Answer 1

If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) = `16/25`

Hint : `k[1/5^0 + 2/5^1 + 3/5^2 + ....]=1`

`Let s = k/5^0+ 2k/5^1+3k/5^2+ ...............`

i.e `s = k + 2k/5+3k/5^2+..........`

∴`1/5s = k/5 + (2k)/5^2 + (3k)/5^3 + ........`

∴ `s - 1/5s = k + k/5 + k/5^2 + k/5^3 + ..........`

∴ `4/5s=k[1 + 1/5 + 1/5^2 + 1/5^3 + ....]`

= `k [1/(1 - 1/5)] = 5k/4`

∴ `s = (25k)/16 = 1`

∴ `k = 16/25`

∴`P(x=0) = k(0+1)5^0 = k = 16/25.`