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CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 5 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Probability Distribution And Its Mean And Variance
Quiz 5
The counter part of probability mass function is:
Report Question
0%
Probability distribution
0%
Probability density function
0%
Distribution function
0%
None of these
Explanation
We know that,
The probability mass function is defined for a discrete random variable.
And for a continuous random variable we define probability density function.
Thus the counter part of probability mass function is
probability density function.
A variable that can assume any possible value between two points is called:
Report Question
0%
Discrete sample space
0%
Discrete random variable
0%
Continuous random variable
0%
Random variable
Explanation
We know that,
A
continuous random variable
is a
random variable
where the variable can take infinitely many values in the given interval.
Thus a
variable that can assume any possible value between two points is called
continuous random variable.
Hence option C is correct.
Amit tosses a fair coin twice, and let
X
be defined as the number of heads he observe. Find probability mass function
P
x
.
Report Question
0%
1
3
0%
1
8
0%
1
4
0%
None off these
Explanation
Given Amit tosses a fair coin twice
∴
Sample space,
S
=
[
H
H
,
H
T
,
T
H
,
T
T
]
⇒
When
P
(
x
)
=
1
it is a PMF
i.e,
1
4
+
2
4
+
1
4
=
4
4
=
1
∴
Probability mass function
P
x
=
1
4
.
Hence, the answer is
1
4
.
Probability distribution of discrete random variable is classified as
Report Question
0%
probability mass function.
0%
posterior mass function.
0%
interior mass function.
0%
continuous mass function.
Explanation
The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values.
It is also sometimes called the probability function or the probability mass function.
Thus option
(
A
)
is correct.
A random variable
X
has the following p.d.f.
X
0
1
2
3
4
5
6
7
P
(
X
=
x
)
0
k
2
k
2
k
3
k
k
2
2
k
2
7
k
2
+
k
The value of
k
is
Report Question
0%
1
8
0%
1
10
0%
0
0%
−
1
Explanation
x
=
7
∑
x
=
0
P
(
X
)
=
1
⇒
P
(
0
)
+
P
(
1
)
+
P
(
2
)
.
.
.
P
(
7
)
=
1
⇒
0
+
k
+
2
k
+
2
k
+
3
k
+
k
2
+
2
k
2
+
7
k
2
+
k
=
1
⇒
10
k
2
+
9
k
−
1
=
0
⇒
10
k
2
+
10
k
−
k
−
1
=
0
⇒
10
k
(
k
+
1
)
−
(
k
+
1
)
=
0
⇒
(
k
+
1
)
(
10
k
−
1
)
=
0
k
=
−
1
or
k
=
1
10
.
But
0
≤
P
(
x
)
≤
1
Hence
k
=
1
10
.
If
x
is a continuous random variable, then
P
(
a
<
x
<
b
)
=
Report Question
0%
P
(
a
≤
x
≤
b
)
0%
P
(
a
<
x
≤
b
)
0%
P
(
a
≤
x
<
b
)
0%
all the three above
Explanation
Consider there are
n
number of points between
a
and
b
where
n
tends to infinite
(
given
x
is a continuous random variable
)
P
(
x
=
a
)
=
P
(
x
=
b
)
=
1
n
. As
n
→
∞
,
1
n
→
0
Therefore
P
(
x
=
a
)
=
p
(
x
=
b
)
→
0
⟹
P
(
a
<
x
<
b
)
=
P
(
a
≤
x
≤
b
)
=
P
(
a
<
x
≤
b
)
=
P
(
a
≤
x
<
b
)
Therefore the correct option is
D
.
A random variable X has the following probability distribution
X
=
x
−
2
−
1
0
1
2
3
P
(
x
)
0.1
0.1
0.2
0.2
0.3
0.1
Then
E
(
x
)
=
Report Question
0%
0.8
0%
0.9
0%
0.7
0%
1.1
Explanation
E
(
x
)
=
∑
x
.
P
(
x
)
X
=
x
−
2
−
1
0
1
2
3
P
(
x
)
0.1
0.1
0.2
0.2
0.3
0.1
x
.
P
(
x
)
−
0.2
−
0.1
0
0.2
0.6
0.3
E
(
x
)
=
∑
x
.
