2} = {_{} }
They being disjoint sets, from set
theory, we know n(E_{1}_{}E_{2}) = n(E_{1})+n(E_{2})
_{} n(E_{1}_{}E_{2})/n(S)= n(E_{1})/ n(S)+n(E_{2})/
n(S)
_{} P(E_{1}_{}E_{2}) = P(E_{1})+P(E_{2})
In general if E_{1}, E_{2},
E_{3} . . . E_{n } are mutually exclusive elements then P(E_{1}_{}E_{2}_{}E_{3} . . . _{}E_{n }) = P(E_{1}) +P(E_{2})+
P(E_{3})+ . . . P(E_{n})
Problem 4 : When a die is thrown, find the
probability that either an odd number or square number occurs
Solution:
S= {1,2,3,4,5,6}, n(S)=6
A= { Odd numbers or square
numbers} = {1,3,4,5}, n(A)=4
_{} P(A)
= 4/6
Problem 5 : The outcome of a random
experiment results in either success or failure. If the probability of success
is thrice the probability of failure, find the probability of success
Solution:
P(A) =Probability of success,
Probability of failure= P(_{})
P(A) = 3P(_{}) =3x, we also know that P(A)+P(_{}) = 1;
_{} P(A)+
1/3P(A)=1
4P(A)= 3 P(A)= 3/4
Problem 6 : Three squares of a chess board
are selected in random. Find the probability of getting two squares of one
colour and the other of different colour.(here square
represents smallest square on which chess is played)
Solution:
Note that chess board has 64 smallest squares.
3 squares can be selected in _{64}C_{3}
ways. Hence
n(S)= _{64}C_{3 }=64*63*62*61!/61!*3!= 64*63*62/2*3=
64*21*31
Number Of black coloured squares
= 32, Number of white coloured squares=32
2 squares can be got in _{32}C_{2}
ways. 1
square can be got in _{32}C_{1}
ways.
If A is number of ways getting two squares of one colour
and the other of different colour, then n(A)= [{_{32}C_{2}}*{_{32}C_{1}}]=[{32*31/2!)}]*{32}=32*31*16
_{} P(A)
= _{32}C_{2}*_{32}C_{1}/_{64}C_{3}=
32*31*16/64*21*31=8/21
Problem 7 : A committee of 5 persons is
selected from 4 men and 3 women. What is the probability that the committee
will have
(i) one man (ii) 2 men (iii)
2 women (iv) at least 2 men
Solution:
Note that total number of people available to form the committee is 7.
Hence number of ways of forming different 5 member
committees = _{7}C_{5}= 7*6/2!=
21= n(S)
(i) one man in the committee:
Since there are only 3 women, we will need 2 men to form a
committee of 5. Hence, the probability of forming a committee having
one man is 0
(ii) 2
men in the committee:
Number of ways committees that can be formed with 2 men
from 4 men =_{4}C_{2}= 4*3/2!=
6
Remaining 3 members are to be women.
Number of ways committees that can be formed with 3 women from 3 women =_{3}C_{3}= 1
_{} The number of committees
that can be formed with 2 men and 3 women = 6*1=6 (1)
_{}Probability = 6/21= 2/7
(iii) 2
women in the committee:
Number of committees that can be formed with 2 women from
3 women = _{3}C_{2}= 3/1!=
3
Remaining 3 members are to be men.
Number of ways committees that can be formed with 3 men from 4 men =_{4}C_{3}=4/1!= 4
_{} The number of
committees that can be formed with 2 women and 3 men = 3*4=12 (2)
_{}Probability = 12/21= 4/7
(iv) at least 2 men in the committee:
Number of committees that can be formed with at least 2
men
= Number of committees that can be formed with 2 men+ Number
of committees that can be formed with 3 men+ Number of committees that can be
formed with 4 men
First we need to find number of committees that can be
formed with 4 men.
Number of committees that can be formed with 4 men from 4
men = _{4}C_{4}= 1
Remaining 1 member has to women.
Number of ways committees that can be formed with 1 woman from 3 women =_{3}C_{1}=3*2!/2!*1!= 3
_{} The number of
committees that can be formed with 1 woman and 4 men = 1*3=3 (3)
_{} Number of committees
that can be formed with at least 2 men = 6+12+3
=21
_{}Probability = 21/21= 1
1.1 Summary of learning
No 
Points studied 
1 
Probability,Event,Favourable
event, exclusive Event 