Probability (Statistics) (Experiment, Event).pdf

Probability is a part of Statistics, Data Analyses

1. Probability
Probability refers to the likelihood of the occurrence
of some random events in future.
Experiment: An experiment is an act that can be
repeated under given conditions.
i.e. in throwing a dice , the possible outcomes are
{1,2,3,4,5,6} but it is not certain which will be result.{1,2,3,4,5,6} but it is not certain which will be result.
Event: One or more specified outcomes from an
experiment is called event.
i.e. tossing a coin and occurs either head (H) or tail
(T) is an event.
Compound event: When two or more events occur in
connection with each other, then their simultaneous

2. occurrence is called a compound event. The
compound event is an aggregate of simple events.
i.e. if A & B are two simple events in the same
sample space then their simultaneous occurrence is
denoted by (AB), (AB) is a compound event.
Sample space: The totality of all possible outcomes
in an experiment is called a sample space and is in an experiment is called a sample space and is
denoted by S.
i.e. in throwing a coin the sample space is S={H, T}
Mutually exclusive event: Two or more events which
cannot occur together such that occurrence of one
event excludes the occurrence of other events is

3. called mutually exclusive event. Two mutually
exclusive events cannot occur together in same
sample space.
Independent event: If the occurrence of an event is
not influenced or affected by the occurrence or not
occurrence of another event, these two events are
said to be independent to each other. In other said to be independent to each other. In other
words, two events A & B are said to be independent
if P(AB)= P(A).P(B)
Equally likely event: Two or more events are said to
be equally likely events when each of the events
has equal chance of occurrence. In other words

4. equally likely event is those which does not occur
more than others.
Exhaustive event: Events are said to be exhaustive
when they include all possible events.
Conditional event: If the occurrence of an event
depends on the occurrence or not occurrence of
another event in the same sample space is called a another event in the same sample space is called a
conditional event.
i.e. If A & B are two events, occurrence of event A
under the condition that B occurs in the same
sample space is conditional event denoted by (A/B)

5. Definitions of Probability
There are mainly two definitions of probability-
i)Mathematical or classical or a-priori definition of
probability.
ii)Statistical or empirical or a-posteriori definition
of probability.
Mathematical definition of probability: If a trial Mathematical definition of probability: If a trial
results in ‘n’ exhaustive, mutually exclusive &
equally likely cases and ‘m’ of them are
favourable to an event A, then the probability of
the happening of A is given by,

6. Favourable no. of outcomes m
P(A)= =
Total no. of outcomes n
This gives the numerical measures of probability.
Limitations of classical definition:
We cannot define this kind of probability ifWe cannot define this kind of probability if
a) the outcomes are not equally likely and
b) The no. of outcomes are infinite.
Statistical definition of probability: If an event A
occur r times in a series of n independent trials
conducted under uniform conditions, the ratio r/n

7. is called relative frequency of the event A in n trials.
The limit of r/n as n tends to infinity is the
probability of event A.
i.eP(A)=Lt (r/n)
n α
No.oftoses No.ofheads obtained Prob. of getting head
10 4 0.40
Limitations of statistical definition:
i) In practice it is not possible to repeat trials under
10 4 0.40
20 9 0.45
50 26 0.52
100 51 0.51
1000 500 0.50

8. the same conditions.
ii) It is difficult to obtain the probability of an event
using this definition.
Laws of probability:
There are two laws of probability namely-
a)Additive law of probability or theorem of total a)Additive law of probability or theorem of total
probability.
b)Multiplicative law of probability or theorem of
compound probability.

9. State and prove the additive law of probability:
Statement:Theprobability of any one of the not
mutually exclusive events A
1 & A
2is the sum of
the individual probabilities of A
1 & A
2minus the
probability of compound event (A
1A
2).
i.e. P(A
1+A
2)=P(A
1)+P(A
2)-P (A
1A
2)
12 1 2 12
Proof: Let us consider two not mutually exclusive
events A
1 & A
2. From the diagram we have,
A
1= A
1A
2+ A
1A
2. So, P(A
1)=P(A
1A
2+ A
1A
2)
= P(A
1A
2)+P( A
1A
2)…..(i)
Similarly, A
2= A
1A
2+ A
1A
2

11. or, P(A
2)= P(A
1A
2)+P( A
1A
2)…..(ii)
Again from the diagram we have,
A
1+A
2= A
1A
2+ A
1A
2+ A
1A
2
So, P(A
1+A
2) = P(A
1A
2)+P( A
1A
2)+ P(A
1A
2)….(iii)
Adding equations (i) & (ii) we have,
P(A)+P(A)=P(AA)+P( AA)+P(AA)+P( AA)P(A
1)+P(A
2)=P(A
1A
2)+P( A
1A
2)+P(A
1A
2)+P( A
1A
2)
= P( A
1A
2)+P(A
1A
2)+P( A
1A
2)+P(A
1A
2)
= P(A
1+A
2) + P(A
1A
2) [From equation (iii)]
or, P(A
1+A
2)= P(A
1)+P(A
2) -P(A
1A
2) (proved)

12. *If the two events A
1 & A
2are independent then,
P(A
1+A
2)= P(A
1)+P(A
2) -P(A
1)P(A
2)
*If the two events A
1 & A
2are mutually exclusive
then, P(A
1+A
2)= P(A
1)+P(A
2)

