Probability · Mathematics · MHT CET

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MCQ (Single Correct Answer)

1

Three numbers are chosen at random from numbers 1 to 20 . The probability that they are consecutive is

MHT CET 2025 5th May Evening Shift
2

Two cards are drawn successively with replacement from fair playing 52 cards. let X denote number of kings obtained when two cards are drawn, then $\mathrm{E}\left(\mathrm{X}^2\right)=$

MHT CET 2025 5th May Evening Shift
3

If $\mathrm{X} \sim \mathrm{B}(35, \mathrm{p})$ such that $7 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$ then the value of $\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=20)}$ is equal to

MHT CET 2025 5th May Evening Shift
4

A student studies for X number of hours during a randomly selected school day. The probability that X can take the values, has the following form, where k is some constant.

$$ \mathrm{P}(\mathrm{X}=x)= \begin{cases}0.2, & \text { if } x=0 \\ \mathrm{k} x, & \text { if } x=1 \text { or } 2 \\ \mathrm{k}(6-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases} $$

The probability that the student studies for at most two hours is

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5

The probability that a non leap year selected at random will contain 52 Saturdays or 53 Sundays is

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6

A fair $n$ faced die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $\mathrm{n}=($ where $\mathrm{n} \in \mathbb{N})$

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7

A fair coin is tossed a fixed number of times. If the probability of getting 5 tails is same as the probability of getting 7 tails, then the probability of getting 3 tails is

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8

The probability that a person is not a sportsperson is $\frac{1}{6}$. Then the probability that out of the 6 members of the family, 5 are sportspersons is

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9

The cumulative distribution function of a discrete random variable X is

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & -4 & -2 & 0 & 2 & 4 & 6 & 8 & 10 \\ \hline \mathrm{~F}(\mathrm{X}=x) & 0.1 & 0.3 & 0.5 & 0.65 & 0.75 & 0.85 & 0.90 & 1 \\ \hline \end{array} $$

then $\frac{P(X \leqslant 0)}{P(X>0)}=$

MHT CET 2025 26th April Morning Shift
10

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

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11

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…

MHT CET 2025 26th April Morning Shift
12

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

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13

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

MHT CET 2025 25th April Evening Shift
14

A player tosses two coins. He wins ₹ 10 , if 2 heads appears, ₹ 5 , if one head appear and ₹ 2 if no head appears. Then variance of winning amount is

MHT CET 2025 25th April Evening Shift
15

Consider the probability distribution

$$ \begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}=x) & \mathrm{K} & 2 \mathrm{~K} & \mathrm{~K}^2 & 2 \mathrm{~K} & 5 \mathrm{~K}^2 \\ \hline \end{array} $$

Then the value of $\mathrm{P}(\mathrm{X}>2)$ is

MHT CET 2025 25th April Evening Shift
16

Let X denote the number of hours you study on a Sunday. It is known that

$$ \mathrm{P}(\mathrm{X}=x)=\left\{\begin{array}{cc} 0.1 & , \text { if } x=0 \\ \mathrm{k} x & , \text { if } x=1 \text { or } 2 \\ \mathrm{k}(5-x) & , \text { if } x=3 \text { or } 4 \\ 0 & , \text { otherwise } \end{array}\right. $$

where k is constant. Then the probability that you study at least two hours on a Sunday is

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17

A pair of fair dice is thrown 4 times. If getting the same number on both dice is considered as a success, then the probability of two successes are

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18

A family has 3 children. The probability that all the three children are girls, given that at least one of them is a girl is

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19

If a random variable X has the following probability distribution of X

$$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0 & \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k}^2 & 2 \mathrm{k}^2 & 7 \mathrm{k}^2+\mathrm{k} \\ \hline \end{array} $$

Then $P(x \geq 6)=$

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20

If $X \sim B\left(6, \frac{1}{2}\right)$, then $P(|X-2| \leqslant 1)=$

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21

A random variable X has following p.d.f. $\mathrm{f}(x)=\mathrm{kx}(1-x), 0 \leqslant x \leqslant 1 \quad$ and $\quad \mathrm{P}(x>\mathrm{a})=\frac{20}{27}$, then $\mathrm{a}=$

MHT CET 2025 23rd April Evening Shift
22

If A and B are independent events such that $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\frac{3}{25}$ and $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{8}{25}$, then $P(A)=$

MHT CET 2025 23rd April Evening Shift
23

A fair coin is tossed 100 times. The chance of getting a head even number of times is

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24

In a game a man wins $Rs \,\, 40$ if he gets 5 or 6 on a throw of a fair die and loses ₹ 20 for getting any other number on the die. If he decides to throw the die either till he gets a five or six or to a maximum of three throws, then his expected gain/loss (in rupees) is

MHT CET 2025 23rd April Morning Shift
25

Bag I contains 3 red and 2 green balls and Bag II contains 5 red and 3 green balls. A ball is drawn from one of the bag at random and it is found to be green. Then the probability that it is drawn from Bag I is

MHT CET 2025 23rd April Morning Shift
26

If a random variable $X$ has p.d.f. $f(x)=\left\{\begin{array}{ll}\frac{a x^2}{2}+b x & , \text { if } 1 \leqslant x \leqslant 3 \\ 0 & , \text { otherwise }\end{array}\right.$ and $f(2)=2$, then the values of $a$ and $b$ are, respectively

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27
Three urns respectively contain 2 white and 3 black, 3 white and 2 black and 1 white and 4 black balls. If one ball is drawn from each um, then the probability that the selection contains 1 black and 2 white balls is
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28

In a box containing 100 apples, 10 are defective. The probability that in a sample of 6 apples, 3 are defective is

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29

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

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30

The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most 2 components out of 4 tested, will survive is

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31

A coin is tossed until one head appears or a tail appears 4 times in succession. The probability distribution of the number of tosses is

MHT CET 2025 22nd April Morning Shift
32

The p.d.f. of a continuous random variable X is $f(x)=\left\{\begin{array}{cl}\frac{x^2}{18} & , \text { if }-3 < x < 3 \\ 0 & \text { otherwise }\end{array}\right.$

