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General
Easy

Question

If m and 2 s are the mean and variance of the random variable X, whose distribution is given by

  1. m equals sigma to the power of 2 end exponent equals 2    
  2. m equals 1 comma sigma to the power of 2 end exponent equals 2    
  3. m equals sigma to the power of 2 end exponent equals 1    
  4. m equals 2 comma sigma to the power of 2 end exponent equals 1    

The correct answer is: m equals sigma to the power of 2 end exponent equals 1

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Related Questions to study

General
Maths-

The distribution of a random variable X is given below

The value of k is

The distribution of a random variable X is given below

The value of k is

Maths-General
General
maths-

The number of different triangles formed by joining the points A, B, C, D, E, F and G as shown in the figure given below is

No. of different triangles equals 7 subscript C subscript 3 end subscript end subscript minus 5 subscript C subscript 3 end subscript end subscript minus 3 subscript C subscript 3 end subscript end subscript equals 24

The number of different triangles formed by joining the points A, B, C, D, E, F and G as shown in the figure given below is

maths-General
No. of different triangles equals 7 subscript C subscript 3 end subscript end subscript minus 5 subscript C subscript 3 end subscript end subscript minus 3 subscript C subscript 3 end subscript end subscript equals 24
General
maths-

The number of ways that the letters of the word ”PERSON” can be placed in the squares of the adjoining figure so that no row remains empty


The number of ways that the letters of the word ”PERSON” can be placed in the squares of the adjoining figure so that no row remains empty


maths-General
General
maths-

The value of open vertical bar table row cell 1 to the power of 2 end exponent end cell cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell row cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell row cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell cell 5 to the power of 2 end exponent end cell end table close vertical bar is

open vertical bar table row cell 1 to the power of 2 end exponent end cell cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell row cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell row cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell cell 5 to the power of 2 end exponent end cell end table close vertical bar{Operate R subscript 3 end subscript rightwards arrow R subscript 3 end subscript minus R subscript 2 end subscript,R subscript 2 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript}
= open vertical bar table row 1 4 9 row 3 5 7 row 5 7 9 end table close vertical bar equals 1 left parenthesis 45 minus 49 right parenthesis minus 4 left parenthesis 27 minus 35 right parenthesis plus 9 left parenthesis 21 minus 25 right parenthesis
= negative 4 plus 32 minus 36 equals negative 8.

The value of open vertical bar table row cell 1 to the power of 2 end exponent end cell cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell row cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell row cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell cell 5 to the power of 2 end exponent end cell end table close vertical bar is

maths-General
open vertical bar table row cell 1 to the power of 2 end exponent end cell cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell row cell 2 to the power of 2 end exponent end cell cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell row cell 3 to the power of 2 end exponent end cell cell 4 to the power of 2 end exponent end cell cell 5 to the power of 2 end exponent end cell end table close vertical bar{Operate R subscript 3 end subscript rightwards arrow R subscript 3 end subscript minus R subscript 2 end subscript,R subscript 2 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript}
= open vertical bar table row 1 4 9 row 3 5 7 row 5 7 9 end table close vertical bar equals 1 left parenthesis 45 minus 49 right parenthesis minus 4 left parenthesis 27 minus 35 right parenthesis plus 9 left parenthesis 21 minus 25 right parenthesis
= negative 4 plus 32 minus 36 equals negative 8.
General
maths-

The determinant open vertical bar table row a b cell a minus b end cell row b c cell b minus c end cell row 2 1 0 end table close vertical bar is equal to zero if a comma b comma c are in

On expanding, negative a left parenthesis b minus c right parenthesis plus 2 b left parenthesis b minus c right parenthesis plus left parenthesis a minus b right parenthesis left parenthesis b minus 2 c right parenthesis equals 0
Þ negative a b plus a c plus 2 b to the power of 2 end exponent minus 2 b c plus a b minus 2 a c minus b to the power of 2 end exponent plus 2 b c equals 0
Þ b to the power of 2 end exponent minus a c equals 0 Þ b to the power of 2 end exponent equals a c.

