Probability +2

 You've probably heard people say things like:
7/3/2013 2VSR
Teen mother
 The chance of rain
tomorrow is 75%.
 Teen mothers who live
with their parents
 He won the lottery!
All of these statements are about probability. We
see words like "chance", "less likely", "probably" since
we don't know for sure something will happen, but we
realise there is a very good chance that it will.

7/3/2013VSR 3
 Probability is a concept which numerically
measures the degree of uncertainty and
therefore is certainty of the occurrence of
events.
 Probability had its origin in the 16th century
when an Italian physician and mathematician
Jerome Cardon (1501 -1576) wrote the first
book on the subject. “Book of Games of
Chance”.
Subsequently, the theory of probability was
developed by Bernoulli, De-Moivar, Fisher and
others.

 The mathematicians is basically concerned
with drawing conclusions (or inference)
from experiments involving uncertainties.
 For these conclusions and inferences to be
reasonably accurate, an understanding of
probability theory is essential.
 In this section, we shall develop the concept
of probability with equally likely outcomes
7/3/2013VSR 4

7/3/2013VSR 5
Experiment
This is any process of observation or
procedure that
 can be repeated (theoretically) an
infinite number of times and
 has a well-defined set of possible
outcomes.

7/3/2013VSR 6
Sample space
This is the set of all possible
outcomes of an experiment.
Event
This is a subset of the sample
space of an experiment
Consider the following
illustrations:
The set of all event is the power
set of S , denoted By 2s .

7/3/2013VSR 7
Experiment 1: Tossing a coin.
Sample space: S = {Head or Tail}
Experiment 2: Tossing a coin twice.
S = {HH, TT, HT, TH}
Experiment 3: Throwing a die.
S = {1, 2, 3, 4, 5, 6}

7/3/2013VSR 8
Experiment 4: Two items are picked, one at a
time, at random from a manufacturing
process, and each item is inspected and
classified as defective or non-defective.
S = {NN, ND, DN, DD} where
N = Non-defective
D = Defective
Some events:
E1 = {only one item is defective} = {ND, DN}
E2 = {Both are non-defective} = {NN}

7/3/2013VSR 9
Definition of a Probability
Suppose an event A can happen in m ways
out of a total of n possible equally likely
ways.
Then the probability of occurrence of the
event (called its success) is denoted by
n
m
Sn
An
AP
)(
)(
)(
→no. of favourable cases
→no. of total outcomes of
the experiment

7/3/2013VSR 10
The probability of non-occurrence of the
event (called its failure) is denoted by
n
m
n
mn
EP 1)(
 Notice the bar above the E, indicating the
event does not occur.
Thus, = 1
In words, this means that the sum of the
probabilities in any experiment is 1.

7/3/2013VSR 11
* Odds in favour of A
)(
)(
AP
AP
* Odds in against A
)(
)(
AP
AP
 Addition theorem for two or more events.
i.e., P(A or B) = P(A) + P(B) – P(A B) .
If A and B are mutually exclusive events then ,
P(A or B) = P(A) + P(B) .

7/3/2013VSR 12
* In this chapter, we shall discuses the concept
of Conditional probability of event. Baye's
theorem, Multiplication rule of probability and
random variable and its probability, Binomial
distribution etc.
Conditional Probability
If E and F are two events associated with the
same sample space of a random experiment, E
given that F has occurred, is given by
0)(,
)(
)(
)( FP
FP
FEP
EorFP

7/3/2013VSR 13
• Properties of conditional probability.
i) P(S or F) = P(F/F) = 1.
ii) If A and B are any two events of a sample
space S and F is an event of S such that P(F) ≠ 0
then,
P[(A B) or F] = P(A or F) + P(B or F) – P(A B)
or F)

7/3/2013VSR 14
P(E F) = P(E).P(F/E)= P(F).P(E/F) .
Provided P(E) and P(F) ≠ 0
# More than two events E, F and
G, then by multiplication rule of
probability
P(E F G) = P(E).P(F/E).P(G/EF).
Where EF = E F.

