Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
  • CUET UG Mathematics Applied
    Previous Year Questions

CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Probability Distribution PYQ

CUET UG Mathematics Applied PYQ
The probability distribution of $X$ is:
 x 01 2 3 4 
 p(X= x) 0.12k k k 2k 
 then var(X) = 





Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Since total probability = 1

$0.1+2k+k+k+2k=1$

$0.1+6k=1$

$6k=0.9$

$k=0.15$

Now probabilities:

$P(0)=0.1$

$P(1)=0.3$

$P(2)=0.15$

$P(3)=0.15$

$P(4)=0.3$

Now,

$E(X)=0(0.1)+1(0.3)+2(0.15)+3(0.15)+4(0.3)$

$=0+0.3+0.3+0.45+1.2$

$=2.25$

Now,

$E(X^2)=0+1(0.3)+4(0.15)+9(0.15)+16(0.3)$

$=0.3+0.6+1.35+4.8$

$=7.05$

Variance:

$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$

$=7.05-(2.25)^2$

$=7.05-5.0625$

$=1.9875$

$= \frac{159}{80}$
CUET UG Mathematics Applied PYQ
If $\mu$ is mean of random variable $X$, with probability distribution:
 x0 1 2 
 P(x = x)4/9 4/9 1/9 
then value of $9\mu+4$ is:






Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Mean:

$\mu=\sum xP(X=x)$

$=0\cdot\frac{4}{9}+1\cdot\frac{4}{9}+2\cdot\frac{1}{9}$

$=\frac{4}{9}+\frac{2}{9}$

$=\frac{6}{9}$

$=\frac{2}{3}$

Now,

$9\mu+4$

$=9\left(\frac{2}{3}\right)+4$

$=6+4$

$=10$
CUET UG Mathematics Applied PYQ
The probability mass function of Random variable $X$ is:

$P(X=x)=(0.6)^x(0.4)^{1-x}, ; x=0,1$

The variance of $X$ is:





Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

For $x=0$:

$P(0)=(0.6)^0(0.4)^1=0.4$

For $x=1$:

$P(1)=(0.6)^1(0.4)^0=0.6$

So $X$ follows Bernoulli distribution with

$p=0.6$

Variance of Bernoulli:

$\mathrm{Var}(X)=p(1-p)$

$=0.6(0.4)$

$=0.24$
CUET UG Mathematics Applied PYQ
A biased dice is thrown once. If $X$ denotes the number appearing on it and the probability distribution is
 $x:$1  23 4 5 6 
 $P(X=x):$ k k/2 2k$8k^2$ 1-5k  $\frac{k}{2}$
where $k>0$.

Then consider the following statements:

A. $P(X=3)$  

B. $P(X\le2)$  

C. $P(X\ge5)$  

D. $P(X=4)$  

E. $P(X=1)+P(X=5)$  

Choose the correct answer from the options given below:





Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Sum of probabilities = 1

$k+\frac{k}{2}+2k+8k^2+(1-5k)+\frac{k}{2}=1$

Simplify

$4k+8k^2+1-5k=1$

$8k^2-k=0$

$k(8k-1)=0$

Since $k>0$

$k=\frac{1}{8}$


Now compute probabilities.

$P(X=1)=\frac{1}{8}$  

$P(X=2)=\frac{1}{16}$  

$P(X=3)=\frac{1}{4}$  

$P(X=4)=\frac{1}{8}$  

$P(X=5)=\frac{3}{8}$  

$P(X=6)=\frac{1}{16}$


Evaluate statements.

A. $P(X=3)=\frac{1}{4}$  

B. $P(X\le2)=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}$  

C. $P(X\ge5)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}$  

D. $P(X=4)=\frac{1}{8}$  

E. $P(X=1)+P(X=5)=\frac{1}{8}+\frac{3}{8}=\frac{1}{2}$


Order (largest to smallest)

$\frac{1}{2}>\frac{7}{16}>\frac{1}{4}>\frac{3}{16}>\frac{1}{8}$

So

$E>C>A>B>D$


CUET UG Mathematics Applied

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

CUET UG Mathematics Applied

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More
Limited Seats
× Aspire MCA Promotion

Game Changer NIMCET Test Series 2026

Boost your preparation with mock tests, analysis and rank-focused practice.

JOIN NOW
Ask Your Question or Put Your Review.

loading...