Unit 3: Probability
Introduction and Importance of
Probability
Probability is one of the important branches of Statistics
that is concerned with random phenomena.
Everyday
concept of “likelihood”, “predictability” and “certainty” are formalized by the
part of statistics called probability. Probability is a statistical method used
to study of the chance of occurrence and non-occurrence of different phenomena.
It is the synonym of the words most likely, chance, probably etc. Hence,
Probability defined as „uncertainty is numerically expressed.
The theory
of probability was developed in the middle of the seventeenth century. It was originated from the problem related to
gambling such as tossing a coin, throwing a die, drawing cards from a pack of
cards etc. Therefore, in the early days it was known as the theory of games and
chance. But nowadays, there is hardly any displine (field) where “probability” has not been used. The theory of probability
is used in almost all displine like physical science, natural science,
biological science, medical science, engineering, social sciences, business,
industry etc.
Definition:
Probability
is a numerical measure (with a value lying between 0 and 1) of the likelihood
or chance that a particular event will occur or not.
Probability
is simply a number lies between zero and one. That is;. It is generally denoted by 'P'.
Basic Terminologies in Probability
Random Experiment
An experiment
is called a random experiment if it performed a large number of times under
essentially homogeneous conditions; the result is not unique but may be any one
of the various possible outcomes. Tossing an unbiased coin and rolling a
uniform die are some examples of random experiment.
Sample Space
The set of
all possible outcomes of a random experiment is called sample space. Each
outcome is thus considered as a sample point in the sample space. It is usually
denoted by S. For example, in tossing an unbiased coin once, there are two
possible outcomes head (H) and tail (T). So, the sample space is given by S =
Trial and Event
Performing a
random experiment is called a trial and outcome or combination of outcomes
(i.e. results) of random experiment is called an event. For example; flipping
an unbiased coin is a trial and getting either head or tail is an event.
Exhaustive Cases
The total number of possible outcomes
of a random experiment is called the exhaustive cases for the experiment. In a
toss of single coin, we can get head (H) or tail (T). Hence, exhaustive number
of cases is 2
Similarly,
in rolling a six faced die, the exhaustive number of cases = 6
Favourable Cases or Events
The number
of outcomes of a random experiment which result in the happening of an event
are termed as the cases favourable to the event. In other words, the number of
outcomes which are in favour of happening of an event are called favourable
cases. For example; in drawing a card from a pack of 52 playing cards, the number
of cases favourable to drawing a queen is 4. Similarly , in a toss of two
coins, the number of cases favourable to the event „exactly one head „ is 2,
viz., HT , TH and for „two heads‟ is one viz., HH.
Mutually Exclusive Events
Two or more events are said to be mutually
exclusive if the happening of any one of them excludes the happening of all
others in the same experiments i.e. if two or more than two events cannot occur
simultaneously at the same time in the same trial. For example in a single tossing
of a coin, we may get either head or tail but not both. Thus, the events head
and tail in a single tossing of a coin are mutually exclusive. Similarly, in
the throw of a die, the six faces numbered 1, 2, 3, 4, 5 & 6 are mutually
exclusive. Thus, no two or more of them can happen simultaneously.
Independent Events
Events are
said to be independent if the occurrence of one event does not affect the
occurrence of the other events and vice versa. In tossing of a coin, the
occurrence of head in first tossing is independent of the occurrence of head in
the second tossing. Similarly, drawing of balls from an urn gives independent
events if the draws are made with replacement.
Dependent Events
Events are
said to be dependent if the occurrence of one event affects the occurrence of
the other events and vice versa. For example, if a pack of cards is playing
without replacemement, the occurrence of a king in first draw affects the
occurrence of other cards in the second draw.
Fundamental Principle of Counting
The counting
rules facilitate to calculate the all possible outcomes in the experiment.
The fundamental principle of counting
states as follows:
If there are
'k' separate parts of an experiment, and the first part can be done in n1
ways, second successive part in n2 ways ....... and kth successive
part in nk ways, then the total number of possible outcomes is given
by n1 × n2 × ... × nk.
Permutation
The literal
meaning of the word permutation is “Arrangement”. Therefore, permutation is the
arrangement of objects taken some or all at time in some order. If there are 'n' objects and they are to be
placed in any definite arrangement or order. The number of permutations of 'n'
different objects taken 'r' objectives at a time is denoted by npr
or P (n, r) and defined as
where, n! = n (n – 1) (n – 2)! r n
Combination
The literal
meaning of the word combination is “Selection”. Therefore, combination is the
selection of objects taken some or all at a time without specific order. Note
that the position of the object is meaningless in combination.
A
combination of 'n' different objects taken 'r' objects at a time, is denoted by
nCr or C (n, r) or and defined to be
nCr for r n
Approaches of Probability
The chance
of occurrence or non-occurrence of any event in a random experiment is called probability.
There are four approaches to define the probability.