Then $\mathrm{P}[|\mathrm{X}|<2]=$

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33

In a single toss of a fair die, the odds against the event that number 4 or 5 turns up is

MHT CET 2025 22nd April Morning Shift
34

Numbers are selected at random, one at a time from the two-digit numbers $00,01,02,-------, 99$ with replacement. An event E occurs only if the product of the two digits of a selected number is 24. If four numbers are selected, then probability, that the event E occurs at least 3 times, is

MHT CET 2025 21st April Evening Shift
35

A random variable, $X$ has p.m.f. $\mathrm{P}(\mathrm{X}=x)=\frac{{ }^4 \mathrm{C}_x}{2^4}, x=0,1,2,3,4$ and $\mu$ and $\sigma^2$ are mean and variance respectively of random variable X , then

MHT CET 2025 21st April Evening Shift
36

$$ \text { The c.d.f. of a discrete random variable } \mathrm{X} \text { is } $$

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & -3 & -1 & 0 & 1 & 3 & 5 & 7 & 9 \\ \hline \mathrm{~F}(\mathrm{X}=x) & 0.1 & 0.3 & 0.5 & 0.65 & 0.75 & 0.85 & 0.90 & 1 \\ \hline \end{array} $$

Then $\frac{P[X=-3]}{P[X<0]}=$

MHT CET 2025 21st April Evening Shift
37

If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are mutually exclusive and exhaustive events of a sample space $S$ such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then $P(A)=$

MHT CET 2025 21st April Evening Shift
38

If a random variable X follows the Binomial distribution $B(10, \quad p)$ such that $5 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$, then the value of $\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=6)}$ is equal to

MHT CET 2025 21st April Morning Shift
39

The following is the probability distribution of X

$$ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1+\mathrm{p}}{5} & \frac{2-2 \mathrm{p}}{5} & \frac{2-\mathrm{p}}{5} & \frac{2 \mathrm{p}}{5} \\ \hline \end{array} $$

$$ \text { For a minimum value of } p \text {, the value of } 5 E(X) \text { is } $$

MHT CET 2025 21st April Morning Shift
40

A random variable X takes values $0,1,2,3$, ........ with probabilities. $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}$ is a constant, then $P(X=1)=$

MHT CET 2025 21st April Morning Shift
41

The probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two 'I's do not come together is

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42

Two numbers are selected at random, without replacement from the first 6 positive integers. Let $X$ denote the larger of the two numbers. Then $\mathrm{E}(\mathrm{X})=$

MHT CET 2025 20th April Evening Shift
43

For $\mathrm{k}=1,2,3$ the box $\mathrm{B}_{\mathrm{k}}$ contains k red balls and $(k+1)$ white balls. Let $P\left(B_1\right)=\frac{1}{2}, P\left(B_2\right)=\frac{1}{3}$ and $\mathrm{P}\left(\mathrm{B}_3\right)=\frac{1}{6} . \mathrm{A}$ box is selected at random and a ball is drawn from it. If a red ball is drawn from it, then the probability that it comes from box $\mathrm{B}_2$ is

MHT CET 2025 20th April Evening Shift
44

A random variable $X$ takes the values $0,1,2,3$, $\qquad$ with probability

$\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant.

Then $\mathrm{P}(\mathrm{X}=0)$ is

MHT CET 2025 20th April Evening Shift
45

A fair coin is tossed 99 times. If X is the number of times head occur then $\mathrm{P}[\mathrm{X}=\mathrm{r}]$ is maximum when $\mathrm{r}=$

MHT CET 2025 20th April Evening Shift
46

If X is a binomial variable with range $\{0,1,2,3,4\}$ and $\mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=4)$ then the parameter ' $p$ ' of the binomial distribution is

MHT CET 2025 20th April Morning Shift
47

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. If X is the random variable of getting queens, then the value of $2 E(X)+3 E\left(X^2\right)$ for the number of queens is

MHT CET 2025 20th April Morning Shift
48

A random variable $X$ has the following probability distribution

$$ \begin{array}{|l|c|c|c|c|c|} \hline \mathrm{X}: & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}): & \mathrm{k} & 2 \mathrm{k} & 4 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\ \hline \end{array} $$

then the value of $\mathrm{P}(1 \leqslant \mathrm{X}<4 \mid \mathrm{X} \leqslant 2)=$

MHT CET 2025 20th April Morning Shift
49

If two numbers $p$ and $q$ are chosen randomly from the set $\{1,2,3,4\}$, one by one, with replacement, then the probability of getting $\mathrm{p}^2 \geq 4 \mathrm{q}$ is

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50

If $X \sim B(n, p)$ then $\frac{P(X=k)}{P(X=k-1)}=$

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51

Let X be a discrete random variable. The probability distribution of X is given below

$$ \begin{array}{|c|c|c|c|} \hline \mathrm{X} & 30 & 10 & -10 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \mathrm{~A} & \mathrm{~B} \\ \hline \end{array} $$

and $\mathrm{E}(\mathrm{X})=4$, then the value of AB is equal to

MHT CET 2025 19th April Evening Shift
52

In a game, 3 coins are tossed. A person is paid $Rs \, 150$ if he gets all heads or all tails and he is supposed to pay ₹50 if he gets one head or two heads. The amount he can expect to win / lose on an average per game in ₹ is

MHT CET 2025 19th April Evening Shift
53

Let $A$ and $B$ are independent events with $\mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{11}{20}$, then $\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)$ is root of the equation

MHT CET 2025 19th April Evening Shift
54
A box contains 9 tickets numbered 1 to 9 both inclusive. If 3 tickets are drawn from the box one at a time, then the probability that they are alternatively either {odd, even, odd} or {even, odd, even} is
MHT CET 2025 19th April Morning Shift
55

The probability distribution of a discrete random variable X is

$\mathrm{X}$ 0 1 2 3 4
$\mathrm{P(X=}x)$ $\mathrm{2k}$ $\mathrm{k}$ $\mathrm{2k}$ $\mathrm{4k}$ $\mathrm{k}$

