Maths-

General

Easy

### Question

#### The distribution of a random variable X is given below

The value of k is

### Hint:

We know that, **Sum of all the probabilities in an event = 1. **Use this to find out the answer.

#### The correct answer is:

#### We know that,

Sum of all the probabilities in an event = 1

P(x= -2) + P(x= -1) + P(x= 0) + P(x= 1) + P(x= 2) + P(x= 3) = 1

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### Related Questions to study

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

#### 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

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#### 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

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#### The value of is

{Operate ,}

=

= .

=

= .

#### The value of is

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{Operate ,}

=

= .

=

= .

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#### The determinant is equal to zero if are in

On expanding,

Þ

Þ Þ .

Þ

Þ Þ .

#### The determinant is equal to zero if are in

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On expanding,

Þ

Þ Þ .

Þ

Þ Þ .

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#### If satisfy then

Þ ,

#### If satisfy then

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Þ ,

chemistry-

#### End product of the following sequence of reaction is :

#### End product of the following sequence of reaction is :

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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)

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

Total number of rectangles

No.of squares

No. of squares

Number of squares

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No. of squares

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

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Total number of rectangles

No.of squares

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

S = 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

S = 1 + 4 + 36 + 576 + …….. +(2008!)

^{2}= 617 + all other terms will have 0 at unit placemaths-

#### 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 =

= 135 + 264 + 265 = 664

= 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 =

= 135 + 264 + 265 = 664

= 135 + 264 + 265 = 664

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 =

in which two nos. are same =

total no. of possible sets= 24 × 4 + 12 × 2 = 120

All sets in which each no. is diff =

^{3}C_{2}. 3! + 3!= 24in which two nos. are same =

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 =

in which two nos. are same =

total no. of possible sets= 24 × 4 + 12 × 2 = 120

All sets in which each no. is diff =

^{3}C_{2}. 3! + 3!= 24in which two nos. are same =

total no. of possible sets= 24 × 4 + 12 × 2 = 120

maths-

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

5 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

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total ways = 5 × 240 = 1200

So this can be done is

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total 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

5 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

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total ways = 5 × 240 = 1200

So this can be done is

= 10 × 3 × 2 × 4 = 240

Now one child can be rejected is = 5 ways

Total ways = 5 × 240 = 1200

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 =

[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 =,

iii) If d = 8 then numbers possible =

iv) If d = 9 then numbers possible =

Add all cases

^{5}C_{3}.^{3}C_{3}[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 =,

iii) If d = 8 then numbers possible =

iv) If d = 9 then numbers possible =

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 =

[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 =,

iii) If d = 8 then numbers possible =

iv) If d = 9 then numbers possible =

Add all cases

^{5}C_{3}.^{3}C_{3}[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 =,

iii) If d = 8 then numbers possible =

iv) If d = 9 then numbers possible =

Add all cases

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 x

(where x

Its now one can get any number of balls

non negative integral solution of

x

will be the number of ways so

_{1}, x_{2}, x_{3}and x_{4}balls are given to them respectively.(where x

_{1}, x_{2}, x_{3}, x_{4}0)Its now one can get any number of balls

non negative integral solution of

x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15will be the number of ways so

^{15 + 4 –1}C_{4–1}=^{18}C_{3}#### 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 x

(where x

Its now one can get any number of balls

non negative integral solution of

x

will be the number of ways so

_{1}, x_{2}, x_{3}and x_{4}balls are given to them respectively.(where x

_{1}, x_{2}, x_{3}, x_{4}0)Its now one can get any number of balls

non negative integral solution of

x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15will be the number of ways so

^{15 + 4 –1}C_{4–1}=^{18}C_{3}maths-

#### One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways of doing this is -

Let the children be A,B.C

A B C →19

Total number of ways

By termination

Permutation

A B C →19

Total number of ways

By termination

Permutation

#### One hundred identical marbles are to be distributed to three children so that each gets atlest 20 and no two get equal number. The number of ways of doing this is -

maths-General

Let the children be A,B.C

A B C →19

Total number of ways

By termination

Permutation

A B C →19

Total number of ways

By termination

Permutation