Mathematical or classical or priori
approach
Statistical or empirical or relative
frequency approach
Subjective approach
Axiomatic approach
Mathematical or Classical or Priori,
approach of Probability
Let n be the
exhaustive, mutually exclusive and equally likely cases (or outcomes) out of
which 'm' are favourable cases to the
happening of an event A. Then the probability of happening (occurrence) of an
event A is given by
P (A) =
or, P (A) = . . . . . . . .(1)
P (A) + P ( ̅) = 1
P ( ̅)
= 1 – p(A)
Statistical or Empirical or Relative
Frequency Approach
This
approach of probability is based on statistical data. If an experiment is
performed repeatedly a large number of times under essentially homogeneous and
identical conditions, then the limiting value of the ratio of the number of times the event
occurs to the total number of trials of the experiment as the number trials
becomes indefinitely large, is called the probability of happening of the
event. It being assumed that the limit is finite and unique.
Suppose an
event A occurs m times in n repetitions of a random experiment. Then the ratio gives the relative
frequency of the event A. In the limiting case when n becomes sufficiently
large, then a number which is called the probability of A.
Symbolically,
P (A) =
Subjective Approach
The subjective probability approach is purely
individualistic in nature. Therefore, this approach of probability is
completely based on the personal beliefs, feelings, experience, judgement,
personal discretion of a person. Since, different persons may assign different
probabilities one cannot arrive at objective conclusions using probabilities
assigned by this subjective method. This method of assigning probability is
generally used by top level authorities on the basis of their discretion
Axiomatic Probability
The modern
theory of probability is based on the axiomatic approach introduced by the
Russian mathematician A.N. Kolmogorov in 1930. The axiomatic definition of
probability includes both classical and empirical definition of probability and
at the same time is the free from their drawbacks. It is also called elementary
properties of probability. It states the following facts:
Definition: Given a sample space of a random experiment, the
probability of occurrence of any event „E‟ is defined as a set function P( ) satisfying the following axioms:
Axiom 1: Axiom of non-negativity
P( ) is defined, is real and
non-negative) i.e. P (E) ≥ 0
Axiom 2: Axiom of certainty
The sum of
the probabilities is unity i.e. P , where S = Sample
space
Axiom 3: Axiom of additivity
If is any finite or
infinite sequence of disjoint events of sample space S, then
P or P
Note: When probability is not given
directly, then it can be categorised into two cases:
Case (i): When one item is selected at a time.
Case (ii): When more than one item is selected
at a time.
Examples related to
Case (i): When one item is selected
at a time.
Probability of getting an event
A, P (A) =
Where, n = Total number of cases m = Favourable number of cases Example 1
A card is
drawn at random from a well-shuffled pack of 52 cards. What is the probability
of getting?
(a)
a
red card,
(b)
a
black king Solution:
There are 52 cards in a pack. Total number of cases (n) = 52 (a) P (a red card) =?
Since, there are 26 red cards.
Favourable number of cases (m) = 26
Required probability of getting a red card is
P (a red card) =
(b) P
(a black king) =?
There are 2 black kings
Favourable number of cases (m) = 2
Required probability of getting a black king
is
P (a black
king) =
Example 2
Twenty balls
are numbered from 1 to 20. If one ball is drawn at random, what is the
probability that the ball drawn is multiple of 4 or 7?
Solution:
Total number of cases (n) = 20
Favourable
number of cases (m) = The number of cases which are multiple of 4 or 7
i.e.
{4, 7, 8, 12, 14, 16, and 20}
i.e. m
= 5+2 = 7
Required
probability that the ball drawn is multiple of 4 or 7 =
Examples related to
Case (ii): When more than one item is
selected at a time.
Probability of getting the event A,
P(A) =
Where, n = Total number of cases = Selection of r
out of total
m =
Favourable number of cases
= Selection of r out of
=
Example 3
A bag
contains 8 red, 4 white and 5 black coloured balls. Three balls are drawn
randomly from a bag. Find the probability that (i) all are red (ii) 2 is red
and 1 white (iii) 2 are red and 1 other (iv) all colour balls.
Solution:
Exhaustive
(total) number of cases (n) = number of cases of selection of 3 balls out of 17
ball
= 17C3 = 680
(i)Favourable
number of cases of drawing all 3 are red balls (m) = 8C3
= 56
P (all are red) =
(ii)
Favourable cases for 2 red and 1 white (m) = 8C2 × 4C1
= 28× 4=
112
P (2 is red and 1 white)
=
(iii)Favourable
cases for 2 are red out of three drawn balls i.e. 2 are red and 1 other
(m) = 8C2 × 9C1
= 28×9 = 252
P (2 are red and 1 other) =
(iv)
Favourable number of cases for all coloured balls
(m) = 8C1 × 4C1
× 5C1 = 8 × 4 × 5 = 160
P (all colour balls) =
PU 2019(Spring), 2016(Spring)
2(a) From a
group of 3 Nepalese, 4 Indians and 5 Americans, a sub –committee of 4 persons
are selected at randomly. Find the probability that the sub-committee will
consist (i) 3 Nepalese and 1 Indian (ii) 1 Nepalese , 1 Indian and 2 American
(iii) 4 American.
