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Probability and Statistics
Chapter 1: #Probability
Chapter 2: #Conditional Probability
Chapter 3: #Baye's theorem
Questions
1. What is the probability that at least two out
of 
people have the same birthday? Assume 365 days
in a year and that all days are equally likely.
Solution: Since
the birthday of any person can fall on any one of the 365 days, the exhaustive
number of cases for birthday of 
persons is 365.
If the birthday of all n persons falls on different days, then the number of the favorable cases is:
Because
in this case the birthday of the first person can fall on any one of 365 days,
the birthday of the second person can fall on any one of the remaining 364
days, and so on. Hence, the probability 
that birthdays of all then n persons are
different is given by
Hence the required probability that at least two persons have same birthday is:
2. (a) Twelve balls are distributed at random among three boxes. What is the probability that the first box will contain 3 balls?
(b)
If n biscuits be distributed among N persons, find the chance that a particular
person receives 
biscuits.
Solution (a) since each ball can
go to any of their boxes, there are 3ways in which a ball can go to any one of
the three boxes. Hence there are 
ways in which 12 balls can be placed in the
three boxes.
Number
of ways in which 3 balls out of 12 can go the first box is ¹²C₃. Now the
remaining 9 balls are to be placed in remaining 2 boxes and this can be done in
ways. Hence the total number of favorable
cases
(b)
Take any one biscuit. This can be given to any one of the N beggars so that
there are N ways of distributing any one biscuit. Hence the total number of
ways in which n biscuits can be distributed at random among N beggars = N.N…(n
times) = 
.
‘r’
biscuits can be given to any particular beggar in 
ways. Now we are left with (n - r) biscuits
which are to be distributed among the remaining (N - 1) beggars and this can be
done in 
ways.
Number
of favorable cases 
Hence
required probability 
3. A and B throw with three dice; if A throws 14, find B’s chance of throwing a higher number.
Solution: To throw higher number than A, B must throw either 15 or 16 or 17 or 18.
Now
a throw amounting to 18 must made up of (6, 6, 6) which can occur in one way;
17 can be made up of (6, 6, 5), which can occur in 
ways;
16
may be made up to (6, 6, 4) and (6, 5, 5) each of which arrangement can occur
in ways;
15
can be made up of (6, 4, 5) or (6, 3, 6) or (5, 5, 5) which can occur in 
,
3 and 1 respectively.
The
number of favorable cases 
In
a random throw of 3 dice, the exhaustive number of cases 
Hence
the required probability 
.
4.
The sum of two non-negative quantities is equal to 2n. Find the chance that
their product is less than 
times their greatest product.
Solution:
Let 
and 
,
be the given quantitates so that 
We
know that the product of two positive quantities whose sum is constant (fixed)
is greatest when the quantities are equal. Thus the product of 
and 
is
maximum whne 
.
Now,
P
[form (*)]
Here,
the required 
5. If 
tickets numbered 
are placed in a bag and three are drawn out,
show that the chance that the sum of the numbers on them is equal to is 
Solution: The total number
of ways of drawing 3 tickets out of 
is given by:
Favorable cases for obtaining a sum of on the three drawn tickets are given below:
cases
cases
cases
cases
.
.
.
Total number of favourable cases
…..(*)
The expression in each bracket of 
is the sum of terms of an Arithmetic Progression (A.P.)
series.
Total number of favourable cases
Hence, the required probability 
.
6. An urn contains 
tickets numbered 
and another contains 
tickets numbered 
If one of the two urns is chosen at random and
a ticket is drawn at random from the chosen urn, find the probabilities that
the ticket drawn bears the number
Solution: (i) Required event can happen in the
following mutually exclusive ways:
(I) First urn is chosen and then a ticket is drawn.
(II) Second urn is chosen and then a ticket is drawn.
Since the probability of choosing any urn is 
,
required probability ' 
' is given by:
(ii) Required probability 
( 
In the 2nd urn there is no ticke with number
3.)
(iii) Required probability 
.
