Maths-
General
Easy

Question

The distribution of a random variable X is given below

The value of k is

  1. fraction numerator 1 over denominator 10 end fraction    
  2. fraction numerator 2 over denominator 10 end fraction    
  3. fraction numerator 3 over denominator 10 end fraction    
  4. fraction numerator 7 over denominator 10 end fraction    

Hint:

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

The correct answer is: fraction numerator 1 over denominator 10 end fraction


    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
    fraction numerator 1 over denominator 10 space end fraction plus space k space plus space 1 fifth plus 2 k plus space 3 over 10 plus k equals space 1
fraction numerator 1 plus space 10 k plus 2 plus 20 k plus 3 plus 10 k over denominator 10 end fraction equals space 1
40 k space plus 6 equals space 10
40 k equals space 10 minus 6
40 k equals 4
k space equals space 1 over 10

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    not stretchy downwards arrow space of 1em not stretchy downwards arrow space of 1em not stretchy downwards arrow  not stretchy downwards arrow
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    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
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    therefore5 toys has to be distributed among 4 children after one of them is excluded. Which means one of them will get 2 toys.
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    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
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    iii) If d = 8 then numbers possible = 7 C subscript 3 times to the power of 5 C subscript 3
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    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
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    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
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    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  open table attributes columnalign right end attributes row 202159 row 3941 end table close curly brackets→19
    text  .................  end text
    323335 right curly bracket not stretchy rightwards arrow 1
    Total number of ways
    equals 1 plus 2 plus midline horizontal ellipsis.. plus 19 minus left parenthesis 17 plus 14 plus 11 plus 8 plus 5 plus 2 right parenthesis
    equals 190 minus 57 equals 133
    By termination
    Permutation
    133 cross times 3 factorial equals 798

    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  open table attributes columnalign right end attributes row 202159 row 3941 end table close curly brackets→19
    text  .................  end text
    323335 right curly bracket not stretchy rightwards arrow 1
    Total number of ways
    equals 1 plus 2 plus midline horizontal ellipsis.. plus 19 minus left parenthesis 17 plus 14 plus 11 plus 8 plus 5 plus 2 right parenthesis
    equals 190 minus 57 equals 133
    By termination
    Permutation
    133 cross times 3 factorial equals 798