Solution:
Total
persons = 3 Nepalese + 4 Indians + 5 Americans =12
Total number
of cases (n) = Selection of 4 persons out 12
= A sub-committee of 4 persons out of 12
= = 495
(i) P(a
sub-committee will consist 3 Nepalese and 1 Indian)= ?
Favourable
number of cases (m) = Number of cases of a sub committee will consist 3
Nepalese and 1 Indian
= = 1 = 4
P (a
sub-committee will consist 3 Nepalese and 1 Indian)=
(ii) P (1
Nepalese, 1 Indian and 2 American) =?
Favourable
number of cases (m) =
= 3 120
P (1
Nepalese, 1 Indian and 2 American) =
(iii) P (a
sub-committee will consist all 4 American) =?
Favourable
number of cases (m) =
= 1
P (a
sub-committee will consist all 4 American)
PU2015 (Fall)
In a company, there are 4 civil engineers, 5 IT officers and
6 finance officers. A committee of 3 members has to be formed from random
selection. What is the probability that committee consists of (i) All civil engineers
(ii) One civil engineer or one civil
engineer & 2 others
(iii)At least one civil engineer Solution:
Total members = 4 + 5 + 6 = 15
Total number
of cases (n) = Number of cases of selection of 3 members out of 15 members
= = 455
(i) P (All
civil engineers) =?
Favourable
number of cases (m)= selection of 3 civil engineers out of 4 civil engineers = = 4
P (All civil engineers) =
(ii) P (One
civil engineer) =?
Favourable
number of cases (m)= selection of 1 civil engineer out of 4 civil engineers and
2 members out 11 others = = 4
P (One civil
engineer)
(iii)
P
( At least one civil engineer ) = ?
Favourable
number of cases (m) = Number of cases of no civil engineer = Selection of 3
members out of 11 non civil engineers = = 165
P( At least
one civil engineer ) = 1 – P(No civil engineer )
= 0.637
Additional
(iv)
P
(each of one kind) =?
or
P (1 civil
engineer, 1 IT officer & 1 finance) =?
Favourable
number of cases (m) = = 4
P (each of
one kind) =
PU 2013 (Fall)
In a
organization, there are 2 civil engineers, 4 IT officers and 2 accounts. A
committee of four members has to be formed. What is the probability that a committee
contains
(i)
one
civil engineer and 3 IT officers ?
(ii)
no
civil engineer ?
(iii) At least one IT officers?
Solution:
Total
members = 2+ 4 + 2 = 8
Since, a
committee of four members has to be formed
Total number
of cases (n) = Number of cases of selection of 4 members out of 8 members
=
= 70
(i)
P
(one civil engineer and 3 IT officers) =?
Favourable number of cases (m) = selection of 1 civil
engineer out of 2 civil engineers and 3 IT officers out of 4 IT officers = = 24= 8 P(one civil
engineer and 3 IT officers) =
(ii)
P(no civil engineer ) = ?
Favourable
number of cases (m) = selection of no civil engineer = = 1
P(no civil
engineer ) =
(iii)
P(At
least one IT officers) = ?
Favourable
number of cases (m) = Number of cases of no IT officers = Selection of 4
members out of 4 non IT officers = = 1
P( At least
one IT officers ) = 1 – P(No IT officers
)
=
0.985
Example 4
Five men in
a group of 20 are graduates. If 3 are chosen out of 20 at random, what is the
probability that
a) all are
graduates b) none of them is graduate
c) at least
one of them being graduate
Solution: Here,
Total number of men = 20
Number of graduates = 5
Number of non-graduates = 20 – 5 = 15
If three are chosen at random
Total number
of possible outcomes (n) = 20C3 = 1140
(a)
P
(all are graduates) =?
Favourable
number of cases (m) = 5C3 = 10
P (all are graduate)
(b)
P
(none of them are graduate) =?
Favourable
number of cases for none of them are graduate (m) = 15C3 = 455
P (none of them are graduate) =
(c)
P
(at least one of them being graduate)
= 1 – P (none them are graduate)
= 0.60
Laws (or rules) of
Probability)
Following
are the laws of probability:
Additive law of probability
Multiplicative law of probability
Additive Law of Probability
(Addition Theorem of Probability)
(At least one, one of them, or)
“Union”
Case I: when events are not mutually
exclusive
If A and B are not mutually exclusive events, then the
probability of occurrence of at least one of them is given by
P (A or B) = P (AB) = P (A) + P (B) – P (AB)
The probability of occurrence of at
least one of them can also be written as
P (AB) = 1 – = 1 –
Demorgan’s
law
&
If A, B and C are three not mutually exclusive events then
the probability of occurrence of at least any one of them is given by
P (A or B or C) = P (ABC) = P (A) + P (B) + P(C) – P (AB) – P (BC) – P (CA) + P (ABC)
Additive Law of Probability (Addition
Theorem of Probability)
Case II: when events are mutually
exclusive
Let A and B
be two mutually exclusive events. Then the probability of the occurrence of
either event A or event B is the sum of their individual probabilities. Hence,
the probability of occurrence of either event A or event B is given by
P (A or B) =
P (AB) = P (A) + P (B)
If A, B and
C are three mutually exclusive events, then the probability of occurrence of
either events A or B or C is given by
P (A or B or
C) = P (ABC) = P (A) + P (B) + P(C)
Example 5
The probability that a company execute will travel by plane
is and that he will
travel by train is. Find the probability of his travelling by plane or train.