If $\mathrm{a}=\mathrm{P}(x<3)$ and $\mathrm{b}=\mathrm{P}(2 \leq \mathrm{X}<4)$, then

MHT CET 2025 19th April Morning Shift
56
If a random variable $X$ has the p.d.f. $f(x)=\left\{\begin{array}{cc}\frac{\mathrm{k}}{x^2+1} & , \text { if } 0< x< \infty \\ 0 & , \text { otherwise }\end{array}\right.$ then c.d.f. of X is
MHT CET 2025 19th April Morning Shift
57
If a random variable $X$ follows the Binomial distribution $\mathrm{B}(33, \mathrm{p})$ such that $3 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$, then the variance of X is
MHT CET 2025 19th April Morning Shift
58

If a discrete random variable X is defined as follows

$\mathrm{P}[\mathrm{X}=x]=\left\{\begin{array}{cl}\frac{\mathrm{k}(x+1)}{5^x}, & \text { if } x=0,1,2 \ldots \ldots . \\ 0, & \text { otherwise }\end{array}\right.$

then $\mathrm{k}=$

MHT CET 2024 16th May Evening Shift
59

Numbers are selected at random, one at a time from two digit numbers $10,11,12 \ldots ., 99$ with replacement. An event $E$ occurs if and only if the product of the two digits of a selected number is 18 . If four numbers are selected, then probability that the event E occurs at least 3 times is

MHT CET 2024 16th May Evening Shift
60

Two friends A and B apply for a job in the same company. The probabilities of A getting selected is $\frac{2}{5}$ and that of B is $\frac{4}{7}$. Then the probability, that one of them is selected, is

MHT CET 2024 16th May Evening Shift
61

If a random variable X has the following probability distribution values

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

Then $P(X \geq 6)$ has the value

MHT CET 2024 16th May Morning Shift
62

A random variable X takes the values $0,1,2,3$ and its mean is 1.3 . If $\mathrm{P}(\mathrm{X}=3)=2 \mathrm{P}(\mathrm{X}=1)$ and $P(X=2)=0.3$, then $P(X=0)$ is

MHT CET 2024 16th May Morning Shift
63

Three persons $\mathrm{P}, \mathrm{Q}$ and R independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by P or Q but not by $R$, is

MHT CET 2024 16th May Morning Shift
64

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one by one, with replacement, then the variance of the number of green balls drawn is

MHT CET 2024 16th May Morning Shift
65
 

Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads, then probability distribution of x is

MHT CET 2024 15th May Evening Shift
66

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five is

MHT CET 2024 15th May Evening Shift
67

If two fair dice are rolled, then the probability that the sum of the numbers on the upper faces is at least 9, is

MHT CET 2024 15th May Evening Shift
68

For the probability distribution

$x :$ 0 1 2 3 4 5
$p(x):$ $\mathrm{k}$ 0.3 0.15 0.15 0.1 2$\mathrm{k}$

The expected value of X is

MHT CET 2024 15th May Morning Shift
69

A random variable X has the following probability distribution

$X$ 1 2 3 4 5
$p(x)$ $\mathrm{k^2}$ $\mathrm{2k}$ $\mathrm{k}$ $\mathrm{2k}$ $\mathrm{5k^2}$

Then $\mathrm{p}(x \geq 2)$ is equal to

MHT CET 2024 15th May Morning Shift
70

The probability, that a year selected at random will have 53 Mondays, is

MHT CET 2024 15th May Morning Shift
71

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

MHT CET 2024 15th May Morning Shift
72

A random variable $X$ has the following probability distribution

$X=x$ 1 2 3 4 5 6 7 8
$P(X=x)$ 0.15 0.23 0.10 0.12 0.20 0.08 0.07 0.05

For the event $E=\{X$ is a prime number $\}$, $F=\{X<4\}$, then $P(E \cup F)$ is

MHT CET 2024 11th May Evening Shift
73

Let $\mathrm{A}, \mathrm{B}$ and C be three events, which are pairwise independent and $\bar{E}$ denote the complement of an event E . If $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$ and $\mathrm{P}(\mathrm{C})>0$, then $\mathrm{P}((\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) / C)$ is equal to

MHT CET 2024 11th May Evening Shift
74

A random variable x takes the values $0,1,2$, $3, \ldots$ with probability $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x$, where k is a constant, then $\mathrm{P}(\mathrm{X}=0)$ is

MHT CET 2024 11th May Evening Shift
75

One hundred identical coins, each with probability p , of showing up heads are tossed once. If $0<\mathrm{p}<1$ and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of $p$ is

MHT CET 2024 11th May Evening Shift
76

The p.m.f. of a random variable X is given by

$$\begin{aligned} \mathrm{P}[\mathrm{X}=x] & =\frac{\binom{5}{x}}{2^5}, \text { if } x=0,1,2,3,4,5 \\ & =0, \text { otherwise } \end{aligned}$$

Then which of the following is not correct?

MHT CET 2024 11th May Morning Shift
77

If three fair coins are tossed, then variance of number of heads obtained, is

MHT CET 2024 11th May Morning Shift
78

If $A$ and $B$ are two independent events such that $\mathrm{P}\left(\mathrm{A}^{\prime}\right)=0.75, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.65$ and $\mathrm{P}(\mathrm{B})=\mathrm{p}$, then value of $p$ is

MHT CET 2024 11th May Morning Shift
79

The probability that a person who undergoes a bypass surgery will recover is 0.6 . the probability that of the six patients who undergo similar operations, half of them will recover is __________.

