Aspire Faculty ID #17307 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

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:

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

$E>C>A>B>D$

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