Choose the correct option from the given alternative: 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) =

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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) = $\frac{16}{25}$ Hint : `k[$\frac{1}{5}$^0 + $\frac{2}{5}$^1 + $\frac{3}{5}$^2 + ....]=1 Let s = k/5^0+ 2k/5^1+3k/5^2+ ............... i.e s = k + 2k/5+3k/5^2+.......... $\therefore$$\frac{1}{5}$s = k/5 + (2k)/5^2 + (3k)/5^3 + ........ $\therefore$ s - $\frac{1}{5}$s = k + k/5 + k/5^2 + k/5^3 + .......... $\therefore$ $\frac{4}{5}$s=k[1 + $\frac{1}{5}$ + $\frac{1}{5}$^2 + $\frac{1}{5}$^3 + ....] = k [1/(1 - $\frac{1}{5}$)] = 5k/4 $\therefore$ s = (25k)/16 = 1 $\therefore$ k = $\frac{16}{25}$ $\therefore$P(x=0) = k(0+1)5^0 = k = $\frac{16}{25}$.`

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