7/3/2013VSR 15
The given events are said to be Equally
Likely, if none of them is expected to occur in
preference to the other.
Two events are said to be Independent, if
the occurrence of one does not depend upon
the other.
Hence events E and F are independent event
if P(EF) = P(E). P(F).

7/3/2013VSR 16
# For two independent events E and F, the
addition theorem becomes,
P(E or F) = P(E) + P(F) – P(E).P(F)
Or P(E or F) ).().(1)(1 FPEPFEP

7/3/2013VSR 17
Let's consider "E1 and E2" as the event that
"both E1 and E2 occur".
If E1 and E2 are dependent events, then:
P (E1 and E2) = P (E1) × P (E2 | E1)
If E1 and E2 are independent events, then:
P (E1 and E2) = P (E1) × P (E2)

7/3/2013VSR 18
For three dependent events E1, E2, E3, we have
P(E1 and E2 and E3)
= P(E1) × P(E2 | E1) × P(E3 | E1 and E2)
For three independent events E1, E2, E3, we
have
P(E1 and E2 and E3) = P(E1) × P(E2) × P(E3)

7/3/2013VSR 19
Two or more events are said to be mutually
exclusive if the occurrence of any one of them
means the others will not occur (That is, we
cannot have 2 events occurring at the same
time).
Thus if E1 and E2 are mutually exclusive
events, then
P(E1 and E2) = 0.

7/3/2013VSR 20
Suppose "E1 or E2" denotes the event that
"either E1 or E2 both occur", then
(a) If E1 and E2 are not mutually exclusive
events:
P(E1 or E2) = P(E1) + P(E2) − P(E1 and E2)
We can also write:
P(E1 ∪ E2) = P(E1) + P(E2) − P(E1 ∩ E2)
A diagram for this situation is as
follows. We see that there is some
overlap between the events E1 and E2.
The probability of that overlap portion
is P(E1 ∩ E2).

7/3/2013VSR 21
(b) If E1 and E2 are mutually exclusive events:
P(E1 or E2) = P(E1) + P(E2)
Our diagram for mutually exclusive events
shows that there is no overlap:

7/3/2013VSR 22
Exhaustive event.
A set of events is said to be exhaustive if the
performance of the experiment results in the
occurrence of at least one of them.
If a set of events E1, E2,……En , then for
exhaustive events
P(E1 E2 ….. En ) = 1.
If E1, E2,……En are mutually exclusive and
exhaustive events and any events E is said to
be compound events, if
0)(,).(
()(
1
1
n
i
i
i
i
n
i
i
EifP
E
E
PEP
EEPEP

7/3/2013VSR 23
• The events E1, E2,……En represent a partition
of the sample space S if they are pair wise
disjoint, exhaustive and have non-zero
probabilities.
a) Ei Ej = , i ≠ j , i, j = 1, 2, 3, ……..n.
b) Ei E2 …….. En = S
c) P( Ei ) > 0 for all i = 1, 2, ………n.

7/3/2013VSR 24
P(A) = P(E1).P(A/E1) + P(E2).P(A/E2) +
…….+ P(En).P(A/En)

7/3/2013VSR 25
Let E1, E2 ......En are n non empty events which
constitute a partition of sample space S, then
n
j
jj
ii
EAPEP
EAPEP
AEP
1
1
)/().((
)/().(
)/(
for any i = 1, 2, 3, ………n.

7/3/2013VSR 26
P(r) , the probability of r successes, is given by:
r
p
rn
qC rnrP )(
Where p = the probability
of successes.
q = the probability of
failure.
p + q = 1 , i.e., q = 1 – p .
Here n, p, q are called the parameters of Binomial
Distribution.

