Maths-
General
Easy

Question

A random variable X has the following distribution

The value of k and P(X<3) are equal to

  1. k equals fraction numerator 1 over denominator 10 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 3 over denominator 5 end fraction    
  2. k equals fraction numerator 1 over denominator 10 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 3 over denominator 10 end fraction    
  3. k equals fraction numerator 3 over denominator 10 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 1 over denominator 10 end fraction    
  4. k equals fraction numerator 1 over denominator 24 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 5 over denominator 12 end fraction    

Hint:

Sum of all probabilities is 1

The correct answer is: k equals fraction numerator 1 over denominator 10 end fraction comma P left parenthesis X less than 3 right parenthesis equals fraction numerator 3 over denominator 10 end fraction


     Given :

    Sum of all probabilities is 1
    P(x=1) + P(x = 2) + P(x = 3) + P(x =4)  = 1
    k + 2k + 3k + 4k = 1
    k = 1 over 10
    Now, P(x < 3) = P(x=1) + P(x =3)
    P(x < 3) = k + 2k = 3k = 3 over 10
    Thus, the value of k  is 1 over 10 and P(X<3) is 3 over 10

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