P
(
x
)
=
−
0.2
−
0.1
+
0.2
+
0.6
+
0.3
Thus
E
(
x
)
=
0.8
If
E
(
X
+
C
)
=
8
and
E
(
X
−
C
)
=
12
then the value of
C
is
Report Question
0%
−
2
0%
4
0%
−
4
0%
2
Explanation
We know that
E
(
X
+
C
)
=
E
(
X
)
+
C
...(i)
Therefore
E
(
X
+
C
)
=
E
(
X
)
+
C
=
8
E
(
X
−
C
)
=
E
(
X
)
−
C
=
12
Subtracting equation ii from i
2
C
=
−
4
C
=
−
2
.
Let the random variable X follow B(6, p). If 16 P(X = 4) = P(X = 2), then what is the value of p?
Report Question
0%
1
3
0%
1
4
0%
1
5
0%
1
6
Explanation
16
P
(
X
=
4
)
=
P
(
X
=
2
)
∴
16
(
6
C
4
p
4
q
2
)
=
6
C
2
p
2
q
4
∴
16
(
6
C
4
p
4
(
1
−
p
)
2
)
=
6
C
2
p
2
(
1
−
p
)
4
∴
16
p
2
=
(
1
−
p
)
2
∴
15
p
2
+
2
p
−
1
=
0
∴
(
3
p
+
1
)
(
5
p
−
1
)
=
0
∴
p
=
−
1
3
or
p
=
1
5
Probability cannot be negative.
So,
p
=
1
5
Answer is option (C)
What is
P
(
Z
=
5
)
equal to ?
Report Question
0%
1
2
0%
1
3
0%
1
4
0%
1
6
Explanation
Given set
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
Z
=
X
+
Y
Where
X
is randomly selected from set of odd numbers and
Y
is randomly selected from set of even numbers
That gives
Z
is odd and greater than one
Now, for
Z
=
5
Possible ordered pairs are
{
(
1
,
4
)
,
(
2
,
3
)
}
Probability of getting these 2 pairs
=
P
(
1
)
×
P
(
4
)
+
P
(
2
)
×
P
(
3
)
=
1
4
×
1
3
+
1
4
×
1
3
=
1
6
What is
P
(
Z
=
10
)
equal to ?
Report Question
0%
0
0%
1
/
2
0%
1
/
3
0%
1
/
5
Explanation
Given set
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
Z
=
X
+
Y
Where
X
is randomly selected from set of odd numbers and
Y
is randomly selected from set of even numbers.
Which gives
Z
as odd and greater than one
Given :
Z
=
10
But
10
is even
So,
P
(
Z
=
10
)
=
0
If
x
is a continuous random variable then
P
(
x
≥
a
)
=
Report Question
0%
P
(
x
<
a
)
0%
1
−
P
(
x
>
a
)
0%
P
(
x
>
a
)
0%
1
−
P
(
x
≤
a
−
1
)
Explanation
Given
x
is continuous random variable , Let us consider that there are
n
points in total, Since
x
is continuous,
n
→
∞
The probability of
x
=
a
is
1
n
→
0
, therefore
P
(
x
=
a
)
=
0
Which implies
P
(
x
≥
a
)
=
P
(
x
>
a
)
+
P
(
x
=
a
)
=
P
(
x
>
a
)
Therefore the correct option is
C
The probability distribution of a random variable is given below :
X
=
x
0
1
2
3
4
5
6
7
P
(
X
=
x
)
0
K
2
K
2
k
3
K
K
2
2
K
2
7
K
2
+
k
Then
P
(
0
<
X
<
5
)
=
Report Question
0%
1
10
0%
3
10
0%
8
10
0%
7
10
Explanation
∑
p
(
X
=
x
i
)
=
1
9
K
+
10
K
2
=
1
10
K
2
+
9
K
−
1
=
0
10
K
2
+
10
k
−
k
−
1
=
0
10
K
(
K
+
1
)
−
1
(
K
+
1
)
=
0
K
=
1
10
P
(
0
<
X
<
5
)
=
p
(
x
=
1
)
+
p
(
x
=
2
)
+
p
(
x
=
3
)
+
p
(
x
=
4
)
=
K
+
2
K
+
2
K
+
3
K
=
8
K
=
8
10
∴
P
(
0
<
X
<
5
)
=
8
10
Let
X
and
Y
be two random variables. The relationship
E
(
X
Y
)
=
E
(
X
)
⋅
E
(
Y
)
holds
Report Question
0%
Always
0%
If
E
(
X
+
Y
)
=
E
(
X
)
+
E
(
Y
)
is true
0%
If
X
and
Y
are independent
0%
If
X
can be obtained from
Y
by a linear transformation
Explanation
We know that the relationship
E
(
X
Y
)
=
E
(
X
)
⋅
E
(
Y
)
holds, when
X
and
Y
are independent variables.
