MHT CET202123 Sep 2021Evening ShiftMathematicsProbabilityActual
A random variable X has the following probability distribution array |r|l|l|l|l|l|l|l|l|l| x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 P ( X = x ) & K & 2 ~K & 3 ~K & 4 ~K & 4 ~K & 3 ~K & 2 ~K & ~K & ~K array Then P (3 < x 6)=
Options
- A3 7
- B4 7
- C13 21
- D8 21
Correct answer
A. 3 7
Step-by-step solution
We know that aligned & k +2 k +3 k +4 k +4 k +3 k +2 k + k + k =1 21 k =1 & k = 1 21 aligned When x=4, P=4 k= 4 21 , When x=5, P=3 k= 3 21 , When x =6, P =2 k = 2 21 P (3 < x 6)== 4+3+2 21 = 9 21 = 3 7