MHT CET 2024 11th May Morning Shift
80

$A$ and $B$ are independent events with $P(A)=\frac{3}{10}$, $\mathrm{P}(\mathrm{B})=\frac{2}{5}$, then $\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)$ has the value

MHT CET 2024 10th May Evening Shift
81

Minimum number of times a fair coin must be tossed, so that the probability of getting at least one head, is more than $99 \%$ is

MHT CET 2024 10th May Evening Shift
82

A random variable X assumes values $1,2,3, \ldots \ldots ., \mathrm{n}$ with equal probabilities. If $\operatorname{var}(X): E(X)=4: 1$, then $n$ is equal to

MHT CET 2024 10th May Evening Shift
83

In a game, 3 coins are tossed. A person is paid ₹ 100$, if he gets all heads or all tails; and he is supposed to pay ₹ 40 , if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

MHT CET 2024 10th May Morning Shift
84

In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability, of getting exactly 4 successes, is

MHT CET 2024 10th May Morning Shift
85

A random variable X has the following probability distribution

$\mathrm{X}$ 1 2 3 4 5 6 7 8
$\mathrm{P(X=}x)$ 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05

For the events $\mathrm{E}=\{\mathrm{X}$ is prime number $\}$

$$\mathrm{F}=\{\mathrm{X}<4\}$$

Then $P(E \cup F)=$

MHT CET 2024 10th May Morning Shift
86
 

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

MHT CET 2024 10th May Morning Shift
87

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

MHT CET 2024 9th May Evening Shift
88

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

MHT CET 2024 9th May Evening Shift
89

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

MHT CET 2024 9th May Evening Shift
90

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

MHT CET 2024 9th May Evening Shift
91

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

MHT CET 2024 9th May Morning Shift
92

A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 \mathrm{p}}{4}, \frac{1-\mathrm{p}}{4}, \frac{1+2 \mathrm{p}}{4}, \frac{1-4 \mathrm{p}}{4}$ respectively, where p varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of X are respectively.

MHT CET 2024 9th May Morning Shift
93

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

MHT CET 2024 9th May Morning Shift
94

If $\mathrm{P}(\mathrm{X}=2)=0.3, \mathrm{P}(\mathrm{X}=3)=0.4, \mathrm{P}(\mathrm{X}=4)=0.3$, then the variance of random variable X is

MHT CET 2024 4th May Evening Shift
95

A man and his wife appear for an interview for two posts. The probability of the husband's selection is $\frac{1}{7}$ and that of the wife's selection is $\frac{1}{5}$. If they appear for the interview independently, then the probability that only one of them is selected, is

MHT CET 2024 4th May Evening Shift
96

The expected value of the sum of the two numbers obtained on the uppermost faces, when two fair dice are rolled, is

MHT CET 2024 4th May Evening Shift
97

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

MHT CET 2024 4th May Evening Shift
98

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

MHT CET 2024 4th May Morning Shift
99

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

MHT CET 2024 4th May Morning Shift
100

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

MHT CET 2024 4th May Morning Shift
101

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability, that a student will get 4 or more correct answers just by guessing, is

MHT CET 2024 4th May Morning Shift
102

Let a random variable X have a Binomial distribution with mean 8 and variance 4 . If $\mathrm{P}(x \leqslant 2)=\frac{\mathrm{k}}{2^{16}}$, then k is equal to

MHT CET 2024 3rd May Evening Shift
103

For the probability distribution

$\mathrm{X:}$ $-2$ $-1$ $0$ $1$ $2$ $3$
$\mathrm{p}(x):$ 0.1 0.2 0.2 0.3 0.15 0.05

Then the $\operatorname{Var}(\mathrm{X})$ is

(Given : $$\left.(0.25)^2=0.0625,(0.35)^2=0.1225,(0.45)^2=0.2025\right)$$

MHT CET 2024 3rd May Evening Shift
104

Two cards are drawn successively with replacement from a well- shuffled pack of 52 cards. Let X denote the random variable of number of kings obtained in the two drawn cards. Then $\mathrm{P}(x=1)+\mathrm{P}(x=2)$ equals

MHT CET 2024 3rd May Evening Shift
105

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then mean of number of kings is

MHT CET 2024 3rd May Evening Shift
106

A random variable x has the following probability distribution. Then value of $k$ is _________ and $\mathrm{P}(3< x \leq 6)$ has the value

$\mathrm{X}=x$ 0 1 2 3 4 5 6 7 8
$\mathrm{P}(x)$ $\mathrm{k}$ $\mathrm{2k}$ $\mathrm{3k}$ $\mathrm{4k}$ $\mathrm{4k}$ $\mathrm{3k}$ $\mathrm{2k}$ $\mathrm{k}$ $\mathrm{k}$

MHT CET 2024 3rd May Morning Shift
107

Let $\mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{2}\right)$, then $\mathrm{P}[|x-4| \leqslant 2]$ is

MHT CET 2024 3rd May Morning Shift
108

A person throws an unbiased die. If the number shown is even, he gains an amount equal to the number shown. If the number is odd, he loses an amount equal to the number shown. Then his expectation is ₹.

MHT CET 2024 3rd May Morning Shift
109

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Let X denote the random variable of number of jacks obtained in the two drawn cards. Then $P(X=1)+P(X=2)$ equals

MHT CET 2024 3rd May Morning Shift
110

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to

MHT CET 2024 2nd May Evening Shift
111

The probability distribution of a random variable X is given by

$\mathrm{X=}x_i$: 0 1 2 3 4
$\mathrm{P(X=}x_i)$ : 0.4 0.3 0.1 0.1 0.1

Then the variance of X is

MHT CET 2024 2nd May Evening Shift
112

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three persons apply for the same house is

MHT CET 2024 2nd May Evening Shift
113

A bag contains 4 Red and 6 Black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with 3 additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red is

MHT CET 2024 2nd May Morning Shift
114

If a discrete random variable X takes values $0,1,2,3, \ldots \ldots$. with probability $\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 5^{-x}$, where k is a constant, then $\mathrm{P}(\mathrm{X}=0)$ is

MHT CET 2024 2nd May Morning Shift
115

Ten bulbs are drawn successively, with replacement, from a lot containing $10 \%$ defective bulbs, then the probability that there is at least one defective bulb, is

