1
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The probability, that a year selected at random will have 53 Mondays, is

A
$\frac{1}{4}$
B
$\frac{3}{28}$
C
$\frac{5}{28}$
D
$\frac{3}{4}$
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

A
$\frac{17}{243}$
B
$\frac{13}{243}$
C
$\frac{11}{243}$
D
$\frac{10}{243}$
3
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable $X$ has the following probability distribution

$X=x$ 1 2 3 4 5 6 7 8
$P(X=x)$ 0.15 0.23 0.10 0.12 0.20 0.08 0.07 0.05

For the event $E=\{X$ is a prime number $\}$, $F=\{X<4\}$, then $P(E \cup F)$ is

A
0.5
B
0.77
C
0.35
D
0.75
4
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{A}, \mathrm{B}$ and C be three events, which are pairwise independent and $\bar{E}$ denote the complement of an event E . If $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$ and $\mathrm{P}(\mathrm{C})>0$, then $\mathrm{P}((\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) / C)$ is equal to

A
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{B}})$
B
$\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\overline{\mathrm{B}})$
C
$\mathrm{P}(\overline{\mathrm{A}})-\mathrm{P}(\mathrm{B})$
D
$\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\overline{\mathrm{B}})$
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