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A person throws two fair dice. He wins Rs. 15 for throwing a doublet

( same numbers on the two dice), wins Rs. 12 when the throw results

in the sum of 9 , and loses Rs. 6 for any other outcome on the throw.

Then the expected gain / loss (in Rs.) of the person is :

• Option 1)

$\frac{1}{2}$  gain

• Option 2)

$\frac{1}{4}$ loss

• Option 3)

$\frac{1}{2}$ loss

• Option 4)

2 gain

Option 2) 1/4 loss

For initial screening of an admission test, 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 problem is :

• Option 1)

$\frac{201}{5}\left ( \frac{1}{5} \right )^{49}$

• Option 2)

$\frac{316}{25}\left ( \frac{4}{5} \right )^{48}$

• Option 3)

$\frac{54}{5}\left ( \frac{4}{5} \right )^{49}$

• Option 4)

$\frac{164}{25}\left ( \frac{1}{5} \right )^{48}$

Probability of Sonning a  the problems out of 50 problems =   (not sonning ) =   The probability that he is unable to solve less than two problems is :  (zero correct ) +  (one correct) {using Binominal Probability)  Option 1)           Option 2) Option 3) Option 4)

If the data $x_{1},x_{2},\cdots ,x_{10}$ is such that the mean of first four of these is $11$, the mean of the remaining six is $16$ and the sum of squares of all of these is $2,000$ ; then the standard deviation of this data is :

• Option 1)

$4$

• Option 2)

$2\sqrt{2}$

• Option 3)

$\sqrt{2}$

• Option 4)

$2$

Variance   standard deviation                                Option 1)   Option 2)    Option 3)     Option 4)

iIf three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :

• Option 1)

$\frac{1}{5}$

• Option 2)

$\frac{3}{20}$

• Option 3)

$\frac{3}{10}$

• Option 4)

$\frac{1}{10}$

Only two equilateral triangle are possible  and    Option 1) Option 2) Option 3) Option 4)

Let a random variable X have a binomial distribution with mean $8$ and variance $4$. If $P\left ( X\leq 2 \right )=\frac{k}{2^{16}}$, then k is equal to :

• Option 1)

$137$

• Option 2)

$17$

• Option 3)

$121$

• Option 4)

$1$

So, the value of  Option 1) Option 2) Option 3) Option 4)

If both the mean and the standard deviation of 50 observations

$x_1,x_2,x_3,.................,x_{50}$ are equal to 16 ,  then the mean

of $(x_1-4)^{2},(x_2-4)^{2},............,(x_{50}-4)^{2}$ is :

• Option 1)

400

• Option 2)

380

• Option 3)

525

• Option 4)

480

50 observations    .....................(1) ......................(2) So, mean value of  Option 1) 400 Option 2) 380 Option 3) 525 Option 4) 480

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 :

• Option 1)

5

• Option 2)

6

• Option 3)

8

• Option 4)

7

So, probability to get atleast one head =      Minimum value of n will be = 7 So, option (4) is correct.  Option 1) 5 Option 2) 6 Option 3) 8 Option 4) 7

If for some $x\epsilon R$, the frequency distribution of the marks

obtained by 20 students in a test is :

 Marks 2 3 5 7 Frequency $(x+1)^{2}$ $2x-5$ $x^{2}-3x$ $x$

then the mean of the marks is :

• Option 1)

3.2

• Option 2)

3.0

• Option 3)

2.5

• Option 4)

2.8

Given that total students = 20 So,  =>  =>   cannot be -ve  So,  Put  correct option (4) Option 1) 3.2 Option 2) 3.0 Option 3) 2.5 Option 4) 2.8

Assume that each born child is equally likely to be a boy or a girl. If two

families have two children each, then the conditional probability that

all children are girls given that at least two are girls is:

• Option 1)

$\frac{1}{11}$

• Option 2)

$\frac{1}{10}$

• Option 3)

$\frac{1}{12}$

• Option 4)

$\frac{1}{17}$

There are 4 children  total number of ways in whcih atleast 2 girls are there Required probabilty =  Option (1) is correct. Option 1) Option 2) Option 3) Option 4)

The mean and the median of the following ten numbers in increasing order  $10,22,26,29,34,x,42,67,70,y$    are  $42\:\:and\:\:35$  respectively , then   $\frac{y}{x}$

is equal to  :

• Option 1)

$9/4$

• Option 2)

$7/2$

• Option 3)

$8/3$

• Option 4)

