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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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