MHT CET 2024 2nd May Morning Shift
116

A fair die with numbers 1 to 6 on their faces is thrown. Let $$\mathrm{X}$$ denote the number of factors of the number, on the uppermost face, then the probability distribution of $$\mathrm{X}$$ is

MHT CET 2023 14th May Evening Shift
117

The p.m.f. of a random variable $$\mathrm{X}$$ is $$\mathrm{P}(x)=\left\{\begin{array}{cl}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \mathrm{n} \\ 0, & \text { otherwise }\end{array}\right.$$, then $$\mathrm{E}(\mathrm{X})$$ is

MHT CET 2023 14th May Evening Shift
118

There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is a negative number, is

MHT CET 2023 14th May Evening Shift
119

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs are selected at random from the lot and are sent to retail store. Then the probability that the store will receive at most one defective bulb is

MHT CET 2023 14th May Evening Shift
120

Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens is

MHT CET 2023 14th May Morning Shift
121

For an initial screening of an entrance exam, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $$\frac{4}{5}$$, then the probability, that he is unable to solve less than two problems, is

MHT CET 2023 14th May Morning Shift
122

$$\text { If } f(x)= \begin{cases}3\left(1-2 x^2\right) & ; 0< x < 1 \\ 0 & ; \text { otherwise }\end{cases}$$ is a probability density function of $$\mathrm{X}$$, then $$\mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}\right)$$ is

MHT CET 2023 14th May Morning Shift
123

Three critics review a book. For the three critics the odds in favour of the book are $$2: 5, 3: 4$$ and $$4: 3$$ respectively. The probability that the majority is in favour of the book, is given by

MHT CET 2023 14th May Morning Shift
124

Two dice are rolled. If both dice have six faces numbered $$1,2,3,5,7,11$$, then the probability that the sum of the numbers on upper most face is prime, is

MHT CET 2023 13th May Evening Shift
125

A random variable $$X$$ has the probability distribution

$$X=x$$ 1 2 3 4 5 6 7 8
$$P(X=x)$$ 0.15 0.23 0.12 0.20 0.08 0.10 0.05 0.07

For the events $$E=\{X$$ is a prime number $$\}$$ and $$F=\{x<5\}, P(E U F)$$ is

MHT CET 2023 13th May Evening Shift
126

A random variable $$X$$ has the following probability distribution

$$\mathrm{X}=x$$ 0 1 2
$$\mathrm{P(X}=x)$$ $$\mathrm{4k-10k^2}$$ $$\mathrm{5k-1}$$ $$\mathrm{3k^3}$$

then P(X < 2) is

MHT CET 2023 13th May Morning Shift
127

Let $$\mathrm{X}$$ be random variable having Binomial distribution $$B(7, p)$$. If $$P[X=3]=5 P[X=4]$$, then variance of $$\mathrm{X}$$ is

MHT CET 2023 13th May Morning Shift
128

If a continuous random variable $$\mathrm{X}$$ has probability density function $$\mathrm{f}(x)$$ given by

$$f(x)=\left\{\begin{array}{cl} a x & , \text { if } 0 \leq x<1 \\ a & , \text { if } 1 \leq x<2 \\ 3 a-a x & , \text { if } 2 \leq x \leq 3 \\ 0 & , \text { otherwise } \end{array}\right.$$,

then a has the value

MHT CET 2023 13th May Morning Shift
129

A card is drawn at random from a well shuffled pack of 52 cards. The probability that it is black card or face card is

MHT CET 2023 13th May Morning Shift
130

An irregular six faced die is thrown and the probability that, in 5 throws it will give 3 even numbers is twice the probability that it will give 2 even numbers. The number of times, in 6804 sets of 5 throws, you expect to give no even number is

MHT CET 2023 12th May Evening Shift
131

A box contains 100 tickets numbered 1 to 100 . A ticket is drawn at random from the box. Then the probability, that number on the ticket is a perfect square, is

MHT CET 2023 12th May Evening Shift
132

Three fair coins with faces numbered 1 and 0 are tossed simultaneously. Then variance (X) of the probability distribution of random variable $$\mathrm{X}$$, where $$\mathrm{X}$$ is the sum of numbers on the upper most faces, is

MHT CET 2023 12th May Evening Shift
133

The p.m.f of random variate $$\mathrm{X}$$ is $$P(X)= \begin{cases}\frac{2 x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3, \ldots \ldots, \mathrm{n} \\ 0, & \text { otherwise }\end{cases}$$

Then $$\mathrm{E}(\mathrm{X})=$$

MHT CET 2023 12th May Morning Shift
134

An experiment succeeds twice as often as it fails. Then the probability, that in the next 6 trials there will be atleast 4 successes, is

MHT CET 2023 12th May Morning Shift
135

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then the probability distribution of number of jacks is

MHT CET 2023 12th May Morning Shift
136

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is

MHT CET 2023 12th May Morning Shift
137

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Then mean of number of tens is

MHT CET 2023 11th May Evening Shift
138

A fair die is tossed twice in succession. If $$\mathrm{X}$$ denotes the number of fours in two tosses, then the probability distribution of $$\mathrm{X}$$ is given by

MHT CET 2023 11th May Evening Shift
139

If $$\mathrm{A}$$ and $$\mathrm{B}$$ are two events such that $$\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}$$, then the value of $$\mathrm{P}\left(\mathrm{A}^{\prime} / \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime} / \mathrm{A}^{\prime}\right)$$ is

MHT CET 2023 11th May Evening Shift
140

Let a random variable $$\mathrm{X}$$ have a Binomial distribution with mean 8 and variance 4. If $$\mathrm{P}(\mathrm{X} \leq 2)=\frac{\mathrm{K}}{2^{16}}$$, then $$\mathrm{K}$$ is

MHT CET 2023 11th May Evening Shift
141

From a lot of 20 baskets, which includes 6 defective baskets, a sample of 2 baskets is drawn at random one by one without replacement. The expected value of number of defective basket is

MHT CET 2023 11th May Morning Shift
142

Three of six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, equals ___________.

