MHT CET202313 May 2023Morning ShiftMathematicsProbabilityActual
A random variable X has the following probability distribution array |l|c|c|c| X =x & 0 & 1 & 2 P ( X =x) & 4 k -10 k ^2 & 5 k -1 & 3 k ^3 array then P ( X < 2) is
Options
- A2 9
- B5 9
- C8 9
- D4 9
Correct answer
C. 8 9
Step-by-step solution
aligned & Since _ x=0 ^2 P ( X =x)=1 & 4 k -10 k ^2+5 k -1+3 k ^3=1 & 3 k ^3-10 k ^2+9 k -2=0 & ( k -1)( k -2)(3 k -1)=0 & k =1 or k =2 or k = 1 3 & aligned & For k =1 or k =2 & P ( X =0) < 0, which is not possible & k = 1 3 aligned & aligned Now, P ( X < 2) & = P ( X =0)+ P ( X =1) & =4 k -10 k ^2+5 k -1 & =9 k -10 k ^2-1 & =9 ( 1 3 )-10 ( 1 9 )-1 & = 8 9 aligned aligned