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PROBABILITY AND STATISTICS
Sample Space:
A set of all possible outcomes of probabilistic exp. or non determinists
exp and experiment. {H, T}, {1, 2, 3, 4, 5, 6}
Random experiment:
If an experiment is conducted any number of times, under essentially
identical conditions there is a set of all possible outcomes associated with it.
If the result is not centaurs or unique and is any one of the several possible out
comes, the exp is called a random trial or a random experiment.
Trial : Single performance is a random exp. tossing a coin, throwing a die.
Outcome: The result of a trial in a random exp.
Event: Every non-empty subset of a sample space of a random exp.
Sample space Eg: Two coins tossed, S.S = {HH, HT, TH, TT}
Three coins
=
{HHH, HHT, HTH, THH, THT, TTT, HTT, TTH}
 If a coin is tossed n times or n coins tossed at a time, the S.S. consists
2n elements.
 In throwing die two times or two dice at a time the S.S. consists
36 elements.
(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1,6)
(2,1), ..
..
..
..
(6,1) ..
..
..
..
..
..
(6, 6)
 If a die is thrown for n times or n dice thrown, the S.S. consists
6n elements.
1
Event: eg: Turning up head (H), turning up tail T. In turning two coins, getting
1H-A = {HT, TH} no head B= {TT}.
2.
In throwing two dice, A event is getting Sum=7
B is getting a sum equal to 11, C is getting Sum 4
A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
B= {(6,5), (5,6)}, {(1,3), (2,2), (3,1)}
Exhaustive Event: All possible events in any trial.
Favourable Cases: The No. of cases favourable to an event in a trial is the
member of outcomes which involve the happening of an event.
Eg: In drawing a card from a pack of 52 cards the favourable case (i) drawing a
queen is 4. (ii) drawing a red card is 26. (iii) The No. of cases favourable in
getting a sum 7, when two dice are thrown is 6.
(1,6), (6,1), (3,4), (4,3), (5,2), (2,5)
Mutually Exclusive events: Events are said to be two events cannot happen at
a time mutually exclusive if the happening if anyone of them prevents the
happening. If all the others (or) if no two or more of them can happen
simultaneously in the same trial.
Eg: In tossing a coin, H turning up and tail turning up are mutually exclusive.
2.
In throwing a die all 6 possible cases are mutually exclusive.
3.
In throwing two dice, getting a sum 2, 3, 4, 5,…..12 are mutually
exclusive.
Note: If two events are mutually exclusive, then the sets are disjoint i.e.
if A and B are mutually exclusive A  B   .
Equally likely events: In a trial one (the out come) event cannot be expected in
preference to the others.
Eg: Tossing a coin, throwing a die.
2
Permutation: If r objects are chosen from a set of n distinct objects, any
ordered arrangements or order of these objects is called a permutation.
npr  nn  1......n  (r  1) 
n!
(n  r )
Eg: If repetitions are not allowed, how many 4 digit numbers can be formed
from digits 1, 2, 3, 4, 5, 6, 7?
7p4 =
7! = 7 x 6 x 5 x 4
3! = 840
If there are 10 balls in a box, in how many ways we can choose 4 balls without
repletion.
10c = 10 x 9 x 8 x 7 = 5040
Combinations: The No. of ways of un-ordered selected objects from a set of n
n cr 
distinct objects is called combination.
n!
r!(n  r )!
Eg: In how many ways 3 students can be selected from 15 students.
15c3 = 455
2)
A student is to answer 5 out of 8 questions in an examination. In how
many ways he/she can choose. (i) if any fine 8 c5  56 ways. (ii) if 1
question is compulsory , 7 c4   became she/he can to see 4 only.
3)
2 out of first 3
Ans 3c2 .5c3
(Mathematical or Classical)
Probability: If an Experiment is performed, P is the No. of exhaustive event
and q is the number of favourable cases of an extent A. Then the probability of
an event A is defined by
Number of favourable cases
=
q n( A)