MHT CET 2023 11th May Morning Shift
143

A binomial random variable $$\mathrm{X}$$ satisfies $$9. p(X=4)=p(X=2)$$ when $$n=6$$. Then $$p$$ is equal to

MHT CET 2023 11th May Morning Shift
144

The three ships namely A, B and C sail from India to Africa. If the odds in favour of the ships reaching safely are $$2: 5,3: 7$$ and $$6: 11$$ respectively, then probability of all of them arriving safely is

MHT CET 2023 10th May Evening Shift
145

If the sum of mean and variance of a Binomial Distribution is $$\frac{15}{2}$$ for 10 trials, then the variance is

MHT CET 2023 10th May Evening Shift
146

In a game, 3 coins are tossed. A person is paid ₹ 7 /-, if he gets all heads or all tails; and he is supposed to pay ₹ 3 /-, if he gets one head or two heads. The amount he can expect to win on an average per game is ₹

MHT CET 2023 10th May Evening Shift
147

A fair die is tossed twice in succession. If $$\mathrm{X}$$ denotes the number of sixes in two tosses, then the probability distribution of $$\mathrm{X}$$ is given by

MHT CET 2023 10th May Evening Shift
148

For a binomial variate $$\mathrm{X}$$ with $$\mathrm{n}=6$$ if $$P(X=4)=\frac{135}{2^{12}}$$, then its variance is

MHT CET 2023 10th May Morning Shift
149

The p.d.f. of a discrete random variable is defined as $$\mathrm{f}(x)=\left\{\begin{array}{l} \mathrm{k} x^2, 0 \leq x \leq 6 \\ 0, \text { otherwise } \end{array}\right.$$

Then the value of $$F(4)$$ (c.d.f) is

MHT CET 2023 10th May Morning Shift
150

A player tosses 2 fair coins. He wins ₹5 if 2 heads appear, ₹ 2 if one head appears and ₹ 1 if no head appears. Then the variance of his winning amount in ₹ is :

MHT CET 2023 10th May Morning Shift
151

Three critics review a book. For the three critics the odds in favor of the book are $$2: 5, 3: 4$$ and $$4: 3$$ respectively. The probability that the majority is in favor of the book, is given by

MHT CET 2023 10th May Morning Shift
152

A man takes a step forward with probability 0.4 and backwards with probability 0.6 . The probability that at the end of eleven steps, he is one step away from the starting point is

MHT CET 2023 9th May Evening Shift
153

A problem in statistics is given to three students A, B and C. Their probabilities of solving the problem are $$\frac{1}{2}, \frac{1}{3}$$ and $$\frac{1}{4}$$ respectively. If all of them try independently, then the probability, that problem is solved, is

MHT CET 2023 9th May Evening Shift
154

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards, then mean of number of queens is

MHT CET 2023 9th May Evening Shift
155

In a Binomial distribution with $$\mathrm{n}=4$$, if $$2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$$, then the variance is

MHT CET 2023 9th May Morning Shift
156

$$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ are three events, one of which must and only one can happen. The odds in favor of $$\mathrm{A}$$ are $$4: 6$$, the odds against $$B$$ are $$7: 3$$. Thus, odds against $$\mathrm{C}$$ are

MHT CET 2023 9th May Morning Shift
157

The probability mass function of random variable X is given by

$$P[X=r]=\left\{\begin{array}{ll} \frac{{ }^n C_r}{32}, & n, r \in \mathbb{N} \\ 0, & \text { otherwise } \end{array} \text {, then } P[X \leq 2]=\right.$$

MHT CET 2023 9th May Morning Shift
158

Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance Var (X) of the probability distribution of random variable $$\mathrm{X}$$, where $$\mathrm{X}$$ is the sum of numbers on the uppermost faces, is

MHT CET 2023 9th May Morning Shift
159

The incidence of occupational disease in an industry is such that the workmen have a $$10 \%$$ chance of suffering from it. The probability that out of 5 workmen, 3 or more will contract the disease is

MHT CET 2022 11th August Evening Shift
160

If $$P(A \cup B)=0.7, P(A \cap B)=0.2$$, then $$P\left(A^{\prime}\right)+P\left(B^{\prime}\right)$$ is

MHT CET 2022 11th August Evening Shift
161

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then mean of number of kings is

MHT CET 2022 11th August Evening Shift
162

The probability that at least one of the events $$E_1$$ and $$E_2$$ occurs is 0.6. If the simultaneous occurrence of $$\mathrm{E}_1$$ and $$\mathrm{E}_2$$ is $$0.2, \mathrm{P}\left(\mathrm{E}_1^{\prime}\right)+\mathrm{P}\left(\mathrm{E}_2^{\prime}\right)=$$

MHT CET 2021 24th September Evening Shift
163

Two dice are thrown simultaneously. If X denotes the number of sixes, then the expectation of X is

MHT CET 2021 24th September Evening Shift
164

The probability distribution of a random variable X is

$$\mathrm{X=x}$$ 1 2 3 ......... $$\mathrm{n}$$
$$\mathrm{P(X=x)}$$ $$\mathrm{\frac{1}{n}}$$ $$\mathrm{\frac{1}{n}}$$ $$\mathrm{\frac{1}{n}}$$ ......... $$\mathrm{\frac{1}{n}}$$

then Var(X) =

MHT CET 2021 24th September Evening Shift
165

A fair coin is tossed for a fixed number of times. If probability of getting 7 heads is equal to probability of getting 9 heads, then probability of getting 2 heads is

MHT CET 2021 24th September Evening Shift
166

If the probability distribution function of a random variable X is given as

$$\mathrm{X=x_i}$$ $$-2$$ $$-1$$ 0 1 2
$$\mathrm{P(X=x_i)}$$ 0.2 0.3 0.15 0.25 0.1

Then F(0) is equal to

MHT CET 2021 24th September Morning Shift
167

If $$\mathrm{P}(\mathrm{A})=\frac{3}{10}, \mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{5}$$, then $$\mathrm{P}(\mathrm{A} / \mathrm{B}) \times \mathrm{P}(\mathrm{B} / \mathrm{A})=$$