Solution: Let A and B be the events that a
company execute will travel by plane and train respectively.
,
P (AB) = P (A) + P (B)
Multiplicative Law of
Probability (or Multiplication Theorem of Probability)
(And, both, as well as, all, three,
two) “Intersection”
Case (i): When events are independent
(a)
If there is no change in total number
(i.e. with replacement)
(b)
If dependent condition is not stated
i.e. conditional case is not given
Let A and B
are two independent events, then the probability of occurrence of both the
events is the product of their individual probabilities.
Symbolically,
P (A and B) = P (AB) = P (A) P (B)
If A, B and
C are three independent events, then
P (ABC) = P (A) .P (B). P(C)
Multiplication Theorem of Probability
(And, both, as well, two, three, all,
)
Case (ii): When the events are
dependent:
(a)
If there is change in total numbers
(i.e. without replacement or not replaced , transferred)
(b)
If dependent condition is stated i.e.
conditional case is given
If A and B
are two dependent events, then the probability of simultaneous happening of two
events A and B is given by
Similarly,
Where, is the conditional
Probability of the occurrence (happening) of event B given that (if) event A
has already occurred (happened).
& is the conditional
Probability of occurrence (happening) of event A given that event B has already
occurred (happened).
If A, B and C are three dependent
events, then
Where, is the conditional
probability of occurrence (happening) of event C given that (if) both events A
and B have already occurred (happened).
Conditional Probability
Definition: Conditional probability is the
probability that an event will occur given that another event has already
occurred. If A and B are two dependent events, then the conditional probability
of event A given that (if) event B has already occurred is given by,
P (A/B) =, provided P (B) 0
Similarly, the conditional probability of event B given that
(if) event A has already occurred is given by P (B/A) = , provided P (A) 0
PU 2014(Fall), 2017(Spring)
A problem in statistics
is given to three students A, B, and C whose chances of solving it are and respectively. If they
independently solve it. Find the probability that
(i)
All
can solve the problem.
(ii)
None
can solve the problem.
(iii)Problem can be solved. (or at least
one of them will solve the problem) (iv) Exactly one can solve the problem
OR
Only one of
them can solve the problem.
(v) A solves
it but B and C cannot.
Solution:
Probability
that A solves a problem i.e. P (A) =
and P (
Probability
that B solves a problem i.e. P (B) =
and P (
Probability
that C solves a problem i.e. P(C) =
and P (
(i)
P(All
can solve the problem) = P (ABC)
= P (A) P (B)P(C)
= 0.25
(ii)
P(None
can solve the problem) = P ()
= P ( ̅) P ( ̅) P ( ̅)
(iii)
P(problem
can be solved)=
=
P (A)+P (B)+P(C)–P (AB)–P (BC)–P(CA)+P ABC)
= P (A) + P (B) + P(C) – P (A) P (B) – P (B)
P(C) – P(C) P (A)+ P (A) P (B) P(C)
= 0.95 are independent events.
Alternatively,
P(problem can
be solved) =
= 1-
P (
= 1 - P ()
=
1 – P ( ̅) P ( ̅) P ( ̅)
(iv)
P
(Exactly one can solve the problem)
= P (A) + P (B) + P(C)
= P (A) P ( ̅) P ( ̅) + P ( ̅) P (B) P ( ̅) + P ( ̅) P ( ̅) P(C)
(v)
P
(A can solve it but B and C cannot)
= P (A)
= 0.041
Example 6
A, B and C
will pass a certain examination in the proportion 2:4:6. What is the
probability that
(i)
at
least two of them will pass the examination?
(ii)
at
least one of the them will pass the examination?
(iii)
all
will pass the examination?
(iv)
none
will pass the examination?
(v)
B
will pass but not A and C?
Solution:
Since, the
pass proportion of A, B and C in a certain examination are in the rato 2:4:6.
Total ratio
= 2 + 4 + 6 = 12. Then
P
P
P
(i)
P
(at least two of them will pass the examination)
= P
= P P P P + P
( Events are mutually
exclusive and independent)
(ii)
P
(at least one of the them will pass the examination) =
= 1– P
= 1 – P ( ̅) P ( ̅) P ( ̅)
= 1– 0.2777
= 0.7222
(iii)
P
= P
= P
(iv)
P
= P
= P P P ( ̅)
= 0.2777
(v)
P
P (
= P ( ̅) P ( ̅)
Example 7
A construction company is bidding for two contracts A and B.
The probability that the company will get contract A is , the probability that the company will get contract B is and the probability
that the company will get both the contracts is. What is the probability that the company will get contract
A or B?