The determinant open vertical bar table row a b cell a minus b end cell row b c cell b minus c end cell row 2 1 0 end table close vertical bar is equal to zero if a comma b comma c are in

maths-General
On expanding, negative a left parenthesis b minus c right parenthesis plus 2 b left parenthesis b minus c right parenthesis plus left parenthesis a minus b right parenthesis left parenthesis b minus 2 c right parenthesis equals 0
Þ negative a b plus a c plus 2 b to the power of 2 end exponent minus 2 b c plus a b minus 2 a c minus b to the power of 2 end exponent plus 2 b c equals 0
Þ b to the power of 2 end exponent minus a c equals 0 Þ b to the power of 2 end exponent equals a c.
General
Maths-

If a not equal to 6 comma b comma c satisfy open vertical bar table row a cell 2 b end cell cell 2 c end cell row 3 b c row 4 a b end table close vertical bar equals 0 comma then a b c equals

A to the power of 3 end exponent equals I
Þ left parenthesis a minus 6 right parenthesis left parenthesis b to the power of 2 end exponent minus a c right parenthesis equals 0 rightwards double arrow b to the power of 2 end exponent minus a c equals 0, equals 3 open square brackets fraction numerator negative 1 plus square root of 3 i over denominator 2 end fraction minus fraction numerator negative 1 minus square root of 3 i over denominator 2 end fraction close square brackets equals 3 square root of 3 i
therefore a c equals b to the power of 2 end exponent rightwards double arrow a b c equals b to the power of 3 end exponent.

If a not equal to 6 comma b comma c satisfy open vertical bar table row a cell 2 b end cell cell 2 c end cell row 3 b c row 4 a b end table close vertical bar equals 0 comma then a b c equals

Maths-General
A to the power of 3 end exponent equals I
Þ left parenthesis a minus 6 right parenthesis left parenthesis b to the power of 2 end exponent minus a c right parenthesis equals 0 rightwards double arrow b to the power of 2 end exponent minus a c equals 0, equals 3 open square brackets fraction numerator negative 1 plus square root of 3 i over denominator 2 end fraction minus fraction numerator negative 1 minus square root of 3 i over denominator 2 end fraction close square brackets equals 3 square root of 3 i
therefore a c equals b to the power of 2 end exponent rightwards double arrow a b c equals b to the power of 3 end exponent.
General
chemistry-

End product of the following sequence of reaction is :

End product of the following sequence of reaction is :

chemistry-General
General
chemistry-

Arrange in the increasing order of boiling points:
i) 
ii) 
iii) 

iv) 

Arrange in the increasing order of boiling points:
i) 
ii) 
iii) 

iv) 

chemistry-General
General
maths-

Number of rectangles in fig. shown which are not squares is

Total number of rectangles
equals to the power of 7 C subscript 2 cross times to the power of 5 C subscript 2 minus No.of squares
equals 210 minus No. of squares
therefore Number of squares  equals 24 plus 15 plus 8 plus 3 equals 50
not stretchy downwards arrow space of 1em not stretchy downwards arrow space of 1em not stretchy downwards arrow  not stretchy downwards arrow
open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 1 cross times 1 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 2 cross times 2 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 3 cross times 3 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 4 cross times 4 end cell row cell text  area  end text end cell end table close parentheses

Number of rectangles in fig. shown which are not squares is

maths-General
Total number of rectangles
equals to the power of 7 C subscript 2 cross times to the power of 5 C subscript 2 minus No.of squares
equals 210 minus No. of squares
therefore Number of squares  equals 24 plus 15 plus 8 plus 3 equals 50
not stretchy downwards arrow space of 1em not stretchy downwards arrow space of 1em not stretchy downwards arrow  not stretchy downwards arrow
open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 1 cross times 1 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 2 cross times 2 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 3 cross times 3 end cell row cell text  area  end text end cell end table close parentheses open parentheses table attributes columnalign left columnspacing 1em end attributes row cell 4 cross times 4 end cell row cell text  area  end text end cell end table close parentheses
General
maths-

Digit at unit place of sum (1!)2 +(2!)2 +(3!)2 ……… +(2008!)2 is

becauseS = 1 + 4 + 36 + 576 + …….. +(2008!)2= 617 + all other terms will have 0 at unit place

Digit at unit place of sum (1!)2 +(2!)2 +(3!)2 ……… +(2008!)2 is

maths-General
becauseS = 1 + 4 + 36 + 576 + …….. +(2008!)2= 617 + all other terms will have 0 at unit place
General
maths-

A person writes letters to 6 friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least 4 of them are in wrong envelopes ?

no. of ways  = blank to the power of 6 straight C subscript 2.4 factorial times open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator 4 factorial end fraction close parentheses times plus to the power of 6 straight C subscript 1.5 factorial open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis minus fraction numerator 1 over denominator 5 factorial end fraction close parentheses times plus times to the power of 6 straight C subscript stack 0 with _ below with _ below.. end subscript times 6 factorial
open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator 6 factorial end fraction close parentheses
= 135 + 264 + 265 = 664

A person writes letters to 6 friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least 4 of them are in wrong envelopes ?

