CUET UG Mathematics Applied Probability Distribution Previous Year Questions (PYQs)
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The probability distribution of $X$ is:
then var(X) =
| x | 0 | 1 | 2 | 3 | 4 |
| p(X= x) | 0.1 | 2k | k | k | 2k |
1
2
3
4
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:
$\mathrm{Var}(X)=E(X^2)-[E(X)]^2$
$=7.05-(2.25)^2$
$=7.05-5.0625$
$=1.9875$
$= \frac{159}{80}$
If $\mu$ is mean of random variable $X$, with probability distribution:
then value of $9\mu+4$ is:
| x | 0 | 1 | 2 |
| P(x = x) | 4/9 | 4/9 | 1/9 |
1
2
3
4
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,
$9\mu+4$
$=9\left(\frac{2}{3}\right)+4$
$=6+4$
$=10$
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:
1
2
3
4
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:
$\mathrm{Var}(X)=p(1-p)$
$=0.6(0.4)$
$=0.24$
A biased dice is thrown once. If $X$ denotes the number appearing on it and the probability distribution is
| $x:$ | 1 | 2 | 3 | 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:
1
2
3
4
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
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CUET UG Mathematics Applied
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and More.
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