1. A Problem in Statistics is given to three students A, B and C whose chances of
1 3
1
solving it are , and respectively .
2 4
4
What is the probability that the problem will be solved if all of them try
independently?
Answer: Let A, B and Cdenote the events that the problem is solved by the
students A, B and C respectively.
1
3
1
P A  , PB   and PC  
Then,
2
4
4
The problem will be solved if at least one of them solves the problem.
Thus it is necessary to find the probability of occurrence of at least one of the
three events A, B, C i.e. P( A  B  C )
P( A  B  C )  P( A)  P( B)  P(C )  P( A  B)  P( B  C )
 P( A  C )  P( A  B  C )
 P( A)  P( B)  P(C )  P( A)  P( B)  P( B).P(C )
 P( A)  P(C )  P( A)  P( B)  P(C )
 A, B, C are independent events
1 3 1 1 3 3 1 1 1 1 3 1
           
2 4 4 2 4 4 4 2 4 2 4 4
1 3 1 3 3 1 3
      
2 4 4 8 16 8 32
16  24  8  12  6  4  3 29


32
32
Or
P( A  B  C ) =1 – P( A  B  C )
 1  P A  B  C 
 1  P A  P B  P C 
 1 3 1
 1  1   1   1  
 2 4 4
1 1 3
3 32  3 29
1   1


2 4 4
32
32
32
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2. Let A and B two events such that P A 
P  A 
3
5
and PB  
4
8
3
5
and PB  
4
8
Show that
a ) P A  B  
b)
3
4
3
5
 P A  B  
8
8
a ) We know
A  ( A  B)

P  A  P  A  B 
3
 P A  B 
4
3
 P A  B  
4
b) we know

A B B

P( A  B)  P( B)
5
P ( A  B)  .............. i 
8
Now, P( A  B)  P( A)  P( B )  P( A  B)  1


P ( A)  P( B)  1  P( A  B )
3 5
  1  P A  B 
4 8
658

 P A  B 
8
3

 P A  B ........... (ii )
8
From i  and ii , we have

3
5
 P A  B  
Proved
8
8
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Application of Bayes Theorem
3. In a bolt factory, machines A, B and C manufacture 25%, 35% and 40%
respectively of the total. Of their output 5, 4, 2 per cent are defective bolts. A bolt
is drawn at random from the product and is found to be defective. What are the
probabilities that it was manufactured by machines A, B and C?
Answer:
Let E1, E2 and E3 denote the events that a bolt selected at random is manufactured
by the machines A, B and C respectively and let E denote the event of its being
defective
Then we have
25
PE1  
 0  25
100
35
P E 2  
 0  35
100
40
P E 3  
 0  40
100
The probability of drawing a defective bolt manufactured by machine A is
 E 
P   5%  0  05
 E1 
 E 
  4%  0  04
Similarly , P
 E2 
 E 
  2%  0  02
P

 E3 
Hence, the probability that a defective bolt selected at random is manufactured by
machine A is given by
 E 
PE1 . P 
E 
 E1 
P 1  
E
 E  3
PE i  P 

i 1
 Ei 
0  25  05

0  25  05  0  35  04  0  40  0  02
0  125
0  0125 125 25




0  0125  0  014  008 0  0345 345 69
and
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Similarly, the probability that a defective bolt random is manufactured by machine B
is given by
 E 

PE 2   P
 E2 
 E2 
P

 E 
 E 
 E 
 E 

  PE 3   P
PE1   P   PE 2   P

E
E
E
 1
 2
 3
0  35  04

0  25  0  05  0  35  04  0  40  0  02
0  014
0  0125  014  008
0  014

0  0345
140

345
28

69
 E 
E 
 E 
Now, P 3   1   P 1   P 2 
 E 
 E 
  E 
 25 28 
1   
 69 69 
53
1
69
69  53

69
16

69
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4. A factory produces certain types of output by three machines. The respective
daily production figures are A = 3,000 units; Machine B = 2,500 units, Machine C
= 4,500 units. Past experience shows that 1 per cent of the output produced by
machine A is defective. The corresponding fractions of defectives for the other
two machines are 1.2 and 2 per cent respectively. An item is drawn at random
from the day’s production run and is found to be defective . What is the
probability that it comes from the output of i) Machine A; ii) Machine B; iii)
Machine C?
Answer: Let E1 , E2 and E3 denote the events that an item selected at random is
manufactured by the machine A, B and C respectively and let E denote the event
of its being defective.
Then we have
3000
3
PE1  

