1
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 \mathrm{p}}{4}, \frac{1-\mathrm{p}}{4}, \frac{1+2 \mathrm{p}}{4}, \frac{1-4 \mathrm{p}}{4}$ respectively, where p varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of X are respectively.

A
$-\frac{7}{4}$ and $\frac{1}{2}$
B
$-\frac{1}{16}$ and $\frac{5}{16}$
C
$-\frac{7}{4}$ and $\frac{5}{16}$
D
$-\frac{1}{16}$ and $\frac{5}{4}$
2
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

A
$\frac{1}{16}$
B
$\frac{9}{16}$
C
$\frac{3}{4}$
D
$\frac{15}{16}$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{P}(\mathrm{X}=2)=0.3, \mathrm{P}(\mathrm{X}=3)=0.4, \mathrm{P}(\mathrm{X}=4)=0.3$, then the variance of random variable X is

A
1.6
B
3.6
C
6.6
D
0.6
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A man and his wife appear for an interview for two posts. The probability of the husband's selection is $\frac{1}{7}$ and that of the wife's selection is $\frac{1}{5}$. If they appear for the interview independently, then the probability that only one of them is selected, is

A
$\frac{1}{7}$
B
$\frac{2}{7}$
C
$\frac{6}{7}$
D
$\frac{4}{7}$
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