Solution:
We have,
The
probability that the company will get contract A or B is given by
P (AB) = P (A) + P (B) – P (AB)
(i)
P
( at least one machine will work without failure )
=
= 1– P
= 1 – P ( ̅) P ( ̅) P ( ̅)
= 1 – = 1– 0.28 = 0.72
(ii)
P
(at least two machine will work without failure)
= P
= P P P P + P
=
(
Events are mutually
exclusive and independent)
(iii)
p
(all machines will work without failure)
= P
= P
=
= 0.03
Example 8
Two brothers: Mr. X and Mr. Y appear in an interview for
getting the scholarship. The scholarship can be provided for two persons. The
probability of getting scholarship by Mr. X is and getting by Mr. Y
is . What is the
probability that,
(a)
both
of them will get scholarship.
(b)
Only
one of them will get scholarship.
(c)None of them will get scholarship.
Solution:
Let P (A) =
Probability that Mr. X will get scholarship
= Probability that Mr. X will not get scholarship P (B) = Probability that Mr. Y will get
scholarship
= Probability that Mr. X will not get scholarship
Given that,
P (A) =
P (B) =
(a) Probability that both of them will get
scholarship,
P (AB) = P (A) P (B)
Events are independent
(b)Probability that only one of them will get
scholarship
= P (AB)
= P (A) + P (B)
= P (A). P ( ̅) + P ( ̅). P (B)
(c )
Probability that non of them will get scholarship
P () = P ( ̅) P ( ̅)
(b)Probability
that only one of them will get scholarship
= P (AB)
= P (A) + P (B)
= P (A). P ( ̅) + P ( ̅). P (B)
(c )
Probability that none of them will get scholarship
P () = P ( ̅) P ( ̅)
(i)Required
probability that both will be alive 25 years
P (AB) = P (A) × P (B) = 0.3 × 0.4 = 0.12
(ii) Probability that only the man will be
alive
P (A) = P (A) × P ( ̅) = 0.3 × 0.6 = 0.18.
(iii)
Probability
that the only woman will be alive
= P (B) = P (B) × P) = 0.4 × 0.7 = 0.28
(iv)
Probability
that none will be alive
P ( ) = P) P ( ̅) = 0.7
(v)
Required
probability that at least one of them will be alive
P(
= 1 - P ( )
= 1-
=
= 0.58
(vi)
Probability
that only one of them will be alive
= P (A or B) = P (A
= P (A) × P ( ̅) + P ( ̅) × P (B)
= 0.3 × 0.6 + 0.7 × 0.4
= 0.46
Example 9
If the probability of machines and working without
failure are 0.2, 0.3 and 0.5 respectively. Find the probability that
(i)
at
least one machine will work without failure.
(ii)
at
least two machine will work without failure.
(iii)
all
machines will work without failure.
Solution:
Let A, B and C denote the events that machines and working without
failure respectively.
Then
P, P
P, P
P, P
(i)
P(at
least one machine will work without failure )
= 1– P
= 1- P
= 1 –P()
= 1 – P ( ̅) P ( ̅) P ( ̅)
= 1 –
= 1– 0.28 = 0.72
(ii)
P
(at least two machine will work without failure)
= P
= P PP+ P
=
(
Events are mutually
exclusive and independent)
(iii)
P
(all machines will work without failure) =
P
= P
=
= 0.03 PU
2018 (Fall)
Define axiomatic approach of probability. In
certain locality of town, a survey of 600 women about the fuels for cooking
revealed that 230 use kerosene oil, 175 women use electricity and 40 women use
both. Find the probability that a women selected use;
(i)
At
least one of the fuel.
(ii)
Neither
of the fuel.
(iii)
Kerosene
oil only.
(iv) Electricity only (v) Exactly one.
Solution:
Let K and E be the events that women in certain locality use kerosene oil and
electricity for cooking respectively.
Total number
of cases (n) = 600
P(K) = = 0.383, P( ̅) = 1-0.383 = 0.617
P( E) = = 0.292, P(̅ ) = 1-0.292 = 0.708
P(K
(i)
P(At
least one of the fuel ) = P(K
P(K P(K) + P( E) - P(K
= 0.383 + 0.292 – 0.067
= 0.608
(ii)
P
(Neither of the fuel) = P (
P ( ) = 1 - P(K
= 1 – 0.608
= 0.392
(iii)
P
(Kerosene oil only) = (K) =?
(K) = P (K) - P (K
= 0.383 - 0.067
= 0.31
(iv)
P(Electricity
only) = (E) = ?
(E) = P(E) - P(K
= 0.292 - 0.067
= 0.225
(V)
P(Exactly one ) = (K) + (E)
= 0.31 +0.225
= 0.535
Example 10 The probability that a manufacturer
will produce „brand X‟ product is 0.13, the probability that he will produce
„brand Y‟ product is 0.28 and the probability that he will produce both brand
is 0.06. What is the probability that the manufacturer who has produced „brand
Y‟ will also have produced „brand X‟?
