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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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:
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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