If the range of random variable
X
=
{
0
,
1
,
2
,
.
.
.
}
and
P
(
X
=
k
)
=
c
(
k
+
1
)
2
k
for
k
=
0
,
1
,
2
,
.
.
.
then
c
=
Report Question
0%
1
2
0%
1
3
0%
1
4
0%
1
5
Explanation
P
(
X
=
0
)
=
c
P
(
X
=
1
)
=
2
c
2
P
(
X
=
2
)
=
3
c
2
2
∑
P
(
X
=
r
)
=
1
⇒
c
2
0
+
2
c
2
1
+
3
c
2
2
+
.
.
.
.
.
.
.
.
=
1
⇒
1
c
2
+
2
c
2
2
+
.
.
.
.
.
.
.
=
1
2
⇒
c
+
c
2
+
c
2
2
+
.
.
.
.
.
.
.
=
1
2
⇒
c
[
1
+
1
2
+
1
2
2
+
.
.
.
.
.
]
⇒
c
[
1
1
−
1
2
]
=
1
2
⇒
c
[
2
]
=
1
2
⇒
c
=
1
4
A
r
.
v
.
X
∼
B
(
n
,
p
)
. If values of mean and variance of
X
are
18
and
12
respectively then total number of possible values of
X
are
Report Question
0%
54
0%
55
0%
12
0%
18
Explanation
M
e
a
n
=
n
p
=
18
V
a
r
i
a
n
c
e
=
n
p
q
=
12
n
p
q
n
p
=
12
18
∴
q
=
2
3
Now,
p
=
1
−
q
=
1
−
2
3
∴
p
=
1
3
Also,
n
p
=
18
⟹
n
(
1
3
)
=
18
⟹
n
=
54
∴
Possible values of
X
are
0
,
1
,
2
,
.
.
.
.
.
54
∴
No. of possible values are
55
.
If
X
is a binomial variate with the range
{
0
,
1
,
2
,
3
,
4
,
5
,
6
}
and
P
(
X
=
2
)
=
4
P
(
X
=
4
)
, then the parameter of
X
is
Report Question
0%
1
3
0%
1
2
0%
2
3
0%
3
4
Explanation
Here n=6
According to the question,
6
C
2
p
2
q
4
=
4
×
6
C
4
p
4
q
2
⟹
q
2
=
4
p
2
⟹
(
1
−
p
)
2
=
4
p
2
⟹
3
p
2
+
2
p
−
1
=
0
⟹
(
p
+
1
)
(
3
p
−
1
)
=
0
⟹
p
=
1
3
,
p
=
−
1
p cannot be negative, hence
p
=
1
3
If the p.d.f of a r.v.
X
is given as
x
i
−
2
−
1
0
1
2
P
(
X
=
x
i
)
0.2
0.3
0.15
0.25
0.1
then
F
(
0
)
=
Report Question
0%
P
(
X
<
0
)
0%
P
(
X
>
0
)
0%
1
−
P
(
X
>
0
)
0%
1
−
P
(
X
<
0
)
Explanation
We know that, probability function
F
for a random variable
X
is given by
F
(
x
)
=
∑
u
≤
x
f
(
u
)
=
P
(
X
≤
x
)
∴
F
(
0
)
=
P
(
X
≤
0
)
=
P
(
X
=
0
)
+
P
(
X
=
−
1
)
+
P
(
X
=
−
2
)
=
0.15
+
0.3
+
0.2
=
0.65
=
1
−
0.35
=
1
−
(
0.25
+
0.1
)
=
1
−
(
P
(
X
=
1
)
+
P
(
X
=
2
)
)
=
1
−
P
(
X
>
0
)
Hence,
F
(
0
)
=
1
−
P
(
X
>
0
)
The value of k when
f
(
x
)
=
1
√
x
,
0
<
x
<
4
, is the p.d.f of r.v.x
Report Question
0%
−
4
0%
−
1
4
0%
4
0%
1
4
Explanation
f
(
x
)
=
k
√
x
where
x
∈
(
0
,
4
)
So,
∫
4
0
k
√
x
d
x
=
1
(Summation of all probability distribution=1)
i.e,
2
k
√
x
]
4
0
=
1
⇒
4
k
=
1
So,
k
=
1
4
Hence, option D is correct.
If r.v
X
: waiting time in minutes for bus and p.d.f of
X
is given by
f
(
x
)
=
{
1
5
,
0
≤
x
≤
5
0
,
o
t
h
e
r
w
i
s
e
then probability of waiting time not more than
4
minutes is
=
...........