MHT CET 2021 24th September Morning Shift
168

The probability distribution of a discrete random variable X is

$$\mathrm{X}$$ 1 2 3 4 5 6
$$\mathrm{P(X)}$$ K 2K 3K 4K 5K 6K

Find the value of $$\mathrm{P}(2<\mathrm{X}<6)$$

MHT CET 2021 24th September Morning Shift
169

A die is thrown four times. The probability of getting perfect square in at least one throw is

MHT CET 2021 24th September Morning Shift
170

A man is known to speck truth 3 out of 4 times. He throws a die and reports that it is 6. Then the probability that it is actually 6 is

MHT CET 2021 23rd September Evening Shift
171

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is

MHT CET 2021 23rd September Evening Shift
172

Let two cards are drawn at random from a pack of 52 playing cards. Let X be the number of aces obtained. Then the value of E(X) is

MHT CET 2021 23rd September Evening Shift
173

A fair coin is tossed 100 times. The probability of getting a head for even number of times is

MHT CET 2021 23rd September Evening Shift
174

If the function defined by $$f(x)=K(x-x^2)$$ if $$0 < x < 1=0$$, otherwise is the p.d.f. of a r.v.X, then the value of $$P\left(X<\frac{1}{2}\right)$$ is

MHT CET 2021 23rd September Evening Shift
175

The probability distribution of the number of doublets in four throws of a pair of dice is given by

MHT CET 2021 23th September Morning Shift
176

For the probability distribution given by following

$$\mathrm{x}$$ 5 6 7 8 9 10 11
$$\mathrm{P(X=x)}$$ 0.07 0.2 0.3 $$\mathrm{k}$$ 0.07 0.04 0.02

Var(X) =

MHT CET 2021 23th September Morning Shift
177

A random variable X has the following probability distribution

$$x$$ 0 1 2 3 4 5 6 7 8
$$P(X=x)$$ K 2K 3K 4K 4K 3K 2K K K

Then $$\mathrm{P}(3<\mathrm{x} \leq 6)=$$

MHT CET 2021 23th September Morning Shift
178

Rooms in a hotel are numbered from 1 to 19. Rooms are allocated at random as guests arrive. The first guest to arrive is given a room which is a prime number. The probability that the second guest to arrive is given a room which is a prime number is

MHT CET 2021 23th September Morning Shift
179

The distribution function $$F(X)$$ of discrete random variable $$X$$ is given by

$$\mathrm{X}$$ 1 2 3 4 5 6
$$\mathrm{F (X=x)}$$ 0.2 0.37 0.48 0.62 0.85 1

Then $$\mathrm{P[X=4]+P[x=5]=}$$

MHT CET 2021 22th September Evening Shift
180

First bag contains 3 red and 5 black balls and second bag contains 6 red and 4 black balls. A ball is drawn from each bag. The probability that one ball is red and the other is black, is

MHT CET 2021 22th September Evening Shift
181

A fair coin is tossed 4 times. If $$X$$ is a random variable which indicates number of heads, then $$\mathrm{P}[\mathrm{X}<3]=$$

MHT CET 2021 22th September Evening Shift
182

If the mean and variance of a binomial distribution are 4 and 2 respectively, then probability of getting 2 heads is

MHT CET 2021 22th September Evening Shift
183

For two events $$\mathrm{A}$$ and $$\mathrm{B}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3}$$, then $$\mathrm{A}$$ and $$\mathrm{B}$$ are

MHT CET 2021 22th September Morning Shift
184

A random variable X has following distribution

$$\mathrm{X = x}$$ 1 2 3 4 5 6
$$\mathrm{P(X = x)}$$ k 3k 5k 7k 8k k

Then P (2 $$\le$$ x < 5) =

MHT CET 2021 22th September Morning Shift
185

A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails, then P (X = 1) =

MHT CET 2021 22th September Morning Shift
186

A random variable X $$\sim$$ B (n, p), if values of mean and variance of X are 18 and 12 respectively, then n =

MHT CET 2021 22th September Morning Shift
187

an urn contains 9 balls of which 3 are red, 4 are blue and 2 are green. Three balls are drawn at random from the urn. The probability that the three balls have difference colours is

MHT CET 2021 21th September Evening Shift
188

It is observed that $$25 \%$$ of the cases related to child labour reported to the police station are solved. If 6 new cases are reported, then the probability that at least 5 of them will be solved is

MHT CET 2021 21th September Evening Shift
189

In a meeting $$60 \%$$ of the members favour and $$40 \%$$ oppose a certain proposal. A member is selected at random and we take $$\mathrm{X}=0$$ if he opposed and $$\mathrm{X}=1$$ if he is in favour, then $$\operatorname{Var} \mathrm{X}=$$

MHT CET 2021 21th September Morning Shift
190

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs selected at random from the lot and sent to retain store, then the probability that the store will receive at most one defective bulb is

MHT CET 2021 21th September Morning Shift
191

A coin is tossed and a die is thrown. The probability that the outcome will be head or a number greater than 4 or both, is

MHT CET 2021 21th September Morning Shift
192

If $$\mathrm{X}$$ is a random variable with p.m.f. as follows.

$$\begin{aligned} \mathrm{P}(\mathrm{X}=\mathrm{x}) & =\frac{5}{16}, \mathrm{x}=0,1 \\ & =\frac{\mathrm{kx}}{48}, \mathrm{x}=2, \quad \text { then } \mathrm{E}(\mathrm{x})= \\ & =\frac{1}{4}, \mathrm{x}=3 \end{aligned}$$

MHT CET 2021 21th September Morning Shift
193

The p.m.f. of a random variable X is $$\mathrm{P(X = x) = {1 \over {{2^5}}}\left( {_x^5} \right),x = 0,1,2,3,4,5}=0$$ then

MHT CET 2021 20th September Evening Shift
194

The variance of the following probability distribution is,

MHT CET 2021 20th September Evening Shift Mathematics - Probability Question 216 English