7. A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn there is at least one ball of each colour.
Solution:
The required event 
that 'in a draw of 4 balls from the box at
random there is at least one ball of each colour', can materialise in the
following mutually disjoint ways:
(i) 1 Red, 1 White, 2 Black balls; (ii) 2 Red, 1 White, 1 Black balls; (iii) 1
Red, 2 White, 1 Black balls.
Hence by addition theorem of probability, the required probability is given by:
8. A problem in Statistics is given to three students 
and 
whose chances of solving it are 
and 
respectively.
What is the probability that the problem will be solved if all of them try independently?
Solution: Let 
denote the events that the problem is solved
by the students respectively. Then
The problem will be solved if at least one of them
solves the problem. Thus we have to calculate the probability of occurrence of
at least one of the three events ,
i.e., 
.
are mutually
independent 
and 
are mutually
independent. 
9. A manager has two assistants and he bases his decision on information supplied independently by each one of them. The probability that he makes a mistake in his thinking is 0.005. The probability that an assistant gives wrong information is 0.3 . Assuming that the mistakes made by the manager are independent of the information given by the assistants, find the probability that he reaches a wrong decision.
Solution: Let us define the
following events:
A : The manager makes a mistake in his thinking.
: The 1st assistant gives him wrong
information.
C : The 2nd assistant gives him wrong information.
In usual notations, we are given :
Assuming that the mistakes made by the manager are
independent of the information supplied independently by each of the two
assistants, we conclude that 
and 
,
and consequently 
and are mutually independent.
[Manager reaches a wrong decision]
Manager reaches a correct decision 
;
because manager will reach a correct decision if he does not make a mistake in
his thinking (i.e., 
happens) and both the assistants supply him
correct information (i.e., 
happens).
[Since the events and are independent.]
10. The odds that a book on Statistics will be favorably reviewed by 3 independent critics are 3 to 2, 4 to 3 and 2 to 3 respectively. What is the probability that of the three reviews:
(i) All will be favorable,
(ii) Majority of the reviews will be favorable,
(iii) Exactly one review will be favorable,
(iv) Exactly two reviews will be favorable, and
(v) At least one of the reviews will be favorable.
Solution: Let and denote respectively the events that the book
is favorably reviewed by first, second and third critic respectively. Then we
are given:
(i) The probability that all the three reviews will be favorable is:
(
and are mutually independent events.)
(ii) The event that majority, i.e., at least 2 reviews are favourable can
materialise in following mutually exclusive ways:
(a) 
happens,
(b) 
happens,
(c) 
happens, and (d) 
happens.
Hence, the required probability is:
(iii) Arguing as in case (ii), the probability that exactly one review will be favorable is
(iv) Similarly, the probability that exactly two reviews will be favorable is:
(v) The probability that at least one of the reviews will be favorable is:
In (ii) to (v) we have used that and are mutually independent.
11. 
and 
alternately cut a pack of cards and the pack
is shuffled after each cut. If starts and the game is continued until one
cuts a diamond, what are the respective chances of and first cutting a diamond?
Solution:
Let 
and 
denote the events of 
and cutting a diamond respectively in the 
th trial. Then, we are given:
If starts the game, he can first cut the diamond
in the following mutually exclusive ways :
(i) 
happens, (ii) 
happens, (iii) 
happens,
and so on. Hence by addition theorem of probability, the probability 
that first cuts a diamond is
given by :
The probability that first cuts a diamond 
.
12. and throw alternatively with a pair of balanced
dice. A wins if he throws a sum of six points before throws a sum of seven points, while wins if he throws a sum of seven points before
throws a sum of six points. If begins the game, show that his probability of
winning is 
.
Solution:
Let denote the event of 's throwing ' 6 ' in the th throw, 
,
and denote the event of 
s throwing ' 7 ' in the th throw, ;
with a pair of dice. Then 
and 
are the complementary events.
' 6 ' can be obtained with two dice in the following ways:
,
i.e., in 5 distinct ways.
' 7 ' can be obtained with two dice as follows:
,
…(**)
If starts the game, he will win in the following
mutually exclusive ways:
(i) happens,
(ii) 
happens
(iii) 
happens, and so on.
Hence by addition theorem of probability, the required
probability of 's winning, (say) 
is given by :
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