 p( E )
p n( S )
3
No. of exhaustive events
n (A) =
no. of elements belonging to A
n (s) =
No. of elements belonging to S, sample S
pq
q
 1   1  p( E )
p
p
p( E ) 
p( E )  p( E )  1
Note: This def fails when (i) The out comes are not equally likely (ii) No. of
outcomes is infinite.
Universal Event: The Entire sample space S is called a universal
(or certain or sure) event.
The nullset  is called the null or impossible event ( )  0
Collective exhaustive events: A list of events A1, A2, …….An are said to be
n
collectively exhaustive if
A
i
S.
i 1
Limitation of class def of probability:
1.
If the outcomes of a random exp are not equally likely
Eg: A student passing in exam is not 50% as success and failure in exam are
not equally likely as tossing of a coin.
2.
Outcomes are infinite:
Eg: Flipping a coin until we get a tail  infinite outcomes. It cann’t give
answer for finding the probability that the tail ever appears before 20th tossing.
Axiomatic Probability:
P(A) is the probability function defined on a power set of s into R, where s is
finite sample space if the following three axiom are satisfied.
(i)
Axiom of positivity :
p( E )  0
4
p(s)  1
(ii)
Axiom of certainity :
(iii)
Axiom of union : If E1, E2 are disjoint subsisting S.
then
p( E1  E2 )  p( E1 )  p( E2 )
The image p(E) of E is called the probability of the event E.
If E1, E2….. En are disjoint subsets of S, there
Note:
p( E1  E2 .....  En )  p( E1 )  p( E2 )  .....  p( E n )
Theorem:
p is a probability function defined on the power set of a sample space s.
Then (i) p ( )  0 (ii) pE   1  pE 
Proceed as E    E and E    ; p( E)  pE   
Proof:
 p( E )  p( E )  p( )  0
(ii)
E  E  s and E  E    p( s)  p( E )  p E  1
* If E1, E2
are two events of a sample space s  E1  E2
Then p( E1 )  p( E2 ) as E1  E2  pE2  E1   0
Addition theorem on probability:
If s is a sample space, and E1, E2 are any events in s then
pE1orE2   pE1  E2   pE1   pE2   pE1  E2 
5
Proof: case (i) If E1  E2  
pE1  E 2   pE1   pE 2 
pE1  E 2   pE1   pE 2 
 pE1   pE 2   0
 pE1   pE 2   p p 
 pE1   pE 2   pE1  E 2 
 pE1  E 2   pE1   pE 2   pE1  E 2 
E1  E1  E2   E1  E2  …(1)
case (ii) If E1  E2  
and E2  E2  E1   E1  E2  …(2)
From (1)

(2) p  E1   p E1  E 2
and E1  E 2 , E1  E2

and E2  E 1
+ pE1  E2  ……. (3)
p  E2   pE2  E1   pE1  E2  ………………. (4)
adding (3) and (4)

 

 E  E  E  E  E  E 
p  E1   pE2   E1  E2   p E2  E1  p E1  E 2  pE1  E2  ………..(5)
we above that E1  E2
2
1

 
1
2
1
2

p( E1  E2 )  E1  E 2  E1  E2  E1  E2   pE1  E2 


 p( E1  E2 )  p E1  E 2  E1  E2 
………….. (6)
pE1   pE2   pE1  E2   pE1  E2   pE1  E2   pE1   p( E2 )  pEpE 
Corollary : If E1, E2 are two mutually exclusive events
Then (i) pE1  E2   pE1   p( E2 )
(ii)

 pE1  E2   p   0

  
p E1  E2  ps  ( E1  E2   p( s)  pE1  E2   1  p E1  p( E2 )
 
 p E 1  p E2 
Theorem: If A, B, C are any three events, then
p( A  B  C )  p( A)  p( B)  p(C )  p( A  B)  p( B  C )  p(C  A)  p( A  B  C )
6
Proof: p( A  B  C)  p( A  B)  C  p( A  B)  p(C)  p( A  B)  C
 p A  B  C   P A  B  C  P A  B  PC   P A  B  C 
 p A  B  PC   P A  C   P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 
 A and B are two events. Find the probability of the event that
(i)
Neither A nor B occurs (ii) A occurs and B doesn’t occur
(iii)
Exactly one of A and B occurs.
(i)  pA  B   pA  B   ps   A  B   P( s)  p A  B 
 1  P( A)  P( B)  P( A  B)  P( A)  P( B)  P( A  B)
(or)
 P( B)  P( A)  P( A  B)
(ii)  P( A  B)  P( A   A  B   P( A)  P( A  B)