Report Question
0%
0.3
0%
0.8
0%
0.2
0%
0.5
Explanation
Given :
f
(
x
)
=
{
1
5
,
0
≤
x
≤
5
0
,
o
t
h
e
r
w
i
s
e
Probability of waiting time not more than
4
is same as probability of waiting time less than
4
minutes and it is given by
P
(
X
<
4
)
=
P
(
X
=
3
)
+
P
(
X
=
2
)
+
P
(
X
=
1
)
+
P
(
X
=
0
)
=
1
5
+
1
5
+
1
5
+
1
5
.... [From the definition of
f
]
=
4
×
1
5
=
0.8
The probability distribution of
X
is
X
0
1
2
3
P
(
x
)
0.3
k
2k
3k
The value of
k
is
Report Question
0%
0.116
0%
0.7
0%
1
0%
0.3
Explanation
∞
∑
x
=
0
P
(
x
)
=
1
3
∑
x
=
0
P
(
x
)
=
1
P
(
0
)
+
P
(
1
)
+
P
(
2
)
+
P
(
3
)
=
1
⇒
0.3
+
k
+
2
k
+
3
k
=
1
⇒
6
k
=
1
−
0.3
⇒
6
k
=
0.7
k
=
0.7
6
;
k
=
0.116
For the following distribution function
F
(
x
)
of a r.v
X
is given
x
1
2
3
4
5
6
F
(
x
)
0.2
0.37
0.48
0.62
0.85
1
Then
P
(
3
<
x
≤
5
)
=
Report Question
0%
0.48
0%
0.37
0%
0.27
0%
1.47
Explanation
x
1
2
3
4
5
6
f
(
x
)
0.2
0.37
0.48
0.62
0.85
1
p
(
x
)
0.2
0.17
0.11
0.14
0.23
0.15
P
(
3
<
x
≤
5
)
=
p
(
x
=
4
)
+
p
(
x
=
5
)
=
0.14
+
0.23
=
0.37
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Report Question
0%
14
3
0%
13
3
0%
12
3
0%
11
3
Explanation
E(X) is nothing but mean of X
→
first sic natural numbers are
1
,
2
,
3
,
4
,
5
,
6
→
We can select two positive no.s in
6
×
5
=
30
ways
→
2
numbers are selected at random among which X denotes the larger of the two no.s
→
But as X is largest among the two ,
it can be
2
,
3
,
4
,
5
or
6
∴
P
(
X
=
2
)
=
P
(larger no. is
2
)
(
1
,
2
)
,
(
2
,
1
)
=
2
30
→
P
(
X
=
3
)
=
P
(Larger no. is
3
)
=
{
(
1
,
3
)
,
(
3
,
1
)
,
(
2
,
3
)
,
(
3
,
2
)
}
=
4
30
→
P
(
X
=
4
)
=
P
(larger no. is
4
)
=
{
(
1
,
4
)
,
(
4
,
1
)
,
(
2
,
4
)
,
(
4
,
2
)
,
(
3
,
4
)
,
(
4
,
3
)
}
=
6
30
→
P
(
X
=
5
)
=
P
(Larger no. is
5
)
=
{
(
1
,
5
)
,
(
5
,
1
)
,
(
2
,
5
)
,
(
5
,
2
)
,
(
3
,
5
)
,
(
5
,
3
)
,
(
4
,
5
)
,
(
5
,
4
)
}
=
8
30
→
P
(
X
=
6
)
=
P
(Larger no. is
6
)
=
{
(
1
,
6
)
,
(
6
,
1
)
,
(
2
,
6
)
,
(
6
,
2
)
,
(
3
,
6
)
,
(
6
,
3
)
,
(
4
,
6
)
,
(
6
,
4
)
,
(
5
,
6
)
,
(
6
,
50
)
}
=
10
30
∴
E
(
X
)
=
mean
=
∑
P
(
X
i
)
.
X
i
=
2
(
2
30
)
+
3
(
4
30
)
+
4
(
6
30
)
+
5
(
8
30
)
+
6
(
10
30
)
=
140
30
=
14
3
The expected value of the number of points, obtained in a single throw of die, is
Report Question
0%
3
2
0%
5
2
0%
7
2
0%
9
2
Explanation
Considering that the dice is a fair dice, the probability of all outcomes is equal.
∴
sample space of outcomes of dice is
{
1
,
2
,
3
,
4
,
5
,
6
}
Probability of each outcome =
1
6
∴
E
[
outcome of a single throw
]
=
1
6
×
(
1
+
2
+
3
+
4
+
5
+
6
)
=
21
6
=
7
2
This is the required answer.