7/3/2013VSR 27
Mean of binomial distribution
Mean = np
Variance V(X) = E(X2) – (E(X))2 .
Mean of probability distribution = XP (X)
Variance = X2P(X) – (mean)2
S.D. = X2P(X) – (mean)2

7/3/2013VSR 28
Variance = npq
Standard Deviation ( )
npqDS ).(.
Recurrence Formula
).(..
)1(
)(
)1( rP
q
p
r
rn
rP

7/3/2013VSR 29
n
i
nni
PxPxpxPixxE
1
2211
.......)(
x1,x2 ……xn are possible values of the random
variable x with probabilities P1 , P2 …..Pn
respectively.

7/3/2013VSR 31
Q.1.A fair die is thrown, what is the probability
that either an odd number or a number greater
than 4 will up. (K.U)
Let S be the sample space throwing of the die
S = {1, 2, 3, 4, 5, 6}. i.e., n(S) = 6
Let A be the event of getting an odd number
B be the number greater than 4 .
A = {1, 3, 5} ; B = {5, 6} ; also A B = {5}
n(A) = 3; n(B) = 2 ; n(A B) = 1 6
1
)(
)(
)(
3
1
6
2
)(
)(
)(
2
1
6
3
)(
)(
)(
Sn
BAn
BAP
Sn
Bn
BP
Bn
An
AP
)()()()( BAPBPAPBAP
3
2
6
123
6
1
3
1
2
1
The required probability = 2/3.

7/3/2013VSR 32
Q.2. A man is known to speak truth 3 out of 4
times. He throws a die and reports that it is a 6.
Find the probability that it is actually 6. (AICBSE)
Let A be the man report that it is a 6.
Let E1 be the event “6 has occurred”
and E2 be that “6 has not occurred”
P(E1 ) = 1/6; P(E1 ) = 1–P(E1 ) = 1–1/6 = 5/6
also P(A/ E1 ) = ¾ . [ i.e., speak 3 out of 4]
P(A/ E2 ) = 1-P(A/ E1 ) = 1-3/4 = ¼

7/3/2013VSR 33
)/().()/().(
)/().(
)/(
2211
11
1
EAPEPEAPEP
EAPEP
AEP
Formulae
8
3
8
24
8
1
24
8
8
1
25
5
8
1
8
1
4
1
.
6
5
4
3
.
6
1
4
3
.
6
1

7/3/2013VSR 34
Q3. An insurance company insured 2000 Scooter
drivers, 3000 Car drivers and 4000 Truck drivers.
The probabilities of their meeting with an
accident respectively 0.04, 0.06 and 0.15. One of
the insured persons meets with an accident, find
the probability that he is a Car driver. (AICBSE)
Let E1 , E2 , E3 , be the Scooter, Car and Truck drivers
respectively.
Let A be person that meets with an accident

7/3/2013VSR 35
9
4
400030002000
4000
)(
9
3
400030002000
3000
)(
3
2
400030002000
2000
)(
3
2
1
EP
EP
EP
100
15
15.0)/(
;
100
6
06.0)/(;
100
4
04.0)/(
3
21
EAP
EAPEAP

7/3/2013VSR 36
Required probability
)
3
/().
3
()
2
/().
2
()
1
/().
1
(
)
2
/().
2
(
)/
2
(
EAPEPEAPEPEAPEP
EAPEP
AEP
43
9
60188
18
100
15
.
9
4
100
6
.
9
3
100
4
.
9
2
100
6
.
9
3

7/3/2013VSR 37
Q.4. Gowrave and Sowrave appear for an
interview for two vacancies. The probability of
Gowrave’s selection is 1/3 and that of
Sowrave’s selection is 1/5. Find the probability
that only one of them is selected. (CBSE)
Let A be Gowrave’s selection
B be the Sowrave’s selection.
P(A) = 1/3 ; P(B) = 1/5.
)( AP Probability that Gowrave not selected
= 1 – P(A) = 1 – 1/3 = 2/3.

