1
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
$\frac{15}{16}$
2
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
$\mathrm{P}(x)$ $\left(\frac{4}{7}\right)^3$ $\frac{12}{7} \cdot\left(\frac{4}{7}\right)^2$ $\frac{9}{7} \cdot\left(\frac{3}{7}\right)^2$ $\left(\frac{3}{7}\right)^3$
3
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A service station manager sells gas at an average of ₹ 100 per hour on a rainy day, ₹ 150 per hour on a dubious day, ₹ 250 per hour on a fair day and ₹ $300$ on a clear sky. If weather bureau statistics show the probabilities of weather as follows, then his mathematical expectation is

Weather Clear Fair Dubious Rainy
Probability 0.50 0.30 0.15 0.05

A
257.5
B
252.5
C
250
D
247.5
4
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability, that the three balls have different colours, is

A
$\frac{1}{3}$
B
$\frac{2}{7}$
C
$\frac{1}{21}$
D
$\frac{2}{23}$
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