X
has three children in his family. What is the probability that all the three children are boys?
Report Question
0%
1
8
0%
1
2
0%
1
3
0%
3
8
Explanation
X has
3
children in his family.
All possible outcomes are if
G
represents girl child and B represent boy child,
{
(
B
B
B
)
,
(
B
B
G
)
,
(
B
G
B
)
,
(
B
G
G
)
,
(
G
B
B
)
,
(
G
B
G
)
,
(
G
G
B
)
,
(
G
G
G
)
}
, hence
n
(
S
)
=
8
Out of which favorable outcomes for all the three chilren being boy,
n
(
E
)
=
1
Hence, the probability that all the three children are boys is
n
(
E
)
n
(
S
)
=
1
8
X
has three children in his family. Probability of atleast two girls in the family is.....
Report Question
0%
1
8
0%
1
2
0%
1
4
0%
3
4
Explanation
X
has
3
children in his family.
All possible outcomes are if
G
represents girl child and
B
represent boy child,
{
(
B
B
B
)
,
(
B
B
G
)
,
(
B
G
B
)
,
(
B
G
G
)
,
(
G
B
B
)
,
(
G
B
G
)
,
(
G
G
B
)
,
(
G
G
G
)
}
, hence
n
(
S
)
=
8
Out of which favorable outcomes for at-least two girls are,
n
(
E
)
=
4
Hence, the probability that there are at-least two girls is
n
(
E
)
n
(
S
)
=
4
8
=
1
2
If
A
and
B
are two non empty sets then
(
A
∪
B
)
C
=
Report Question
0%
A
C
∪
B
C
0%
A
C
∩
B
C
0%
A
∪
B
C
0%
A
C
∪
B
Explanation
(
A
∪
B
)
C
=
A
C
∩
B
C
The probability distribution of a random variable
X
is given below:
x
1
2
3
4
5
6
P
(
X
=
x
)
a
a
a
b
b
0.3
If mean of
X
is
4.2
, then
a
and
b
are respectively equal to
Report Question
0%
0.3
,
0.2
0%
0.1
,
0.4
0%
0.1
,
0.2
0%
0.2
,
0.1
Explanation
Given, mean of
x
is
4.2
Thus that
a
+
2
a
+
3
a
+
4
b
+
5
b
+
1.8
=
4.2
⇒
6
a
+
9
b
=
2.4
.....(1)
Also
a
+
a
+
a
+
b
+
b
+
0.3
=
1
⇒
3
a
+
2
b
=
0.7
.....(2)
Multiply equation (2) by
2
,
6
a
+
4
b
=
1.4
.....(3)
Subtracting equations (1) and (3), we get
5
b
=
1.0
⇒
b
=
0.2
Substituting this value in equation (1), we get
6
a
+
9
(
0.2
)
=
2.4
⇒
6
a
+
1.8
=
2.4
⇒
a
=
0.1
Therefore,
a
=
0.1
,
b
=
0.2
X
has three children in his family. What is the probability of two or more boys in the family?
Report Question
0%
1
8
0%
1
2
0%
1
4
0%
3
8
Explanation
X
has
3
children in his family.
All possible outcomes are if
G
represents girl child and
B
represent boy child,
{
(
B
B
B
)
,
(
B
B
G
)
,
(
B
G
B
)
,
(
B
G
G
)
,
(
G
B
B
)
,
(
G
B
G
)
,
(
G
G
B
)
,
(
G
G
G
)
}
, hence
n
(
S
)
=
8
Out of which favorable outcomes for two or more boys are,
n
(
E
)
=
4
Hence, the probability that there are two or more boys in the family is
n
(
E
)
n
(
S
)
=
4
8
=
1
2
X
has three children in his family. The probability of one girl and two boys is......
Report Question
0%
1
8
0%
1
2
0%
1
4
0%
3
8
Explanation
X
has
3
children in his family.
All possible outcomes are if G represents girl child and
B
represent boy child,
{
(
B
B
B
)
,
(
B
B
G
)
,
(
B
G
B
)
,
(
B
G
G
)
,
(
G
B
B
)
,
(
G
B
G
)
,
(
G
G
B
)
,
(
G
G
G
)
}
, hence
n
(
S
)
=
8
Out of which favorable outcomes for one girl and two boys are,
n
(
E
)
=
3
Hence, the probability that there are one girl and two boys is
n
(
E
)
n
(
S
)
=
3
8
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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