1
MHT CET 2025 26th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The following is p.d.f. of continuous random variable X

$$ \mathrm{f}(x)= \begin{cases}\frac{x}{8} & , \text { if } 0 < x < 4 \\ 0 & , \text { otherwise }\end{cases} $$

Then $F(0.5), F(1.7)$ and $F(5)$ is respectively

A
$\frac{1}{64}, 1,0.18$
B
$0.0156,0.18,1$
C
$0.18,0.0156,1$
D
$1,0.0156,0.18$
2
MHT CET 2025 26th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

A box contains 8 red and $x$ number of green balls. 3 balls are drawn at random, if the probability that 3 balls being red is $\frac{7}{15}$, then number of green balls is…

A
2
B
4
C
3
D
5
3
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

A doctor assumes that patient has one of three diseases $\mathrm{d} 1, \mathrm{~d} 2$ or d 3 . Before any test he assumes an equal probability for each disease. He carries out a test that will be positive with probability 0.7 if the patient has disease $\mathrm{d} 1,0.5$ if the patient has disease d 2 and 0.8 if the patient has disease d3. Given that the outcome of the test was positive then probability that patient has disease d2 is

A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{1}{5}$
D
$\frac{1}{7}$
4
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The probability that a student is not a swimmer is $\frac{1}{5}$. The probability that out of 5 students selected at random 4 are swimmers is

A
$\left(\frac{4}{5}\right)^4$
B
$\left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)$
C
$\left(\frac{4}{5}\right)^5 \times \frac{1}{5}$
D
$\left(\frac{4}{5}\right)^3 \times \frac{1}{5^2}$
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