(iii)  P( A  B)  ( A  B)  P A  B

 ( A  B)  ( A  B)  
 p( A)  P( A  B)  P( B)  P( A  B)  P( A)  P( B)  2 P( A  B)
 1  p( A)  1  P( B)  2 P( A  B)  2  P( A)  P( B)  2(1  P( A  B)
 2  P( A)  P( B)  2  2 P( A)  2 P( B)  2 P( A  B)
 P( A)  P( B)  2 P( A  B)
Problems: (1) What is the probability that a card drawn at random from the pack
of playing cards may be either a Queen or a King?
4
4
  0 … p( E1  E2 )  0
52 52
Sol : Let s be the
(2)
Three electronic bulbs are fitted in a room. Three bulbs are chosen at
random from 10 bulbs having 6 good bulbs. What is the chance that the
room is lighted?
Sol:
6c3 x4c0 x6c2 x4c1 x6c1 x4c2 x
10c3

116
120
7
(3)
In a group there are 3 men and 2 women. Three persons are selected at
random from this group. Find the probability that one man and two
women (or) two men and one women are selected.
p( E2 )  3c 2 x2c 1
p(s)  5c 3  10
Sol:
p( E1 )  3c 1 x2c 1
(4)
60 boys and 20 girls are there in a class. Half of the boys and half of the
girls of the class play cricket. Find the probability of the selected person
to be a “boy” or “a girl who plays cricket”.
60
,
80
10
,
80
Sol:
p( E1 )  60C 1 
(5)
An integer is chosen at random from the first 200 positive integers. What
p( E2 )  10C 1 
E1  E2  
is the probability that the integer chosen is divisible by 6 or 8?
Sol: Divisible by 6=200/6 = 33, div days = 200/8=25
Divisible by 6 and 8 = 200/24=8
 For

events = 25+33-8 = 50, exhaustive events = 200
Required probability = 50/200 = 1/4
1
2
If p( A)  ,
1
p( B)  ,
3
p( A  B) 
1
, then find
5
(i) p ( A  B) , (ii) p( Ac  B) , (iii) p( A  B c ) , (iv) p( Ac  B c ) , (v) p( Ac  B c )
19/30
2/15
3/15
11/30
( A  C )c  1   A  B   1 
1 4

5 5
 A card is drawn from a well shuffled pack of cards. What is the probability
that it is either a spade or an ace.
All =
3)
4
13
, Spade = ,
52
52
E  E2 
1
52
Three students A, B, C are in running race. A and B have the same
probability of winning and each is twice as likely to win as C. Find the
probability that B or C units.
Sol: p( A)  p( B), p( A)  2 p(C ), p( B)  2 p(C )
8
Let s  A  B  C .
 (2  2  1) P ( c )  1, p (c) 
P( S )  1  P( A)  P( B)  P(C )
1
;
5
p B  C  
BC 
(4)
p ( A) 
2
 p( B)
5
3
5
From a city 3 news papers A,B,C, are being published A is read by 20%.
B is read by 16%, C is read by 14% both A and B are read by 8% both
A and C hereby are read by 2%. What is the percentage of the population
that rent at least one paper.
Problem:
If A  B, P.T (i) pAc  B  p( B)  p( A)
Ac  B  B   A  B   A  B, A  B  A

 Ac  B  B  A
(2)

P Ac  B  P( B)  p( A)
If A and B are mutually exclusive events, there prone that p( A)  p  B c 
Sol: A  B  
P( A  B)  P( A)  P( B).  1  P( A)  1  P( B)
 P( A)  P( B c )
(3)
If
P( A  B) 
(i) P(B)
4
,
5
1
1
P ( B c )  , p ( A  B)  Find
3
5
(ii)
P( B c )  1  p ( B).P ( B)  1 
(4)
(iii) P( Ac  B)  pB   A  B 
P( A)
1 2

3 3
Two dice are thrown. Let A be the event that the sum of the points on the
faces is 9. Let B be the event hat at least one number is 6. Find the
probabilities of the following event (i) A B
(iii) A B c
n( B ) 
11
,
36
(iv)
Ac  B
n( A  B ) 
(ii) A B
(v) Ac  B c .nn (( sA))36 (3,6), (6,3), (5,4), (4,5) 
4
36
2
.
36
9
(5)
3
5
If A and B are two events and P( A)  ,
Then P.T.
(i) P( A  B) 
3
5
(ii)
A  A B,
(ii)
A  B  B  P A  B   P( B)  P A  B  
P( A  B)  P( A)  P( B)  P( A  B)  P
 P( A  B) 