$7/3$

Option 1) Option 2) Option 3) Option 4)

Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac{1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit,is:

• Option 1)

$\frac{25}{192}$

• Option 2)

$\frac{7}{32}$

• Option 3)

$\frac{1}{192}$

• Option 4)

$\frac{25}{32}$

(target is hit)  (No one hit the target) Option 1) Option 2)   Option 3) Option 4)

If the standard deviation of the numbers  $-1,0,1,k$  is  $\sqrt{5}$ where  $k> 0,$ then $k$ is equal to :

• Option 1)

$2\sqrt{6}$

• Option 2)

$2\sqrt{\frac{10}{3}}$

• Option 3)

$4\sqrt{\frac{5}{3}}$

• Option 4)

$\sqrt{6}$

Option 1)                Option 2) Option 3) Option 4)

The minimum number of times one has to toss a fair coin so that the probability of observation at least one head is at least 90% is :

• Option 1)

$5$

• Option 2)

$3$

• Option 3)

$4$

• Option 4)

$2$

Let x fair coin tossed 'n' times the P( at least one head)   ( all tail)   Option 1) Option 2) Option 3) Option 4)

A student scores the following marks in five tests : 45,54,41,57,43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:

• Option 1)

$\frac{10}{\sqrt{3}}$

• Option 2)

$\frac{100}{3}$

• Option 3)

$\frac{10}{3}$

• Option 4)

$\frac{100}{\sqrt{3}}$

Let score of 6th test be x then,  x=48 correct option (i) Option 1) Option 2) Option 3) Option 4)

The mean and variance of seven observations are $8 \; and \; 16$, respectively. If $5$ of the observations are $2,4,10,12,14,$ then the product of the remaining two observations is :

• Option 1)

$49$

• Option 2)

$48$

• Option 3)

$45$

• Option 4)

$40$

Mean Variance Given Option 1) Option 2) Option 3)   Option 4)

Let $A$ and $B$ be two non-null events such that $A\subset B$. Then, which of the following statements is always correct ?

• Option 1)

$P(A\mid B)\leq P(A)$

• Option 2)

$P(A\mid B)\geqslant P(A)$

• Option 3)

$P(A\mid B)=1$

• Option 4)

$P(A\mid B)=P(B)-P(A)$

and non null events (given) Since Option 1) Option 2) Option 3)   Option 4)

In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs.50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

• Option 1)

$0$

• Option 2)

$\frac{400}{3}gain$

• Option 3)

$\frac{400}{9}loss$

• Option 4)

$\frac{400}{3}loss$

Multiplication Theorem of Probability - If A and B are any two events then    - wherein where      The law of Total Probability - Let S be the sample space and E1, E2, ......En be n mutually exclusive and exhaustive events associated with a random experiment. - wherein where A is any event which occurs with E1, E2, E3......En.   p (success)=   p ( 5 or 6 )  =1/3 Option 1)      Option...

Let S = {1,2,.........,20}. A subset B of S is said to be ''nice'' , if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ''nice'' is:

• Option 1)

$\frac{7}{2^{20}}$

• Option 2)

$\frac{4}{2^{20}}$

• Option 3)

$\frac{5}{2^{20}}$

• Option 4)

$\frac{6}{2^{20}}$

Probability of occurrence of an event - Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as  - wherein Where n repeated experiment and E occurs r times.               Possibilities of numbers that can be removed from  (7), (1,6) , (2,5) , (3,4) , (1,2,4)        Option 1)Option 2)Option 3)Option 4)

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $(\frac{mean \: of\: X}{standard \: deviation\: of\: X})$  is equal to:

• Option 1)

$\frac{4\sqrt3}{3}$

• Option 2)

4

• Option 3)

$3\sqrt2$

• Option 4)

$4\sqrt3$

ARITHMETIC Mean - For the values x1, x2, ....xn of the variant x the arithmetic mean is given by  in case of discrete data. Standard Deviation - In case of discrete frequency distribution    P(white ball) =  q=         and    n =  16 mean (X) = np =                           = 12 Standard deviation (X) =  Ans.   Option 1)  Option 2)  4Option 3)  Option 4)

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :

• Option 1)

$3$

• Option 2)

$1$

• Option 3)

$7$

• Option 4)

$5$

ARITHMETIC Mean - For the values x1, x2, ....xn of the variant x the arithmetic mean is given by  in case of discrete data. -     Variance - In case of discrete data  -    Option 1)Option 2)Option 3)Option 4)
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