10000 10
2500 1
P E 2  

10000 4
4500
9
P E 3  

10000 20
The probability of drawing a defective item manufactured by machine A is
 E 
P   1%  0  01
 E1 
 E 
  1  2%  0  012 and
Simillarly , P
 E2 
 E 
  2%  0  02
P

 E3 
Hence, (i) the probability that the item comes from the output of machine, A is
given by
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 E 
PE1   P 
E 
 E1 
P 1  
 E 
 E 
 E 
 E 

  PE 3   P
PE1   P   PE 2   P

E
E
E
 1
 2
 3
3
 0  01
10

3
1
9
 0  01   0  012 
 0  02
10
4
20
0  003

0  003  0  003  0  009
0  003

0  015
1

5
02
Similarly ,
for machine B,
 E 

PE 2   P
E
 E2 
 2
P

E
 E 
 E
 E 


  PE 3   P 
PE1   P   P
 E1 
 E2 
 E3 
1
 0  012
4

3
1
9
 0  01   0  012 
 0  02
10
4
20
0  003

0  003  0  003  0  009
0  003

0  015
0  003 3 1

  02
0  015 15 5
 For machine C ,
 E 
 E 
E
 P   1   P 1   P 2
 E
  E
 E3 
 1  0  2  0  2



1 0  4
06
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5. There are three machines producing 10,000; 20,000 and 30,000 springs per week
These machines are known to produce 5% ; 4% and 2% defective springs
respectively. One spring is taken at random from a week’s production of three
machines. What is the probability that it is defective? If the spring is defective,
what is the probability that it is produced by the second machine?
Answer: Let the A, B and C be the 1st , 2nd and 3rd machines producing 10,000,
20,000; 30,000 springs per week.
Let E1, and E2 and E3 denote the events that a spring selected at random is
manufactured by the machines A, B and C respectively and let E denote the
event of its being defective.
10,000 1
Then PE1  

60,000 6
20,000 1
P E 2  

60,000 3
30,000 1
P E 3  

60,000 2
The probabilities of drawing a defective spring manufactured by machine A, B
and C are
 E 
P   5%  05
 E1 
 E
P
 E2

  4%  04

 E 
  2%  02
P

 E3 
Hence, the probability that a defective spring selected at random is manufactured by
 E 
PE1   P 
E 
 E1 
machine A is given by P 1  
E
 E  3
PE i   P 

i 1
 Ei 
1
 0  05
6

1
1
1
 05   0  04   0  02
6
3
2
0.05
0  05 5
6


  0  26
0  05  0  08  0  06 0  19 19
6
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Simitarly ,
1
 0  04

3


 1  0  05  1  0  04  1  0  02
2
3
6
0  08
0  04
0  08 8
3
  0  42
 6 

0  05  0  08  0  06 0  19 0  19 19
6
6
1
 0  02
 E3 
2
P

 E  1  0  05  1  0  04  1  0  02
2
3
6
0  01

0  05  0  08  0  06
6
6
0  01
 0  01 

0  19
0  19
6
0  06 6
  0  32

0  19 19
E
P 2
 E
 The probability that it is manufactured by the second machine is
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6
 0  32 
19
6. There groups of workers contain 3 men and 1 woman, 2 men and 2 women, 1 man
and 3 women respectively. One worker is selected at random from each group.
What is the probability that the Group selected consists of 1 man and 2 women?
Answer: The required event of getting 1 man and women can materialise in the
following way:
Group no.
(i)
(ii)
(iii)
1
m
w
w
11
w
m
w
111
w
w
m

by addition theorem of probability,
the required probability = P (i) +P(ii) +P(iii) ……… (1)
3
Now, the probability of selecting a man from 1st group =
4
2
the probability of selecting a man from 2nd group =
4
3
the probability of selecting a man from 3rd group =
4
The three events of selecting children are independent of each other.

by compound probability theorem,
3 2 3 18 9
Pi     

4 4 4 64 32
Similarly ,
Pii  
1 2 3 6
3
  

4 4 4 64 32
1 2 1 2
1
P(iii )    

4 4 4 64 32

from (1) , the required probability =
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9
3
1 13



 0  4063
32 32 32 32
7. A committee of 4 people is to be appointed from 3 officers of production
Department, 4 officers of the purchase Department, 2 officers of the Sales
Department and 1 chartered accountant. Find the probability of forming the
committee in the following manner:
i) There must be one from each category.
ii) It should have at least one from the purchase department.
iii) The chartered accountant must be in the committee.
Answer: There are 3+4+2+1 = 10 persons in all and a committee of 4 people can be
formed out of them in 10 c 4 ways
10!
 Exhaustive no. of cases = 10 c 4 
4!16!
10.9.8.7.6!

