AP EAMCET201823 Apr 2018Evening ShiftMathematicsProbabilityActual
A random variable X has the following distribution array lllllllll array l Values of X(x) array & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 (X=x) 0 & k & 2 k & 2 k & 3 k & k^2 & 2 k^2 & 7 k^2+k array
Options
- A9 10
- B( 9 10 )^2
- C( 3 10 )
- D1 10
Correct answer
B. ( 9 10 )^2
Step-by-step solution
aligned & Since, P (X=x_i )=1 & 0+K+2 K+2 K+3 K+K^2+2 K^2 & +7 K^2+K=1 & 9 K+10 K^2=1 & K= 1 10 , [ K>0] & aligned So, aligned & P(0 < K < 6)=K+2 K+2 K+3 K+K^2 & =8 K+K^2= 8 10 + 1 100 = 81 100 = ( 9 10 )^2 aligned