13. Multiplicative law of probability
Statement: The two dependent events A & B,
probability of joint occurrence of A & B is
P(AB)=P(A).P(B/A) ,or P(AB)= P(B).P(A/B)
Proof: Let n denote the total number of equally likely
outcomes of an experiment. n
1outcomes are
favourable to the event A, noutcomes are favourable to the event A, n
2outcomes are
favourable to the event B and n
12outcomes are
favourable to both A & B. Then by definition we
have, P(A)=n
1/n, P(B)=n
2/n, P(B/A)=n
12/n
1and
P(A/B)=n
12/n
2
Now, P(AB)=n
12/n = (n
1/n).(n
12/n
1)

15. or, P(AB)= P(A).P(B/A)
Similarly, P(AB)=n
12/n = (n
2/n).(n
12/n
2)
or, P(AB)= P(B).P(A/B) (Proved)
* When the events are independent then,
P(AB)= P(A).P(B/A)=P(A).P(B)

16. Problems(Probability)
* There are 20 tickets numbered 1,2,….,20. A
ticket is chosen at random. Find the probability
that the serial number of the ticket will be i)
multiple of 2 or 5 ii) multiple of 3 or 5.
Solution: i) Multiple of 2 are-2,4,6,8,10,12,14,16,
18, 20. Multiple of 5 are-5,10,15,20.
Let A be the event which is multiple of 2 & B be
the event which is multiple of 5.
So, P(A)=10/20 & P(B)=4/20.
2 cases are favourable to A & B, So P(AB)=2/20

17. So,P(A+B)=P(A)+P(B)-P(AB)=10/20+4/20-2/20
=12/20=3/5.
ii) Multiple of 3 are-3,6,9,12,15,18.
Let C be the event which is multiple of 3
So, P(C)=6/20So, P(C)=6/20
1 case is favourable to B & C, So P(BC)=1/20
So, P(B+C)=P(B)+P(C)-P(BC)=4/20+6/20-1/20
= 9/20

18. * 4 accountants,2 economists, 3 statisticians & 1
physician are working in an organization. A 4
member team is to be formed for a certain
purpose. Find the probability that the team will
include i) at least 1 statistician, ii) the
physician.physician.
Solution:Total no. of employees= 4+2+3+1=10
4 members can be selected
10
C
4=210 ways.
So, total no. of outcomes=210
i) Let A be the event that at least 1 statistician
will include in the team.

19. So probability that there is no statistician in the
team= 1-P(A). So if there is no statistician 4
member team will be
7
C
4= 35 ways.
So, 1-P(A)= 35/210=1/6, So, P(A)=1-1/6=5/6.
ii) If physician can include then 3 members are
chosen another 9 members.chosen another 9 members.
So P(a physician & other 3 members)
1
C
1X
9
C
3
=
210
= 2/5

20. * Students A & B can individually solve 75% &
50% problems respectively of a book. What is
the probability that either A or B can solve any
randomly selected problem of that book?
Solution:As given, P(A)=75/100=3/4
and P(B)=50/100=1/2.
According to additive law of probability,
P(A+B)= P(A)+P(B)-P(A).P(B)
[Since P(AB)=P(A).P(B)]
=3/4+1/2-3/4*1/2=7/8

21. * An urn contains 6 white & 5 red balls. Three balls
are drawn at random. Find the probability that all
the three balls are white.
Solution: Total no.ofballs=6+5=11
11x10x9
Total no.ofoutcomes=
11
C
3= = 165Total no.ofoutcomes=C
3= = 165
3x2
6x5x4
Favourable no. of outcomes=
6
C
3= =20
3x2

22. Favourable no.ofoutcomes
So, required probability=
Total no. of outcomes
= 20/165= 4/33
* An urn contains 17 balls of which 8 are white, 6 are
red & 3 are blue. Two balls are drawn at random. red & 3 are blue. Two balls are drawn at random.
Find the probability that i) the balls are of same
colour & ii) the balls are different in colour.
17x16
Solution: Total no.ofoutcomes=
17
C
2= =136
2x1

23. Let, A denote the event that both the balls are white,
B denote the event that both the balls are red & C
denote the event that both the balls are blue.
8
C
2
6
C
2
Here, P(A)= =28/136, P(B)= = 15/136
136
3
C
2 136
P(C)= =3/136P(C)= =3/136
136
Since A, B & C are mutually exclusive, so
P(A+B+C)=P(A)+P(B)+P(C)= 28/136+15/136+3/136
=46/136=23/68
ii) Both the balls are different in colour, then the

24. Combination will be-a) 1ball is white & 1 ball is red
or b) 1ball is white & 1 ball is blue or c) 1ball is red
& 1 ball is blue. So,
P(1ball is white & 1 ball is red)+P(1ball is white & 1
ball is blue)+P(1ball is red & 1 ball is blue)
8
C
1x
6
C
1
8
C
1x
3
C
1
6
C
1x
3
C
1C
1xC
1 C
1xC
1C
1xC
1
= + +
136 136 136
8x6 8x3 6x3
= + +
136 136 136

25. 48+24+18
=
136
=90/136
=45/68