Download CPT Section D, Quantitative Aptitude, Chapter 13 Prof

CPT Section D, Quantitative Aptitude, Chapter 13
Prof(Dr.) P.R.Vittal
Random experiment
Sample space
Events
Null event (Impossible event)
1.Classical probability
2.Statistical definition
Modern definition(axiometric approach)
If all the possible outcomes are equally
likely
Then the probability of an event A is
P(A)=(Number of favourable
outcomes)/Total number of outcomes)
If n is very large then
Number of Ways
P( A) = lim
n →∞ Total number of Outcomes
For any event A,
Axiom-1 : P(A)>0
Axiom-2 : p(S)=1
Axiom-3 : For any two disjoint events A and B
P(AUB)=P(A)+P(B)
1.Mutually exclusive events
2.Mutually exclusive and exhaustive events
3.Independent events
Two coins are tossed.
(i)Write down the sample
space,
(ii)Find the probability of
getting exactly one head and
(iii)Find the probability of
getting atleast one head.
(i)S={HH,HT,TH,TT}
p(one head)=(2/4)=1/2
(ii)p(atleast one head)=3/4
Three coins are tossed.
(i)Write down the sample space,
(ii)Find the probability of getting exactly two heads,
(iii)Find the probability of getting atleast one head
(i)s={HHH,HTT,THT,TTH,THH,HTH,HHT,TTT}
(ii)p(exactly two heads)=3/8
(iii)p(atleast one head)=7/8
P(atleast one head)=1-p(no head)
When ten coins are tossed
P(at least one head)=1-p(no head)
1
= 1 − 10
2
1023
=
1024
Two events are said to be mutually exclusive if the
occurrence of one prevents the occurrence of the
other
Examples
1.when a coin is tossed head and tail are mutually exclusive
2.when a die is thrown odd number and even number are
mutually exclusive
3. If a card is drawn from a pack of cards getting ace and club
are not mutually exclusive
A set of events are said to be mutually
exclusive and exhaustive if
• i)all the events are pairwise mutually exclusive
ii)no other event is possible
NOTE
• If A,B,C are pairwise mutually exclusive and exhaustive
then
• P(A)+P(B)+P(C)=1
Two events are said to be independent if the
occurrence of one does not influence of the other
EXAMPLES
(i)when a coin is tossed twice, head with the first
and head with second coin are independent
(ii)one card from two pack of cards. One ace and
the other king are independent
1. If two events are independent they cannot be mutually exclusive
2. If two events are mutually exclusive they cannot be independent
3. If A and B are mutually exclusive events
P(AUB)=P(A)+P(B)
4. If A and B are in independent events then P(AnB)= P(A) P(B)
5. A,B,C are said to be completely independent then
P(A∩B)= P(A).P(B), P(B∩C)=P(B).P(C), P(A∩C)=P(A).P(C),
P(A∩B∩C)=P(A).P(B).P(C)
Odds in favour of an
event is the ratio
between the number of
success of the event to
number of failure of the
event
If a and b are the
number of
success and
failure of an
event A then
odds in favour of
A is a:b similarly
odds against A is
b:a
EXAMPLES
Odd is
favour of an
event A is
2:3 then
P(A)=2/5
Odds
against an
event B is
5:4 then
P(B)=4/9
In a random
experiment
outcomes
may be
numbers or
qualities
like head
and tail
In case of
quantitative
data we can
associate
numbers
with them
.Then we
get a
random
variable
Definition: A
random
variable is a
real valued
function. It
can take
any real
number as
its value.
There are
two types of
random
variables
• i) Discrete RV
• ii) Continuous
RV
A RV is
discrete if it
can take
only
countable
number of
values
The outcomes of a throw of a die is
discrete random variable.
A random variable is continuous if it can
take any real value.
The height a person chosen is a
continuous random variable.
Suppose X is a discrete random variable taking discreteset of values
x1 ,x 2 ,x 3 ,.....x n ,.... p(x=x i ) is called probability function if the following conditions
are satisfied :
1
( i ) p( xi ) ≥ 0, ( ii ) ∑ p( xi ) =
i
Suppose X is a continuous Random
Variable. F(x) is called probability density
function if the following conditions are
satisfied:
Every Random variable has certain important
charecteristics.