maths-General
no. of ways  = blank to the power of 6 straight C subscript 2.4 factorial times open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator 4 factorial end fraction close parentheses times plus to the power of 6 straight C subscript 1.5 factorial open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis minus fraction numerator 1 over denominator 5 factorial end fraction close parentheses times plus times to the power of 6 straight C subscript stack 0 with _ below with _ below.. end subscript times 6 factorial
open parentheses 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator 6 factorial end fraction close parentheses
= 135 + 264 + 265 = 664
General
maths-

The total numbers of integral solutions for (x,y,z) such that xyz = 24 is

24 can be broken as (1, 1, 24),(1, 2, 12),(1, 3, 8),(1, 4, 6),(2, 3, 4) & (2, 2, 6)
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same = blank cubed straight C subscript 2 times fraction numerator 3 not stretchy rightwards arrow over leftwards arrow factorial over denominator 2 factorial end fraction plus fraction numerator 3 factorial over denominator 2 factorial end fraction times equals times 12
total no. of  possible sets= 24 × 4 + 12 × 2 = 120

The total numbers of integral solutions for (x,y,z) such that xyz = 24 is

maths-General
24 can be broken as (1, 1, 24),(1, 2, 12),(1, 3, 8),(1, 4, 6),(2, 3, 4) & (2, 2, 6)
All sets in which each no. is diff = 3C2. 3! + 3!= 24
in which two nos. are same = blank cubed straight C subscript 2 times fraction numerator 3 not stretchy rightwards arrow over leftwards arrow factorial over denominator 2 factorial end fraction plus fraction numerator 3 factorial over denominator 2 factorial end fraction times equals times 12
total no. of  possible sets= 24 × 4 + 12 × 2 = 120
General
maths-

Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy

therefore5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.
So this can be done is
equals fraction numerator 5 straight C subscript 2 times cubed straight C subscript 1 cross times squared straight C subscript 1 cross times to the power of 1 straight C subscript 1 cross times 4 factorial over denominator 3 factorial end fraction times open square brackets table attributes columnalign left columnspacing 1em end attributes row cell text  asgroup  end text end cell row cell text  willbe  end text 2 comma 1 comma 1 comma 1 end cell end table close square brackets
= 10 × 3 × 2 × 4 = 240
Now one child can be rejected is 5 straight C subscript 1 = 5 ways
thereforeTotal ways = 5 × 240 = 1200

Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy

maths-General
therefore5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.
So this can be done is
equals fraction numerator 5 straight C subscript 2 times cubed straight C subscript 1 cross times squared straight C subscript 1 cross times to the power of 1 straight C subscript 1 cross times 4 factorial over denominator 3 factorial end fraction times open square brackets table attributes columnalign left columnspacing 1em end attributes row cell text  asgroup  end text end cell row cell text  willbe  end text 2 comma 1 comma 1 comma 1 end cell end table close square brackets
= 10 × 3 × 2 × 4 = 240
Now one child can be rejected is 5 straight C subscript 1 = 5 ways
thereforeTotal ways = 5 × 240 = 1200
General
maths-

A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are

Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,6 straight C subscript 3 times to the power of 4 straight C subscript 3
iii) If d = 8 then numbers possible = 7 C subscript 3 times to the power of 5 C subscript 3
iv) If d = 9 then numbers possible = 8 straight C subscript 3 times to the power of 6 straight C subscript 3
Add all cases

A seven digit number is in form of abcdefg (g, f, e, etc. are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are

maths-General
Cases : i) If d = 6 then seven digit numbers possible are = 5C3. 3C3
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0 can be used]
ii) If d = 7 then numbers possible =,6 straight C subscript 3 times to the power of 4 straight C subscript 3
iii) If d = 8 then numbers possible = 7 C subscript 3 times to the power of 5 C subscript 3
iv) If d = 9 then numbers possible = 8 straight C subscript 3 times to the power of 6 straight C subscript 3
Add all cases
General
maths-

Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-

First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
(where x1, x2, x3, x4 greater or equal than 0)
Its now one can get any number of balls
therefore non negative integral solution of
x1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3

Number of ways in which 25 identical balls can be distributed among Ram, Shyam, Sunder and Ghanshyam such that atleast 1, 2, 3, and 4 balls are given to Ram, Shyam, Sunder and Ghanshyam respectively, is-

maths-General
First of all select 1 + 2 + 3 + 4 = 10 balls out of 25 identical balls and distribute them as desired. It can happen only in one way. Now let x1, x2, x3 and x4 balls are given to them respectively.
(where x1, x2, x3, x4 greater or equal than 0)
Its now one can get any number of balls
therefore non negative integral solution of
x1 + x2 + x3 + x4 = 15
will be the number of ways so
15 + 4 –1C4–1 = 18C3