(6)
1
2
P( A  B) 
1
2
3
3
 P A  B   P A  B  
5
5
(i)
P( A)  P( A  B) 
P( B) 
1
2
11
10
3 1
1
  P A  B   1   P A  B 
5 2
10
1
1
 P A  B  
10
2
The probabilities that students A, B, C, D solve a problem are 1/3, 2/5,
1/5 and 1/4 respectively. If all of them try to solve the problem. What is
the prob. that the problem is solved.
(after conducted probability)
Conditional Events: If E1, E2 are events of a sample space S and if E2 occurs
after the occurrence of E1, then the event of occurrence of E2 after the event E1
is called conditional event of E2 given E1. 1
It is denoted by E2/E1. (Similarly use define E1/E2)
Eg:
Two unbiased dice are thrown. If the sum of the numbers thrown on there
is 7, the event of getting 1 on any one of them is cond. Event.
(2)
A die is thrown 3 times. The event of getting the sum of the numbers
thrown is 15, when it is known that the first throw was a 5 is a cond. Event.
10
Conditional probability:
If E1 and E2 are two events in a sample space s and p(E1 )  0 , then the
probability of E2 after the event E1, has occurred is called the conditional prob.
of the event E2 given. E1, and denoted by p (
E2
E
) are define p( 2 ) 
E1
E1
 E  E2 
 .
p 1
 p( E1 ) 
In this case E1 is sample space (s), E2 is subset of s Results. If A,B,C are three
events of a sample space, we have
(1)
 A  P A  A P( A)
p  

 1; P( A)  0
p( A)
P( A)
 A
(2)
0
   P A   
p  

 0; P( A)  0
p( A)
P( A)
 A
(3)
A  B  A  C  B  C  P A  C   PB  C 

(4)
P  A  C  P B  C 
 A
B

 P   P  PC   0
A(C )
P(C )
C 
C 
A, B, are mutually exclusive events and p(B)>0
0
B
 A
 P  
 0 and P   0 If P( A)  0
 A
 B  P( B)
(5)

 A P A B
P ( A  B )  0 , P ( B )  1  P  
P( B)
B
=
(6)

P( A)  P( A  B)
P( A)

1  P( B)
1  P( B)
P A  P( B)  P A  B  P A  B  P( A)  P( B)  P A  B  1
 P A  P( B)  1  P A  B  
P( A)  P( B)  1 P( A  B)

P( B)
P( B)
 A  P( A)  P( B)  1
 P  
P( B)
B
(7)

 A
 A
P .P( B)  P .P( B)  P A  B
B
B

 P A  B  P( A)  P A  B  P( A)
11


(8)
 A  P A  B P ( S )  P A  B  1  P A  B 
P  


1  P( B)
1  P( B)
PB
B
(9)
A, B are mutually exclusive and P A  B  0
 P A  B  0 And P A  B  P( A)  P( B)
A  P A  ( A  B  PB  A PC  A


Now P

P A  B 
 A B 
P  A
P( A)
B
C 
P   P 
 A
 A
=
Multiplication Theorem of probability (or) Theorem of compound probabilities
In a random experiment if PE1   0, pE2   0
 E2 
E 
, PE2  E1   PE2 .P 1 
 E1 
 E2 
Then PE1  E2   pE1 .P
Proof: Let s be the sample space associated with rand exp.
PE1   0, PE2   0
Let E1, E2 be two events of s.
Since PE1   0 , by the def of con. Prob of E2 given E1, we have
 E  PE1  E2 
E 
P 2  
 PE1  E2   P( E1 ).P 2 
P( E1 )
 E1 
 E1 
 E1  PE2  E1 
E  E1 
 
P 2
( E2 )
P( E2 )
 E2 
Since PE2   0 , we have P
E 
PE2  E1 P( E2 ).P 1 
 E2 
Note: If A. B are two events  0 then


