1
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If a discrete random variable X takes values $0,1,2,3, \ldots \ldots$. with probability $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 5^{-x}$, where k is a constant, then $\mathrm{P}(\mathrm{X}=0)$ is

A
$\frac{7}{25}$
B
$\frac{16}{25}$
C
$\frac{18}{25}$
D
$\frac{19}{25}$
2
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Ten bulbs are drawn successively, with replacement, from a lot containing $10 \%$ defective bulbs, then the probability that there is at least one defective bulb, is

A
$1-\left(\frac{1}{10}\right)^{10}$
B
$1-\left(\frac{3}{10}\right)^{10}$
C
$1-\left(\frac{9}{10}\right)^{10}$
D
$1-\left(\frac{7}{10}\right)^{10}$
3
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A fair die with numbers 1 to 6 on their faces is thrown. Let $$\mathrm{X}$$ denote the number of factors of the number, on the uppermost face, then the probability distribution of $$\mathrm{X}$$ is

A
$$\mathrm{X=}x$$ 1 2 3 4
$$\mathrm{P(x=}x)$$ $$\frac{1}{6}$$ $$\frac{1}{2}$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$
B
$$\mathrm{X=}x$$ 1 2 3 4
$$\mathrm{P(x=}x)$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$ $$\frac{1}{2}$$
C
$$\mathrm{X=}x$$ 1 2 3 4
$$\mathrm{P(x=}x)$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$
D
$$\mathrm{X=}x$$ 1 2 3 4
$$\mathrm{P(x=}x)$$ $$\frac{1}{6}$$ $$\frac{1}{6}$$ $$\frac{1}{2}$$ $$\frac{1}{6}$$
4
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The p.m.f. of a random variable $$\mathrm{X}$$ is $$\mathrm{P}(x)=\left\{\begin{array}{cl}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \mathrm{n} \\ 0, & \text { otherwise }\end{array}\right.$$, then $$\mathrm{E}(\mathrm{X})$$ is

A
$$\frac{\mathrm{n}+1}{6}$$
B
$$\frac{2 \mathrm{n}+1}{6}$$
C
$$\frac{\mathrm{n}+1}{3}$$
D
$$\frac{2 n+1}{3}$$
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