7/3/2013VSR 38
)(BP Probability that Sowrave not selected
= 1 – P(B) = 1 – 1/5 = 4/5
Probability that only one of them is selected
5
2
15
2
15
4
5
1
.
3
2
5
4
.
3
1
)().()().( BPAPBPAP

7/3/2013VSR 39
Q.5. In a factory which manufactures
bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts of
their output 5, 4 and 2 percent are respectively
defective bolts. A bolt is drawn at random from
the total production and if found to be
defective. Find the probability that it is
manufactured by the machine B.
(CBSE)
Let E1 , E2 and E3 be the events of manufacturing the
bolt by machine A, B and C respective.

7/3/2013VSR 40
Let A be the event that bolt drawn is defective
100
2
)/(;
100
4
)/(;
100
5
)/(.,. 321
EAPEAPEAPei
Required probability
69
28
345
140
100
40
.
100
2
.
100
35
.
100
4
100
25
.
100
5
100
35
100
4
)()./()()./()()./(
)()./(
)/(
332211
22
2
EPEAPEPEAPEPEAP
EPEAP
AEP

7/3/2013VSR 41
Q.6. A four digit number is formed using the
digits 1, 2, 3, 5 with no repetitions. Find the
probability that the number is divisible by 5 . (CBSE)
Number of favourable cases m = four digit
numbers, which are divisible by 5 = 6.
The no. of exhaustive cases n = 4C4 = 4! =24.
Required probability = m/n = 6/24 = ¼

7/3/2013VSR 42
Q.7. A lot contains 50 defective and 50 non-
defective bulbs. Two bulbs are drawn at
random, one by one, with replacement. The
events A, B and C are defined as the first
bulbs is defective, the second bulbs is non-
defective, the two bulbs are both defective or
non-defective, respectively. Determine
whether A, B and C are pair wise independent.
Sample space S = {DD, DN, ND, NN} .
Where D = Defective bulb and N = Non-defective bulb

7/3/2013VSR 43
Thus A = {DD, DN} , B = {DN, NN} , C = {DD, NN}.
P(A) = 2/4 = ½ , P(B) = 2/4 = ½ , P(C) = 2/4 = ½ .
P(A B) = P(DN) = ¼ ; P(B C) = P(NN) = ¼
P(A C) = P(DD) = ¼
Now P(A). P(B) = ½ ½ = ¼ = P(A B)
i.e., A and B are independent
Similarly B and C and also A and C are independent
Hence A, B, C are pair wise independent.

7/3/2013VSR 44
Q.8. An urn contains m white and n black
balls. A ball is drawn at random and is put
back into the urn along with k additional balls
of the same colour as that of the ball drawn. A
ball is again drawn at random. What is the
probability that the ball drawn now is white.
Let W1 be : Ball drawn is white in the first draw.
W2 be : Ball drawn is white in the second draw.
and B1 be : Ball drawn is black in the first draw.
Now, P(W2) = P(W1) . P(W2 / W1) + P(B1) P(W2 / B1)

7/3/2013VSR 45
nm
m
knmnm
knmm
knmnm
mnmkm
knm
km
nm
m
knm
km
nm
m
))((
)(
))((
..
2

7/3/2013VSR 46
Q.9. A card is drawn out from a well shuffled
pack of 52 cards. If E is the event “the card
drawn out is a king or queen” and F is the event
“the card drawn out is a queen or an ace”, find
the probability P(E/F). (CBSE)
P(F) = P( card is queen or ace) = 8/52 = 2/13.
P(E F) = P(card is queen) = 4/52 = 1/13

7/3/2013VSR 47
Q.10. If P(A) = 3/5 and P(B) = 1/5 find P(A B) if
A and B are independent events.
Given that A and B are independent events.
P(A B) = P(A).P(B)
= (3/5) (1/5)
= 3/25

7/3/2013VSR 48
Q.11. If P(A) = 1/2 and P(B) = p ,P(A B) = 3/5 if
A and B are mutually exclusive events.
Given that A and B are mutually exclusive
events.
P(A B) = P(A) + P(B)
i.e., 3/5= ½ + p
p = 3/5 – ½
= 1/10.