Solution: Let X and Y be the events that a
manufacturer will produce brand X and brand Y respectively.
P(X) = 0.13,
P(Y) = 0.28, P (,
The probability that
the manufacturer who has produced „brand Y‟ will also have produced „brand X‟
is
Example 11
In a certain
school, 20% students failed in English, 15% students failed in Mathematics and
10% students failed in both English and Mathematics. A student is selected at
random. If he failed in English, what is the probability that he also failed in
Mathematics.
Solution:
Let E and M
be the events that denote the students failed in English and failed in
Mathematics respectively. Then
P
(E) = 20% = 0.2
P
(M) = 15% = 0.15
P
(EM) = 10% = 0.10
If one
student is selected at random, the Probability that if he failed in English
then he also failed in Mathematics is given by
P (M/E) =
PU 2016 (Fall)
Q.No.2 (a)
An oil
exploration company currently has two active projects, one in Asia and the
other in Europe. Let A be the event that the Asian project is successful and B
be the event that the European project is successful. Suppose that A and B are
independent events with
P (A) = 0.4 and P (B) = 0.7
(i)If the
Asian project is not successful, what is the probability that the European
project is also not successful? Explain your reasoning.
(ii)
What
is the probability that at least one of the two projects will be successful?
(iii)Given that at least one of the two
projects is successful, what is the probability that only the Asian project is
successful?
Solution:
P (A) =
0.4, P ( ̅ ) = 1 - P (A) = 1- 0.4 = 0.6
P (B) =
0.7, P ( ̅ ) = 1 - P (B) = 1- 0.7 = 0.3
(i) If the
Asian project is not successful, the probability that the European project is
also not successful is given by
P(
A and B are
independent events
= P (AB) = P (A) P (B)
P ( ̅ / ̅ ) = 0.3
P ( ̅ / ̅ ) =
This result concludes that if A and B are two independent
events then conditional probability of
event ̅ given ̅ is equal to
unconditional probability of event i.e. P( ̅ / ̅ ) =P( ̅ )
(ii) The probability that at least one of the two
projects will be successful is given by
P (A P(A) + P( B) - P(A
= P(A) + P( B) - P( A) P(
B)
= 0.4 + 0.7 – 0.4
= 0.82 A and B are
independent events
P (AB) = P (A) P (B)
(iii) Given
that at least one of the two projects is successful , the probability that
only the Asian project is successful is
given by
P (A/A
PU 2015 (Spring)
Q.N0. 2(b)
The HAL
Corporation wishes to improve the resistance of its personal computer to disk
driver and key board failures. At present, the design of computer is such that
disk –drive failures occurs only one third as often as keyboard failure. The
probability of simultaneous disk-drive and key board failure is 0.05.
(i)
If
the computer is 80% resistant to disk and / or keyboard failure, how must the
disk-driver failure probability be?
(ii)
If
the keyboard is improved so that it fails only twice as often as the
disk-driver (and the simultaneous failure probability is still 0.05), will the
disk drive failure probability from part (i) yield a resistance to disk-drive
and/or keyboard failure higher or lower
than 90%
Solution:
Let D and K
be the events that the computer has disk-driver failure and key-board failure
respectively.
By given
condition
The disk
–drive failures occurs only one third as often as keyboard failure
P(D) = P(K)
P (K) = 3 P(D)
& The
probability of simultaneous disk-drive and key board failure is 0.05.
i.e. P(D
(a) Since,
the computer is 80% resistant to disk and / or keyboard failure
P ( ) = 0.80
P(D
= 1 – 0.80
= 0.20
P (D) =?
Now,
P (D P (D) + P (K) - P (D
or, 0.20 = P(D) + 3 P(D)- 0.05 or,
0.25 = 4 P(D)
or,
P(D) = 0.0625
The
probability of disk driver failure is 0.0625
(ii) P(D) =
0.0625
P(K) = 2
P(D)= 2 = 0.125
P(D
P(D P(D) + P( K) - P(D
= 0.0625 + 0.125 -0.05
= 0.1375
P(D=0.1375
Hence, the
probability of resistance to disk drive and/or keyboard failure is
P( ) = 1- P(D
= 1 -0.1375
= 0.8625
= 86.25% which is lower than
90%
Applications of Conditional
Probability
Bayes's Theorem (Rule for the Inverse
Probability)
Statement: Let be n mutually exclusive (disjoint) and exhaustive events of
sample space S with probabilities P for i= 1, 2, 3,. . .
.,n. For any arbitrary event A which is a subset of the given sample space S
such that, then the probability that it was preceded by the particular
event is given by
, i=1,2,3, . . . ,n
Proof:
Let S be the sample space and it can
be partioned (divided) into various mutually exclusive (disjoint) subsets with P for i= 1, 2, 3,. . .
.,n . Let A be any arbitrary event which is subset of sample space S and having
some intersecting part with mutually exclusive and exhaustive events .
