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

Consider a group of 5 boys and 7 girls. The number of different teams, consisting of 2 boys and 3 girls that can be formed from this group if there are two specific girls A and B , who refuse to be the members of the same team, is

A
350
B
300
C
200
D
500
2
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
$\frac{15}{16}$
3
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y+\sqrt{x^2-y^2}}{x}$ is

A
$\sin ^{-1} y=\log x+c$, where c is a constant of integration.
B
$\frac{y}{x}=\sin ^{-1} x+\mathrm{c}$, where c is a constant of integration.
C
$\frac{y}{x}=\sqrt{x^2-y^2}+\mathrm{c}$, where c is a constant of integration.
D
$\sin ^{-1}\left(\frac{y}{x}\right)=\log x+\mathrm{c}$, where c is a constant of integration.
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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