B
P A  B  1  P A  B   1  P( A).P 
 A
Multiplication theorem can be extended to three events as
 E3 
E 
E 
  PE1 .P 2   P 2 
PE1  E2  E3   PE1  E2   E3   PE1  E2 .P
 E1  E2 
 E1 
 E1 
12
Compound Event: When two or more events occur in conj with each other,
their joint occurrence is component.
Eg: If 2 balls are drawn from a bag containing 4 green, 6 black, 7 white balls,
the even of drawing 2 green or 2 white balls is a compound event.
Eg2: When a compound die and tossed, the event of getting utmost 4 on the die
and tossed, the event of getting utmost 4 on the die and head on the coin is a
comp event, and separately independent events.
(3)
From a pack, a spade is drawn not replaced if again a club is drawn, they
are compare even but separately they are dependent events.
End Event: If the occurrence of the event E2 is not affected by the occurrence
or non occurrence of the event E1 then the event E2 is said to be independents of
E1 and
E 
P 2   P( E2 )
 E1 
If P( E1 )  0 , P( E2 )  0 and E2 is independent of E1 and E1, is independent of E2.
In this case we say that E1, E2 are mutually independent (or) Simply
Independent. If the occerance of the event E2 is effected by the occurrence of
E1m then the events E1, E2 are dependent and
E 
P 2   P( E2 )
 E1 
Theorems: If A and B are Events of A is independents of B then B is
independent of A.
A
Proof: A doesn’t depend on B. There P   P( A)
B
 A  P A  B 
 P  
 P A  B   P( A).P( B)
P( B)
B
or
P
A B
B
 P   P ( B )  B
P( B)
 A
13
If A, B, C are pain use independent, i.e. A and B are ind, B care and C and A on
int then P A  B  P A.P(B)PB  C   P(B).P(C), P A  C   P( A)BC
Then If A and B are independent events. Then Ac and Bc are also independent
events.
Proof: Given A and B are independent events
 P A  B  P( A).P( B)


P Ac  B c  P.1  P( A  B)  1  P( B)  P A  B 
 1  P( A)  P( B)  P( A).P( B)  (1  P( A)(1  P( B))  P( Ac ).P( B c )
Theorem: If A and B are independent events. Then A & Bc are independent.
P( A  B c )   P( A).P( B c )  P( A).(1  P( B))  P( A).P( B c )
Proof:
Theorem: If A, B, C are mutually independent events then AUB, C are also
independent.
P A  B  C  P A  C   P( B  C)  P( A).P(C)  P( B).P(C)
 P( A).P( B).P(C)  P(C)P( A)  P( B)  P( A).P( B)
 P(C).P A  B
Sol: Let
P(A) =
Prob. that A solves the problem =
1
3
P(B) =
Prob. that B solves the problem =
2
5
P(C) =
Prob. that C solves the problem =
1
5
P(D) =
Prob. that D solves the problem =
1
4
The probability that the problem is solved = P(AUBUCUD)



 1  A  B  C  D  1  P A  B  C  D
1
 1  P( A) P( B) P(C ) P( D)

( A, B, C and D are cindyed events)
 1  1  P( A)1  P( B)1  P(C)1  P( D)
2343
6 19
 1  2  1  1 
 1  1  1  1  1    1 
 1

3554
25 25
 3  5  5  4 
14
(2)
Two bolts are drawn from a box containing 4 good and 6 bad bolts. Find
the probability that the second bolt is good if the first bold is found to be
bad.
Sol: There are 6 bad and 4 good bolts.
The probability that the first one is bad = 6/10
The probability that the second is good if first is bad
(3)
64 4
  . 
 10  9 15
A class has 10 boys and 5 Girls. Three students are selected at random
one after other. Find the probability that
(i)
First two are boys and third is girl
(ii)
First and third of same sex and second is of apposite sex.
Sol : (i)
The prob. that the first one is a boy = 10/15
The prob. that the Second one is a boy = 9/14, 10/15
The prob. that the two boys and third girl =
(ii)
(4)

10 9 5 15
. . 
15 14 13 91
First and third of some sex and second is appoint sex
B
G
B
G
G
B

10 5 9 5 10 4
x x  . .
15 14 13 15 14 13
A and B throw with a pair of dice. A wins if he throws to before B throws
7 and 6. B wins if he throws 7 before A throws 6. If A begins, show that
his chance of winning is 30/61.
Sol: Two dice are thrown. The possibilities of getting a sum 6 are
(1,5), (5,1), (3,3), (4,2), (2,4)
Sample space contests of 36 elements
 The prob. of getting sum 6 = 5/36
The possibilities of getting 7 one (2,5), (5,2), (3,4), (4,3) (6,1), (1,6)