Then from
figure
A =
P(A)=P
Since, , , . . . . . .. , are mutually exclusive
(disjoint) events. Then, by addition theorem of probability
P(A)= P+ P+P
P(A) = . . . . . . .(i)
For any
particular event , the conditional probability P is given by
P = P (A) P
or, P
, i= 1, 2, 3,. . .
.,n [From (i)]
Where,
P (E1), P(E2) ...., P(En)
which are already known before conducting an experiment is termed as priori
probabilities.
The
conditional probabilities P (E1/A), P (E2/A),. . . . .
.,P (En/A) which are computed after conducting the experiment are
termed as posteriori probabilities
Example 12
Assuming that a factory
has two machines and Past record showed
that machine produces 30% of the items of output and machine produces 70% of the items. Further, 5% and 1% of the items
produced by machine and respectively were defective. If a defective item is selected
at random what is the probability that it was produced by machine ?
Solution: Let and be the events of
producing items of output by machines and respectively.
= 0.30
= 0.70
Let A be the
event of producing defective item. Then,
= 0.05
= 0.01
= ?
By Bayes‟s
theorem
The required
probability that the defective item was produced by machine is
Example 13
A company
buys tyres from two suppliers, 1 and 2. Supplier 1 has a record of delivering
tyres containing 10% of the defectives, whereas supplier 2 has a defective rate
of only 5%. Suppose 40% of the current supply came from supplier 1. If a tyre
is taken from this supply and observed to defective, find the probability that
it is from a supplier 1.
Solution:
Let and be the events of
delivering tyres from suppliers 1 and 2 respectively.
= 0.40
=1-0.40 = 0.60
Let A be the
event of delivering defective tyres. Then,
= 0.10 = 0.05
= ?
By Bayes‟s
theorem
0.571429
The required
probability that the defective tyre is from a supplier 1 is 0.571429
PU2019 (Spring)
2. b) In a
certain factory machines I, II, III are all producing springs of the same
length of their production machines I, II, III produces 2%, 1% and 3% defective
springs respectively of the total production of springs in the factory, I
machine produces 35%, machines II produces 25% and machine III produces 40%. If
one spring is selected at random from total springs in a day, find:
i.
the probability that it is defective ii. the conditional probability that it
was produced by machine III.
Solution:
Let and be the events of
producing springs of the same length by machines I, II and III respectively.
Then
= 0.35
= 0.25
= 0.40
Let A be the
event of producing defective springs by three machines. Then,
= 0.02 = 0.01
= 0.03
(i)
Probability
that defective springs produced by
machines is
= 0.35
= 0.0215
(ii)
By
Bayes‟s theorem
The probability that defective spring was
produced by machine III is given by
PU: 2018(Spring)
2. a) A firm has 80% of its service calls
made by a contractor and 10% of these calls result in customer complaints. The
other 20% of the service calls are made by their own employees, and these calls
have 5% complaint rate. Find the probability of a complaint. Also, using Bayes
theorem to find the probability that a complaint was from a customer whose
service was provided by the contractor.
Solution: Let and be the events that a
firm‟s service calls made by a contractor and by their own employees
respectively.
= 0.80
= 0.20
Let A be the
event that a complaint from customer. Then
= 0.10
= 0.05
The
probability of a complaint is given by
= 0.80
= 0.09
By Baye‟s
theorem
The
probability that a complaint was from a Customer whose service was provided by
the contractor is given by
PU: 2017 (Fall), 2013(Spring)
2. a) State
Bayes theorem. In a class of 75 students, 15 students were considered to be
very intelligent, 45 as medium and rest below the average. The probability that
a very intelligent student fails in examination is 0.005, the medium students
failing has probability 0.05 and corresponding probability of a below average
is 0.15. If a student is known to have passed the examination, what is the
probability that he is below the average?
Solution: Statement: Let be n mutually exclusive (disjoint) and exhaustive events of
sample space S with probabilities P for i= 1, 2, 3,. . .
.,n. For any arbitrary event A which is a subset of the given sample space S
such that, then the probability that it is was preceded by the
particular event is given by
, i=1,2,3, . . . ,n
Let and be the events that
denote for very intelligent, medium and below the average students
respectively.
Let A be the
event that a student have passed the examination. Then,
= 1- 0.005 = 0.995
= 1- 0.05 = 0.95
=1- 0.15 = 0.85
If a student
is known to have passed the examination, the probability that he is below the
average is given by
Where,
= 0.20
= 0.939
PU 2018 (Fall)
2 (a) The
contents of urns I, II and III are as follows: 1 white, 2 black and 3 red balls
; 2 white balls, 1 black and 1 red balls and 4 white , 5 black and 3 red balls.
One urn is chosen at random and two balls are drawn. They happen to be white
and red. What is the probability that they come from urns I, II or III?
Solution:
Let and be the events of
selection of urns I, II and III respectively.
