1
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let X be a discrete random variable. The probability distribution of X is given below

$$ \begin{array}{|c|c|c|c|} \hline \mathrm{X} & 30 & 10 & -10 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \mathrm{~A} & \mathrm{~B} \\ \hline \end{array} $$

and $\mathrm{E}(\mathrm{X})=4$, then the value of AB is equal to

A
$\frac{3}{10}$
B
$\frac{2}{15}$
C
$\frac{1}{15}$
D
$\frac{3}{20}$
2
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

In a game, 3 coins are tossed. A person is paid $Rs \, 150$ if he gets all heads or all tails and he is supposed to pay ₹50 if he gets one head or two heads. The amount he can expect to win / lose on an average per game in ₹ is

A
100
B
0
C
200
D
-100
3
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $A$ and $B$ are independent events with $\mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{11}{20}$, then $\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)$ is root of the equation

A
$4 x^2-7 x+3=0$
B
$4 x^2+7 x+3=0$
C
$4 x^2-3 x-7=0$
D
$6 x^2-5 x+1=0$
4
MHT CET 2025 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is
A
$\frac{5}{17}$
B
$\frac{4}{17}$
C
$\frac{5}{16}$
D
$\frac{5}{18}$
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