7/3/2013VSR 49
Q.12. A husband and his wife appear for an
interview for two posts. The probability of
husband’s selection is 1/7 and that of wife’s
selection is 1/5 . What is the probability that
only one of them is selected.?
Let A and B be the events of husband and wife
respectively.
Here P(A) = 1/7 and P(B) = 1/5
5
4
5
1
1)(
7
6
7
1
1)(
BP
andA

7/3/2013VSR 50
Here A and B are independent events

7/3/2013VSR 51
Q.13. A problem in Mathematics is given to
three students whose chances of solving it are
½, 1/3, ¼. What is the probability in the
following cases ?
1) That the problem is solved. (Kerala)
2) only one of them solved correctly (CBSE)
3) at teast one of them may solve it.
Let A, B, C be the three event when the problem in
maths is solved by the three students.

7/3/2013VSR 52
4
3
4
1
1)(;
3
2
3
1
1)(;
2
1
2
1
1)(
.
4
1
)(;
3
1
)(;
2
1
)(
CPBPAP
CPBPAP
1) The probability that the problem is solved
= Probability that the problem is solved by at least
one student.
4
3
4
1
1
4
3
3
2
2
1
1
)().().(1 CPBPAP

7/3/2013VSR 53
2) The probability that only one solves it correctly
)().().()().().()().().( CPBPAPCPBPAPCPBPAP
24
11
12
1
8
1
4
1
4
1
3
2
2
1
4
3
3
1
2
1
4
3
3
2
2
1

7/3/2013VSR 54
3) The probability that atleast one of them may
solve the problem
4
3
4
1
1
4
3
3
2
2
1
1
)().().(1 CPBPAP

7/3/2013VSR 55
Q.14. In a bolt factory, machines A, B and C
manufacture respectively 25%, 35%, 40% of the
total. Of their output 5, 4 and 2% are defective.
A bolt is drawn at random from the product.
1). What is the probability that the bolt drawn
is defective?
2) If the bolt drawn is found to be
defective, find the probability that it is a
product of machine B?.

7/3/2013VSR 56
100
2
)/(;
100
4
)/(;
100
5
)/(
100
40
)(;
100
35
)(;
100
25
)(
CDPBDPADandP
CPBPAHereP
Where D denotes defective bolt.
0345.0
100
2
.
100
40
100
4
.
100
35
.
100
5
.
100
25
)/()()/()()/()()().1 CDPCPBDPBPADPAPDP

7/3/2013VSR 57
2) By Baye’s theorem
69
28
345
140
100
2
.
100
40
100
4
.
100
35
100
5
.
100
25
100
4
.
100
35
)/().()/().()/().(
)/().(
)/(
CDPCPBDPBPADPAP
BDPBP
DBP

7/3/2013VSR 58
Q.15. If the probability that person A will be alive in
20 years is 0.7 and the probability that person B will
be alive in 20 years is 0.5, what is the probability that
they will both be alive in 20 years?
These are independent events, so
P(E1 and E2) = P(E1) P(E2)
= 0.7 0.5 = 0.35

7/3/2013VSR 59
Q.16. A fair die is tossed twice. Find the
probability of getting a 4 or 5 on the first toss
and a 1, 2, or 3 in the second toss.
P(E1) = P(4 or 5) = 2/6 = 1/3
P(E2) = P(1, 2 or 3) = 3/6 = 1/2
They are independent events, so
P(E1 and E2) = P(E1) P(E2) = 1/3 1/2 = 1/6