Let A be the
event of drawing a white ball and a red ball. Then
By Bayes‟s
theorem
The required probabilities that white and red
balls come from urns I, II or III
respectively are given by
Where,
= 0.238
PU 2016 (Spring)
2(c) A
manufacturing firm produces steel pipes
in three plants with daily production volumes of 500, 1000 and 2000 units
respectively. According to the past experience, it is known that the fraction
of defective outputs produced by the three plants are respectively 0.005, 0.008
and 0.010. If pipe is selected from a day‟s total production and found to be
defective. From which plant the defective pipe is expected to have been
produced?
Solution: Let and be the events
producing steel pipes by three plants I, II and III respectively. Then
Let A be the
event that the defective steel pipes produced by three plants. Then,
= 0.005
= 0.008
= 0.010
= ?= ?, = ?
By Bayes‟s theorem, the required probabilities that the
defective steel pipes are produced from plants
I, II and
III respectively are given by
The
probability that the plants produce defective pipe is
= 0.00871
>
>
therefore
plant III produced more defective
pipes than plants
I , II.
Example 14
The chances
of X, Y and Z becoming manager will be of a certain company are 4:2:3. The
probabilities that bonus scheme will be introduced if X, Y and Z become manager
are 0.3, 0.5 and 0.8 respectively.
(i) What is the probability that bonus
scheme will be introduced?
(ii)If the bonus scheme has been
introduced, what is the probability that X is appointed as the manager? Solution:
Let and be the events of X, Y
and Z becoming managers respectively. Then Total ratio = 4+2+3 = 9
Let A be the
event of introducing bonus scheme. Then,
= 0.3
0.5
0.8
= ?= ?, = ?
The
probability that the bonus scheme will be introduced is given by
By Bayes‟s theorem,
If bonus
scheme has been introduced, then the probability that X is appointed as the
manager is given by
PU: 2013 (Fall)
2. b) A
company has two plants to manufacture the screws. Out of 10000 screws, plant I
manufactures 80% of the screws and plant II manufactures 20%. At plant I, 85%
screws are rated standard quality. At plant II, only 65% screws are rated
standard quality.
i.
One
screw is selected at random from the whole consignment and was found standard
quality, What is the probability that the selected screw was manufactured by
plant I?
ii.
What
is the probability that the selected screw being defective?
Solution:
Let be the events of
manufacturing screws by plant I and plant II respectively.
=0.80
= 0.20
Let A be the
event that screws are rated standard quality.
Then,
= 0.85
0.65
(i) If screws are rated standard quality,
the probability that the selected screw was manufactured by plant I is given
by
=
Where,
= 0.80
= 0.81
(ii)
The
probability that the selected screw being defective
= 1- 0.839
= 0.161
PU 2015
(Spring)
2. a) Two
production plants A and B, make wire cables that are sent to a common
distributor. 40% of the cables sent to the distributor come from plant A, and
the remaining 60% from plant B. Among the cables produced at Plant A, 95% meet
the strength specifications; among the cables produced at plant B, 98% meet the
strength specification.
i.
The
distributor selects one cable at random from among all cables in stock. If the
cable selected is found to meet the strength specifications, what is the
probability that the cable was produced at plant A?
ii.
Let
A be the event that a cable is manufactured at plant A; similarly, B be the
event that a cable is manufactured at plant B. Let Y be the event that a cable
meets the strength specifications; and let N be the event that a cable does not
meet the strength specifications. (i) Are events A and Y independent? Give
reasons.
Solution:
Let be the events of
producing wire cables by plant A and plant B respectively.
=0.40
= 0.60
Let A be the event that the cables produced meet the strength
specifications.
= 0.95
= 0.98
(i) If the cable selected is found to meet the strength
specifications, the probability that the cable was produced at plant A is given
by
=
P(A) =
=
=
(ii)
A
=
the event that a cable is manufactured at plant A
B
=
the event that a cable is manufactured at plant B
Y = the event that a cable meets the strength
specifications
N = the
event that a cable does not meet the strength specifications.
P (A) = 0.40
P(Y) =
0.968
P (A) P(Y) =
0.400.968 = 0.3872
P(Y P(A)P(Y/A)
= 0.40
= 0.38
Since, P(Y P(A) P(Y)
Hence, events A and Y are not independent
PU 2016
(Fall)
2. b) Three
road construction firms, X, Y and Z, bid for a certain contract. From past
experience, it is estimated that the probability that X will be awarded the
contract is 0.40, while for Y and Z the probabilities are 0.35 and 0.25. If X
does receive the contract, the probability that the work will be satisfactorily
completed on time is 0.75. For Y and Z these probabilities are 0.80 and 0.70.
i.
What
is the probability that the work will be completed satisfactorily?
ii.
It
turns out that the work was done satisfactorily, what is the probability that Y
was awarded the contract?
Solution:
Let and be the events of firms
X, Y and Z awarded the contract respectively. Then
= 0.40
= 0.35
= 0.25
Let A be the
event that the work will be completed satisfactorily.
= 0.75
0.80
0.70
= ?= ?, = ?
(i)
The
probability that the work will be completed satisfactorily.
= 0.40
= 0.755
By Bayes‟s theorem
(ii)
If
the work was done satisfactorily, the probability that Y was awarded the
contract is given by
= 0.370