MHT CET 2024 4th May Evening Shift | Probability Question 75 | Mathematics

For an entry to a certain course, a candidate is given twenty problems to solve. If the probability that the candidate can solve any problem is $\frac{3}{7}$, then the probability that he is unable to solve at most two problem is

$\frac{256}{49}\left(\frac{4}{7}\right)^{18}$

$\frac{1966}{49}\left(\frac{4}{7}\right)^{18}$

$\frac{1710}{49}\left(\frac{4}{7}\right)^{18}$

$\frac{1726}{49}\left(\frac{4}{7}\right)^{18}$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

-0

A random variable has the following probability distribution

$\mathrm{X:}$	0	1	2	3	4	5	6	7
$\mathrm{P}(x):$	0	$\mathrm{2p}$	$\mathrm{2p}$	$\mathrm{3p}$	$\mathrm{p^2}$	$\mathrm{2p^2}$	$\mathrm{7p^2}$	$\mathrm{2p}$

Then the value of p is

$\frac{1}{10}$

$\frac{1}{30}$

$\frac{1}{100}$

$\frac{3}{20}$

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

-0

Let A and B be two events such that the probability that exactly one of them occurs is $\frac{2}{5}$ and the probability that A or B occurs is $\frac{1}{2}$, then the probability of both of them occur together is

0.1

0.2

0.01

0.02

MHT CET 2024 4th May Morning Shift

MCQ (Single Correct Answer)

-0

A random variable $X$ has the following probability distribution

$\mathrm{X:}$	1	2	3	4	5
$\mathrm{P(X):}$	$\mathrm{k^2}$	$\mathrm{2k}$	$\mathrm{k}$	$\mathrm{2k}$	$\mathrm{5k^2}$

Then $\mathrm{P(X > 2)}$ is equal to

$\frac{7}{12}$

$\frac{23}{36}$

$\frac{1}{36}$

$\frac{1}{6}$

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