The prob. of getting A 
5
36
15
Not winning 1 
5 31

36 36
Probability of winning of B 
1
6
Not winning = 1  
1
6
5
6
1 31
6 36
B wins if A fails to throw 6 in fut throw = .
A gets the second chance if A and B fail to throw in the first chance.
Chance of A to win in the second throw =
31 5 5
. .
36 6 36
B’s chance of winning in second throw

31 5 31 1
. .
36 6 36 6
and so on.
A’s probability of winning
B’s probability of winning
5
36
31 5 5
. x
36 6 36
31 1
.
36 6
31 5 31 1
. . .
36 6 36 6
F.A.
F.A. F.B. F.A. W.B
31 5 31 5 5
. . . x
36 6 36 6 36
31 5 31 5 31 1
. . . . .
36 6 36 6 36 6
F.A. F.B. F.A. F.B. W.A.
F.A. F.B. F.A. F.B. F.A. W.B.
31 5 31 5 31 5 5
. . . . . .
36 6 36 6 36 6 36
31 5 31 5 31 5 31
. . . . . .
36 6 36 6 36 6 36
A Chance Wins = All probabilities in the fut throw


 5 6 x36 30
5 
1
 .
x


36 1  31 x 5  36 61
61
 36 6 
Theorem of total probability (or) the rule of elimination
Statement: Let E1, E2……Ek be in a portion of the sample space s with
PE1   0 FOR I= 1,2,….K, then for any event A of s,
16
P A   PE1  A
i 1
Proof : Given that E1, E2…. Ek constitute the sample space
K
S  S   Ei and Ei  E j   for any i= dj
i 1
 E1’s
are mutually disjoint sets
Let A be any event of S.
k

 k 
 A  S  A  A  A    Ei E i   A  UEi 


 Ei 
 A  E1  E2U ....UEk    A  E1    A  E2 ....   A  E1 
(By distribution law)
the sets A  E1 , A  E2 ...... A  Ek are all mutually exclusive
Applying addition theorem
P(A) = PA  E1  ........U  A  Ek 
 P A  E1   P A  E2   .....  P A  Ek 
Applying again multiplication them.
P( A)  PE1 .P A / E1   .....  PEk .P( A / Ok )
P( Ei ).P( A / Ei ) Hence the theorem.
(or) Formula for the probability of causes.
Bayes Theorem: Suppose E1, E2,…….En are mutually exclusive events of a
sample spaces S, I
P( E i )  0 for i=1………n and let A be any arbitrary event of S p(A)>0. then
P ( E j / A) 
P ( A / Ej ).P( E j )
k
 P( Ei ).P( A / Ei )
i 1
Proof: Given E1, E2, En are mutually exclusive events of a sample space s.
Ei  E j  
n
for any i and j
and also s   Ei
i 1
given A be any arbitrary event of sample space s. Now A  S  A
17
n 
 A  A  U E1    A  E1    A  E n 
 i 1 
 P( A)   P A  Ei 
n
i 1
we know that by cond. Prob. P( E j / A) 
 P( A) 

1.
P A  E j 
P( E / A)
P A  E j 
P( A)
....(2)
P.P ( A / E j ).P ( E j )
P ( E j / A)
Box A contains 9 cards numbered 1 to 9 and box B contains five cards
numbered. 1to 5. A box is chosen at random and a card is drawn; if the
card shows an even number another card is drawn from the same box if
the card shows an add number a card is drawn from the other box.
(i) What is the Pr. that both Cardo show even number
(ii) If both cards show even numbers, what is pr. that they come from box A.
(iii) What is the prob that both cards are add.
Sol: pr. of choosing a box = 1/2
Box I
Pr. of choosing odd
Box II
=
5/9
The pr. of choosing an odd = 3/5
Pr. of choosing an even =
4/9
The pr. of choosing an even = 2/5
If card is even, another is drawn from If card is odd another in chosen from
same box.
other box= 5/9.
Therefore there are 8 cards. So Even number = 4/9
choosing odd number = 5/8
If card is odd another is choose for If even is chosen the prob. changes
other box = 3/5
even =1/4
Even number = 3/8
Odd number = 3/4
18
Both are even
1 4 3 1 2 1
 . . . . .
2 9 8 2 5 4