Two important charecteristics are
(i)mean of X ,
(ii)Variance of X or Standard deviation of X.
The mean of x is also characteristic x and
is denoted by
E ( x) = x1 p1 + x 2 p 2 + ..... + x n p n
∞
E ( x) =
∫ xf
−∞
( x ) dx
(i ) E (c ) = 0
(ii ) E ( x + c ) = E ( x ) + c
(iii ) E (ax + b) = aE ( x) + b
(iv ) E ( x + y ) = E ( x) + E ( y )
(v)E(xy) = E(x)E(y) if x and y are independent
Expected value of X is affected by change of origin
and change of scale.
Another Property : E(x - µ ) = E ( x − x ) = 0
Definition :
V ( x) = E ( x − µ ) = E ( x ) − µ
Pr operties of var iance :
2
2
2
(i )V (c) = 0
(ii )V ( x + c) = V ( x)
(iii )V (ax + b) = a V ( x)
(iv ) If x and y are independent then
V ( x + y ) = V ( x) + V ( y )
2
If A and B are disjoint events then
• P(AUB)=P(A)+P(B)
If A and B are any two events
then
• P(AUB)=P(A)+P(B)-P(A∩B)
If A and B are independent events
• P(AB)=P(A)P(B)
If A and B are two dependent events
then
• P(A∩B)=P(A).P(B/A) = P(B).P(A/B) where
• P(A/B)=P(A∩B)/P(B) if P(B)≠0
(a)2/9
(b)1/4
(c)4/9
(d)1/3
Answer: a
Answer :
Sample space
S={[1,3],[2,4],[3,5],[4,6],[3,1],[4,2],[5,3],
[6,4]}
P(difference is 2 )=8/36 = 2/9
Answer : (a)
(a)3/10
(b)2 /5
(c)1/5
(d)1/4
Answer:c
Solution
:p(A)=1/2,p(A’)=1/2,p(B)=3/5,p(
B’)=2/5
P(neither of them solves the
problem)=p(A’)p(B’)=(1/2).(2/5)
=1/5
Answer: (c)
a)2 /5
(b)1/5
(c)3/5
(d)3/11
Answer: d
Solution :Since the total probability is 1,
p(A)+p(B)+p(C)=1. Hence we have
2p(B)+p(B)+(2/3)(p(B)=1
P(B)=3/11
Answer: (d)
(a)1/2
(b)1/4
(c)1/6
(d)1
Answer:c
Solution :
There are 2 vowels and 2
consonants
P(2 are vowels )=1/6
Answer: (c)
(a)1/2
(b)1/4
(c) 1/8
(d)1/16
Answer: b
P(Both are
black)=5/10.5/10=1/4
Answer : (b)
(a)5/6
(b)7/12
(c)11/12
(d)3/4
Answer: c
Solution :
P(sum is atleast 4)=1-p(sum is
less than 4)
=1-(3/36 )= 11/12
Ans: (c)
(a)0.27
(b)0.35
(c)0.29
(d)0.22
Answer: c
Solution :
Number of numbers divisible by 7 =142
and divisible by 11 is 90 and divisible by
77 is 12
Then p(Number divisible by 7 or 11) =
142/1000 +90/1000 -12/1000 =220/1000
=0.22
Answer : (d)
(a)0.4
(b)0.6
(c)0.45
(d)0.35
Answer: a
Solution :
P(A)=3/10 , p(B)=1/4
,p(A’)=7/10 , p(B’)=3/4
P(they contradict each
other)=p(A)p(B’)+p(A’)p(B)=
0.4
Answer: (a)
(a)11/20
(b)11/16
(c)5/16
(d)9/20
Answer: b
Solution :
P(AUB)=P(A)+p(B)-p(A)p(B)
¾ =1/5 +p(B)-1/5 p(B).On
simplifying we get
P(B )=11/16
Answer : (b)
(a)1/20
(b)9 /20
(c)4/5
(d)11/20
Answer:c
Solution :
P(B/A)= P(A∩B)/P(A)
=1/5/1/4=4/5
Answer : (c)
(a)3
(b)4
(c)3.5
(d)2.5
Answer:c
Solution :
X:1 2 3 4 5 6
