MHT CET202120 Sep 2021Evening ShiftMathematicsProbabilityActual
The p.m.f of a random variable X is P(X=x)= 1 2^5 ( array l 5 x array ), x=0,12345 then =0 otherwise,
Options
- AP ( X 2) < P ( X 3)
- BP ( X 2)> P ( X 3)
- CP ( X 2)=2 P ( X 3)
- DP ( X 2)= P ( X 3)
Correct answer
D. P ( X 2)= P ( X 3)
Step-by-step solution
aligned & P ( X = x )= 1 32 ^5 C _ x , where x =0,1,2,3,4,5 & =0, otherwise & P ( X 2)= P ( X =0)+ P ( X =1)+ P ( X =2) & = 1 32 [ ^5 C ₀+ ^5 C ₁+ ^5 C ₂ ]= 1 32 (1+5+10)= 16 32 = 1 2 & P ( X 3)= P ( X =3)+ P ( X =4)+ P ( X =5) & = 1 32 [ ^5 C ₃+ ^5 C ₄+ ^5 C ₅ ]= 1 32 (10+5+1)= 16 32 = 1 2 aligned