1
1
32
2



12 20 12 x 20 5
(ii)
 A  P A  E  1 4 3
p  
 . .
P E 
2 9 8
E
(iii)
both are odd  x x  x x 
(2)
Determine (i) P(B/A)
1 4 3
2 9 8
1 3 5
2 5 9
1
3
(ii) pA / Bc  If A and B are events with
1
1
1
P( A)  , P( B)  , P A  B  
3
4
2
P( A)  P( B)
p
 A  B  , P A  B    P A  B   1  1  1  7  6  1
P  A
3 4 2
12
12

1 1 1
/ 
12 3 4
(ii)
1 1
3 4

x
P A B
P A   A  B  3 12 12 3
c
P( A / B ) 



1
1
P( B c )
1  P( B)
1
4
4
(3)
(iii)
(4)
A can hit 1 target once in 5 shots

P(F)=81, P
c

PF  S  0.18 2


P F 
0.18 9
B can hit 2 target once in 3 shots
C can hit 1 target once in 4 shots
Prob. that 2 shots fit the target
P(A) = 1/5, P(B)=2/3,
P(C)=1/4
The probability that thus shots hit the target

c
= P  A  B   C c   A  C   B   B  C   Ac

19
These are independent and mutually exclusive also
(5)
=
P(A).P(B).P(Cc)+ P(A).P(C).P(Bc)+P(B).P(B).P(C).P(Ac)
=
1 2 3 1 1 1 2 1 4 15 1
. .  . .  . . 

5 3 4 5 4 3 3 4 5 60 4
A box I contains 5 Red balls. A box is chosen at contains 3 Red balls 6
white balls. A box is chosen at random and a ball is drawn and put into A
ball is drawn from second box. Find the prob. that both the ball are of
some colour
Sol: I, II 1/2, 1/2
(5)
A.B and C in order toss on coin. The first one to toss a head coins the
game. What are their probabilities of winning, assuming that the game
may continue indefinitely?
Sol: Supp A toss head
He will coin probability = 1/2
Suppose he can’t. The game will go to B.
Suppose he can toss head the probability
= A’s not winning. B’s winning = 1/2. 1/2 = 1/4
If not the choice will go to C, if he can toss a heal the prob. will be A’s
not win. B’s not. A’s win = 1/2. 1/2 . 1/2 =1/23.
4
7
1 1 1
 Probability of A’s winning         ..
2 2 2


3
6
 1
1  1 1
1  1 8 4
 1        ....  1    . 
1  2 7 7
2   2   2 
 2 

 23 
III ly B’s prob. of winning = 2/7, C’s prob. = 1/7
20
(6)
The probabilities of passing in subject A, B, C, D, are 3/4, 2/3, 4/5 and
1/2 respectively. So qualify the examination a student should pass in A
and two subjects among the three. What is the prob. of qualifying in that
exam.
Sol: Let the prob. of passing the subject A be P(A) = 3/4
Let the prob. of passing the subject B be P(B) = 2/3
Let the prob. of passing the subject C be P(C) = 4/5
Let the prob. of passing the subject D be P(D) = 1/2
There are four possibilities to qualify the examination
(i)
to pass in ABC and fail in D
to pass in ACD and fail in B
to pass in ABD and fail in C
and pass in all subjects
Sum of all the above prob = Required probably.

 
 
 P A  B  C  Dc  P A  Bc  C  D  P A  B  C c  D

 P A  B  C  D
All the events are independent
 P A  B  C  D c  
11
20
 P( A).P( B).P(C ).P( D c )
(7)
What is the probability of getting two queens, if we draw two cards from
a pack of 52 cards. If (i) with replacement (ii) without replacement.
Sol: (i) With replacement (The events are independents)
Prob. of drawing one Queen = 4/52
Card is replaced and drawn again
The prob. of second drawing = 4/52
 Required prob. 4/52. 4/52 = 1/169
21
(ii)
Without replacement (Events are dependent)
prob. of 1st draw = 4/52
not replace three will be 51 cards
Prob. of second drawing: 3/51

Reg. prob = 4/52. 3/51= 1/221
P(A) . P(B/A)
#
An assembly plant receives its voltage regulators from three different
suppliers are 0.6% from supplier B1, 0.3% from supplier B2 and 0.1 from
supplier B3. If 95% of the voltage regulators from B1, 80% of those from B2,
and 65%. of those from B3 performing according to specifications, what is the
probability that any one voltage regulator received by the plant will perform
according to specification.
Sol: If A denotes the event that a voltage regulator received by the plant
performs according to specification and B1, B2 and B3 one events that it comes
form respective suppliers, we can with.
22