1
MHT CET 2024 10th May Morning Shift
MCQ (Single Correct Answer)
+2
-0
 

There are three events $\mathrm{A}, \mathrm{B}, \mathrm{C}$, one of which must and only one can happen. The odds are 8:3 against $\mathrm{A}, 5: 2$ against B and the odds against C is $43: 17 \mathrm{k}$, then value of k is

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

Four persons can hit a target correctly with probabilities $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is

A
$\frac{1}{5}$
B
$\frac{3}{5}$
C
$\frac{2}{5}$
D
$\frac{4}{5}$
3
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

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

A
$\frac{5}{16}$
B
$\frac{11}{16}$
C
$\frac{12}{16}$
D
$\frac{15}{16}$
4
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A bag contains 4 red and 3 black balls. One ball is drawn and then replaced in the bag and the process is repeated. Let X denote the number of times black ball is drawn in 3 draws. Assuming that at each draw each ball is equally likely to be selected, then probability distribution of $X$ is given by

A
$x$ 0 1 2 3
$\mathrm{P}(x)$ $\left(\frac{4}{7}\right)^3$ $\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$ $\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$ $\left(\frac{3}{7}\right)^3$
B
$x$ 0 1 2 3
$\mathrm{P}(x)$ $\left(\frac{3}{7}\right)^3$ $\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$ $\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$ $\left(\frac{4}{7}\right)^3$
C
$x$ 0 1 2 3
$\mathrm{P}(x)$ $\left(\frac{3}{7}\right)^3$ $\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$ $\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$ $\left(\frac{4}{7}\right)^3$
D
$x$ 0 1 2 3
$\mathrm{P}(x)$ $\left(\frac{4}{7}\right)^3$ $\frac{12}{7} \cdot\left(\frac{4}{7}\right)^2$ $\frac{9}{7} \cdot\left(\frac{3}{7}\right)^2$ $\left(\frac{3}{7}\right)^3$
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