P : 1/6 1/6 1/6 1/6 1/6 1/6
E(x)=∑xp= 21/6 =3.5
Ans : (c)
(a)3/35
(b)1/2
(c)41/70
(d)29/70
Answer:c
Solution :
P(A)=2/7 , P(B)=3/10
P(AUB)=2/7 + 3/10 =
41/70
Answer: (c)
In a group of students , 30% failed in
Economics and 40% failed in mathematics
and 50% failed in both. The probability that a
student selected has passed in Mathematics
given that he has passed in Economics is
•
•
•
•
(a)3/7
(b)4/7
(c)5/7
(d)5/6
% passed in Economics=P(EC)=70,
P(Mathematics)=60 , P(both)=p(EC∩M)=50
P(M/EC )=50/70 =5/7
Ans : (c)
(a)Mutually exclusive
(b)independent
(c)Exhaustive
(d)mutually independent and exhaustive
Answer; b
Solution :
P(A∩B)= P(A)+P(B)P(AUB)
= ½ +1/4 -5/8
=1/8=P(A).P(B)
A and B are
independent
Answer : (b)
A random variabe has the
following probability mass
function
• X : -2
• P(x) : 1/3
3
1/2
Then E(2x+5) is
•
•
•
•
(a)6
(b)1
(c) 7
(d)5
1
1/6
Solution :
E(x)=-2/3 + 3/2 +1/6
=1
E(2x+5)=2E(x)+5=7
Answer : (c)
(a)3
(b)5/3
(c) 25/3
(d)3/5
Answer:c
Solution :
3x-5y+7=0
X =5y/3 -7/3
SD (x)=(5/3) x5 =25/3
Answer: (c)
(a)52
(b)38
(c)10
(d)45
Answer:d
Solution :
v(x)=5 , Let y=3x+7 then we have
V(y)=3^2 v(x) =9x5=45
Answer : (d)
(a)21/6
(b)343 /8
(c)21/8
(d)1/2
Answer:b
Solution :
E(xyz) = E(x)E(y)E(z)
since x,y,z are
independent random
variables.
E(xyz)=7/2 7/2 7/2 =
343/8
Ans : (b)
(a)1/10
(b)1/5
c) 1/15
(d)1/2
Answer:a
Solution :
P(A∩B)=P(A)+P(B)P(AUB)
=1/2 +1/3 -4/5
= 1/30
P(A/B)=P(AB)/P(B)
=1/30/1/3 = 1/10
Answer : (a)
(a)1/2
(b)1/6
(c)1/20
(d)1/12
Answer:c
Solution :
There are 3 vowels
and 3 consonants .
P(they are vowels)=
=1/20
Ans: (c)
(a)5/6
(b)2/3
(c) 11/12
(d)3/4
Answer:c
Solution :
P(A)=1/2 ,p(B)=1/4,p(C)=1/6
Then p(AorBorC)=1/2 +1/4+1/6 =
11/12
Answer: (c)
(a)7/ 8
(b)9/10
(c)11/12
(d)5/6
Answer: a
Solution :
P(AB)=P(A)+P(B)-P(A)P(B)
4/5 = 1/3 +p(B)-1/3 P(B)
P(B)=7/8
Answer : (a)
(a)21
(b)3.5
(c)18
(d)24
Answer:a
Solution :
Expected vaue of their
sum =7/2 x 6 = 21
Answer : (a)
(a)36
(b)27
(c)65
(d)None of these
Answer:a
Solution :
Expected value =∑xp =
4/7 x 63 = 36
Ans : (a)
(a)Rs300
(b)Rs290
( c) Rs200
(d)None of these
Answer: b
Solution :
Expected value =∑xp =
3/10 x 1000 = 300
Expected gain = 30010 = 290
Answer: (b)
(a)2/7
(b)3 /7
(c)4/7
(d)None of these
Answer:b
Solution :
A leap year contains 52 weeks plus 2
days.The least 2 dyas can be
[s,m],m,tu],[tu,wed],[wed,thurs],[thurs,fri]
,[fri,sat],[sat,sun]
P(53 sun or mon )=3/7
Answer: (b)
(a)3/8
(b)1/2
(c)1/8
(d)1/4
Answer:b
Solution :
Sample space =
S={HHH,HTT,THT,TTH,THH,HTH,HHT,TTT}
P(atleast 2 heads)==1/2
Answer : (b)