7/3/2013VSR 60
Q.17. It is known that the probability of
obtaining zero defectives in a sample of 40 items
is 0.34 whilst the probability of obtaining 1
defective item in the sample is 0.46. What is the
probability of
(a) obtaining not more than 1 defective item in a
sample?
(b) obtaining more than 1 defective items in a
sample?
(a) Mutually exclusive, so
P(E1 or E2) = P(E1) + P(E2) = 0.34 + 0.46 = 0.8
(b) P(more than 1) = 1 − 0.8 = 0.2

7/3/2013VSR 61
Q.18. Find the mean and variance of the random
variable X, where probability distribution is given
by the following table:
X -2 -1 0 1 2 3
P(X) 0.10 0.20 0.30 0.20 0.15 0.05
Mean = XP(X)
- 2 0.10 + -1 0.20 + 0 0.30 + 1 0.20
+ 2 0.15 + 3 0.05. = 0.25.
Variance = X2P(X) – (mean)2
= - 22 0.10 + -12 0.20 + 02 0.30 + 12 0.20
+ 22 0.15 + 32 0.05 – (0.25)2 = 1.85 – 0.0625
= 1.7875

7/3/2013VSR 62
Q.19. The SD of a binomial distribution (q + p)16
is 2 , its mean is ?
SD = npq = 2 ., squaring npq = 2 .
But n = 16 ; we know q = 1 – p
16p(1 – p) = 4 ; 4p(1 – p) = 2
4p2 – 4p + 1 = 0
(2p – 1)2 = 0 ; p = ½
We know mean = np
= 16 ½
= 8.

7/3/2013VSR 63
Q.20. In a binomial distribution mean = 5 and
variance = 4 , then the number of trials is ?
Mean = np = 5. ; Variance = npq = 4 .
npq/np = 4/5 q = 4/5.
p = 1 – q = 1 – 4/5 = 1/5
Now, n p = 5
i.e., n 1/5 = 5
n = 25

7/3/2013VSR 64
Q.21.How many words can be formed from the
letter of the word “COMMITTEE”.
We have the formulae
!!!
!
321
nnn
n
Total no. of letters = 9 ; M = 2, T = 2, E = 2
3
321
)!2(
!9
!2!.2!.2
!9
!!!
!
nnn
n

7/3/2013VSR 65
Q.22. In a leap year , the probability of having
53 Friday or Saturday is ?
In a non-leap year there are 365 days, 52 complete
weeks and 1 day.
That can be
Monday, Tuesday, Wednesday, Thursday, Friday and
Saturday.
P(Friday) = 1/7 ; P( Saturday) = 1/7
The required probability = 1/7 + 1/7 = 2/7 .

7/3/2013VSR 66
Q.23. A card is chosen at random from a deck of
52 cards. It is then replaced and a second card is
chosen. What is the probability of choosing a
Jack and an Eight.
P(Jack) = 4/52 ; P(8) = 4/52
Multiplication rule
P(Jack and Eight) = 4/52 4/52
= 16/2704
= 1/169.

7/3/2013VSR 67
Q.23. Obtain binomial distribution, if n = 6,
p = 1/5
We have p = 1/5 and n = 6.
We know q = 1 – p
= 1 – 1/5 = 4/5 .
Binomial distribution = (q + p)n
=

7/3/2013VSR 68
Q.24. The probability that a bulb produced by a
factory will fuse after 150 days of use is 0.05.
What is the probability that out of 5 such bulbs
(i) None ; (ii) not more than one ; (iii) at least one
will fuse after 150 days of use.
Let X represent the number of bulbs that will fuse after
150 days of use in an experiment of 5 trials.
The trials are Bernoulli trials.
It is given that, p = 0.05 ; q = 1 – p = 1 – 0.05 = 0.95.
X has a binomial distribution with n = 5 and p = 0.05

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(i) P (none) = P(X = 0)
Using this formulae

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(ii) P (not more than one) = P(X ≤ 1)

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(iii) P (at least one) = P(X ≥ 1)

 The End of Probability
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