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

Let $X=\left[\begin{array}{l}\mathrm{a} \\ \mathrm{b} \\ \mathrm{c}\end{array}\right], \mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A X=B$, then the value of $2 a-3 b+4 c$ will be

A
0
B
$-$4
C
6
D
4
2
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to

A
$\frac{80}{243}$
B
$\frac{40}{7243}$
C
$\frac{32}{625}$
D
$\frac{128}{625}$
3
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The unit vector which is orthogonal to the vector $5 \hat{i}+2 \hat{j}+6 \hat{k}$ and is coplanar with the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

A
$\frac{2 \hat{i}-6 \hat{j}+\hat{k}}{\sqrt{41}}$
B
$\frac{2 \hat{i}-5 \hat{j}}{\sqrt{29}}$
C
$\frac{-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{10}}$
D
$\frac{2 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{69}$
4
MHT CET 2024 2nd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The probability distribution of a random variable X is given by

$\mathrm{X=}x_i$: 0 1 2 3 4
$\mathrm{P(X=}x_i)$ : 0.4 0.3 0.1 0.1 0.1

Then the variance of X is

A
1.76
B
2.45
C
3.2
D
4.8
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