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KCET2019MathematicsProbability

A random variable ' ( X ) ' has the following probability distribution array |l|l|l|l|l|l|l|l| X & 1 & 2 & 3 & 4 & 5 & 6 & 7 P ( X ) & k -1 & 3 k & k & 3 k & 3 k ² & k² & k²+ k array Then the value of ( k ) is

Options

  1. A( -2 )
  2. B( 1 10 )
  3. C( 1 5 )
  4. D( 2 7 )

Correct answer

C. ( 1 5 )

Step-by-step solution

(C) P_ i =1K-1+3 K+k+3 K+3 K²+K²+K²+K=15 K²+9 K-2=05 K²+10 K-K-2=05 K(K+2)-1(K+2)=0(5 K-1)(K+2)=0 ~K = 1 5 ,-2( ~K =-2 is not possible).

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