4.3.2.1.6!
10.9.8.7

 210
4 . 3 .2
(i)
Favourable no. of cases for the committee to consist of 4 members, one from
each category 3 c1  4 c1  2 c  1  4  3  2  1  24
1

(ii)



24
8

210 70
P[ Committee has at least one purchase officer = 1 – P (Committee has no
purchase officer)
In this case, all the members are to be selected from production, Sales
departments, and chartered Acctt.
6!
6.5.4!
i.e.
total officer, = 3+2+1 = 6 and this can be done in 6 c 4  i.e.
4! 2!
4! 2!
= 15 ways.
15
1

P (committee has no purchase officer ) =
210 14
P (Committee has at least one purchase officer)
the required probability =
1
(iii)
1 13

14 14
Favourable number of cases that the committee consists of a chartered
accountant as a member and three others are
9!
1  9 c3 
6! 3!
9.8.7.6!

 84 ways
6! 3.2.1
84 2


the required probability =
210 5
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8. A company has 8 screw machines but only 3 spaces are available in the
production area for the machines. In how many different ways can the eight
machines be arranged in the three spaces available ?
Answer: We know,
n!
n
Pr 
, where n is total number of objects
n  r !
r  the number of objects / places available
Here , total of number of machines, n  8
number of spaces availabe, r  3
 the total number of possible ways in which 8machines can be arranged in the three
available spaces
 8 c3

8!
8  3 !

8!
5!

8 .7 .6 .5 !
5!
 8 .7 .6
 336
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9. A candidate is selected for interview of management trainees for 3 companies.
For the first company, there are 12 candidates, for the second there are 15
candidates and for the third there are 10 candidates. What are the chances of
getting job at least at one of the companies?
Answer: The probability that the candidate gets the job at least at one company 
1 – probability that the candidate does not get the job in any company
Probability that the candidate does not get the job in the 1st company
1 11
1

12 12
Probability that the candidate does not get the job in the 2nd company
1
1 14

15 15
Probability that the candidate does not get the job in the 3rd company
1
1
9

10 10
Since the events are independent, the probability that the candidate does not get any
1 14 9
 
12 15 10
1386

 0  77
1800
job in any of the three companies 
 the required probability = 1  0  77
= 0.23
10. A sales manager has 7 field representatives working under him. A local
consulting firm at a fee of Tk. 1500 per man is conducting a three – day seminar
on sales to which the sales manager would like to send all the seven of his field
representatives. However, his budget will allow him to send only three men. How
many different ways are there for him to compose this group of three men?
Answer: The no. of possible combination of three men selected from a set of 7 men is
7 ! 7.6.5.4 !
 7 c3 

 35
4! 3 ! 4! 3.2 !
Example:
U 
P( A). P 
 A
 A
P  
 U  P( A). P U   PB . P U 
 
 
 A
B
55 3

100
100

55 3
45 4



100 100 100 100
 55  03

 55  03  45  04
 4783
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11. Out of 5 mathematicians and 7 Statisticians, a Committee consisting of 2
mathematicians and 3 statisticians is to be formed. In how many ways can this be
done if
a) any mathematician and any statistician can be included
b) one particular statistician must be on the committee
c) particular mathematician cannot be on the committee?
Answer: a) 2 mathematicians out of 5 can be selected in 5 c 2 ways
3 statisticians out of 7 can be selected in 7 c3 ways
Total no. of possible selections = 6 c 2

7
c3  10  35  350
b) 2 mathematicians out of 5 can be selected in 5 c 2 ways
2 statisticians out of 6 can be selected in 6 c 2 ways
Total no. of possible selections = 5 c2  6 c  10  15  150
2
c) 2 mathematicians out of 4can be selected in 4 c 2 ways
3 statisticians out of 7 can be selected in 7 c3 ways

Total no. of possible selections = 4 c 2
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
7
c3  6  35  210
12.
A husband and wife appear in an interview for two vacancies in the same post .
1
1
The probability of husband’s is
and that of wife’s selection is . What is the
7
5
probability that
a) both of them will be selected
b) only one of them will be selected
c) none of them will be selected.
1
Answer: The probability of husband’s selection, P( H) is
7
1
The probability of wife’s selection, P( W) is
5
1 1 1
 a) the probability that both of them will be selected, =  
7 5 35
b) the probability that only one will be selected
 1  a   c 
 1 24 
1   
 35 35 
25 35  25
1

35
35
10 2


35 7
1 6
c) The probability that husband will not be selected = 1  
7 7
1 4
The probability that wife will not be selected = 1  
5 5
6 4 24
 The probability that none of them will be selected =  
7 5 35
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