MHT CET 2021 20th September Evening Shift
195

If the sum of mean and variance of a binomial distribution for 5 trials is 1.8, then probability of a success is

MHT CET 2021 20th September Evening Shift
196

Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is

MHT CET 2021 20th September Evening Shift
197

Two dice are rolled simultaneously. The probability that the sum of the two numbers on the dice is a prime number, is

MHT CET 2021 20th September Morning Shift
198

A random variable X has the following probability distribution

$$\mathrm{X=x}$$ 0 1 2 3 4 5 6 7
$$\mathrm{P[X=x]}$$ 0 $$\mathrm{k}$$ $$\mathrm{2k}$$ $$\mathrm{2k}$$ $$\mathrm{3k}$$ $$\mathrm{k^2}$$ $$\mathrm{2k^2}$$ $$\mathrm{7k^2+k}$$

then F(4) =

MHT CET 2021 20th September Morning Shift
199

Rajesh has just bought a VCR from Maharashtra Electronics and the shop offers after sales service contract for Rs. 1000 for the next five years. Considering the experience of VCR users, the following distribution of maintenance expenses for the next five years is formed.

Expenses 0 500 1000 1500 2000 2500 3000
Probability 0.35 0.25 0.15 0.10 0.08 0.05 0.02

The expected value of maintenance cost is :

MHT CET 2021 20th September Morning Shift
200

If $$X \sim B(4, p)$$ and $$P(X=0)=\frac{16}{81}$$, then $$P(X=4)=$$

MHT CET 2021 20th September Morning Shift
201

The odds in favour of getting sum multiple of 3 , when pair of dice are thrown is

MHT CET 2020 19th October Evening Shift
202

If $X$ is a.r.v. with c.d.f $F(x)$ and its probability distribution is given by

$X=x$ $-1.5$ $-0.5$ 0.5 1.5 2.5
$P(X=x)$ 0.05 0.2 0.15 0.25 0.35

then, $F(1.5)-F(-0.5)=$

MHT CET 2020 19th October Evening Shift
203

The odds in favour of drawing a king from a pack of 52 playing cards is

MHT CET 2020 16th October Evening Shift
204

Out of 100 people selected at random, 10 have common cold. If five persons selected at random from the group, then the probability that at most one person will have common cold is

MHT CET 2020 16th October Evening Shift
205

The pdf of a continuous r.v. $$X$$ is given by $$f(x)=\frac{x}{8}, 0 < x < 4=0$$, otherwise, then $$P(X \leq 2)$$ is

MHT CET 2020 16th October Evening Shift
206

If a die is thrown at random, then the expectation of the number on it is

MHT CET 2020 16th October Evening Shift
207

The probability that bomb will miss the target is 0.2. Then, the probability that out of 10 bombs dropped exactly 2 will hit the target is

MHT CET 2020 16th October Morning Shift
208

The letters of the word 'LOGARITHM' are arranged at random. The probability that arrangement starts with vowel and end with consonant is

MHT CET 2020 16th October Morning Shift
209

The p.d.f of c.r.v $$X$$ is given by $$f(x)=\frac{x+2}{18}$$, if $$-2

MHT CET 2020 16th October Morning Shift
210

If the p.m.f of a. r.v. $$X$$ is given by

$$P(X=x)=\frac{{ }^5 C_x}{2^5}$$

if $$x=0,1,2, \ldots \ldots . .5=0$$,

0 , otherwise,

then which of the following is not true?

MHT CET 2020 16th October Morning Shift
211

Let $X$ be the number of successes in ' $n$ ' independent Bernoulli trials with probability of success $p=\frac{3}{4}$. The least value of ' $n$ ' so that $P(X \geq 1) \geq 0.9375$ is ......

MHT CET 2019 3rd May Morning Shift
212

The probability that three cards drawn from a pack of 52 cards, all are red is

MHT CET 2019 3rd May Morning Shift
213

$$\begin{aligned} &\text { The pdf of a random variable } X \text { is }\\ &\begin{aligned} f(x) & =3\left(1-2 x^2\right), & & 0< x<1 \\ & =0 & & \text { otherwise } \end{aligned} \end{aligned}$$

The $P\left(\frac{1}{4}< x<\frac{1}{3}\right)=\ldots$

MHT CET 2019 3rd May Morning Shift
214

A player tosses 2 fair coins. He wins Rs. 5 if 2 heads appear, Rs. 2 If 1 head appear and Rs. 1 if no head appears, then variance of his winning amount is

MHT CET 2019 3rd May Morning Shift
215

In a bionomial distribution, mean is 18 and variance is 12 then $p=$ ...........

MHT CET 2019 2nd May Evening Shift
216

The p.d.f of a random variable $x$ is given by

$$\begin{aligned} & f(x)=\frac{1}{4 a}, \quad 00) \\ & =0 \text {, otherwise } \end{aligned}$$

and $P\left(x<\frac{3 a}{2}\right)=k P\left(x>\frac{5 a}{2}\right)$ then $k=$ ..............

MHT CET 2019 2nd May Evening Shift
217

If three dices are thrown then the probability that the sum of the numbers on their uppermost faces to be atleast 5 is

MHT CET 2019 2nd May Evening Shift
218

It is observed that $25 \%$ of the cases related to child labour reported to the police station are solved. If 6 new cases are reported, then the probability that atleast 5 of them will be solved is

MHT CET 2019 2nd May Morning Shift
219

A bag contains 6 white and 4 black balls. Two balls are drawn at random. The probability that they are of the same colour is ...........

MHT CET 2019 2nd May Morning Shift
220

A random variable X has following probability distribution

$X=x$ 1 2 3 4 5 6
$P(X=x)$ K 3K 5K 7K 8K K

Then $P(2 \leq X<5)=\ldots \ldots$

MHT CET 2019 2nd May Morning Shift
221

If the c.d.f (cumulative distribution function) is given by $F(x)=\frac{x-25}{10}$, then $P(27 \leq x \leq 33)=\ldots \ldots$

MHT CET 2019 2nd May Morning Shift