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Question Number 192105 by Spillover last updated on 08/May/23

	Answered by mehdee42 last updated on 09/May/23

	$a) l e m a : i f 0 < x < 1 \Rightarrow \underset{i = 0}{\sum^{\infty}} {i x}^{i} = \frac{x}{{(1 - x)}^{2}} & \underset{i = 0}{\sum^{\infty}} i^{2} x^{i} = \frac{x (1 + x)}{{(1 - x)}^{3}}$ $t h u s E (x) = Σ x f (x) = \frac{2}{3} \underset{i = 0}{\sum^{\infty}} i {(\frac{1}{3})}^{i} = \frac{1}{2} ✓$ $b) V a r (x) = E (x^{2}) - E^{2} (x)$ $E (x^{2}) = Σ x^{2} f (x) = \frac{2}{3} \underset{i = 0}{\sum^{\infty}} i^{2} {(\frac{1}{3})}^{i} = \frac{2}{3} \times \frac{4}{9} \times \frac{27}{8} = 1$ $\Rightarrow V a r (x) = 1 - \frac{1}{4} = \frac{3}{4} ✓$ $c) P (0) + P (1) + P (2) + P (3) = \frac{2}{3} (0 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}) = \frac{26}{81} ✓$
	Commented by Spillover last updated on 12/May/23

	$t h a n k s$
	Answered by Spillover last updated on 12/May/23

	$F r o m f a c t o r i a l g e n e r a t i n g f u n c t i o n$ $\emptyset_{x} (t) = E (t^{x})$ $E (t^{x}) = \sum_{x}^{\infty} \frac{2}{3} t^{x} {(\frac{1}{3})}^{x} \frac{2}{3} \underset{x = 0}{\sum^{\infty}} t^{x} {(\frac{1}{3})}^{x} = \frac{2}{3} \underset{x = 0}{\sum^{\infty}} {(\frac{t}{3})}^{x}$ $\frac{2}{3} \underset{x = 0}{\sum^{\infty}} {(\frac{t}{3})}^{x} = \frac{2}{3} (1 + \frac{t}{3} + \frac{t^{2}}{3^{2}} + \frac{t^{3}}{3^{3}} + . . .)$ $l e t y = \frac{t}{3}$ $E (t^{x}) = \frac{2}{3} (1 + y + y^{2} + y^{3} + . . .)$ $f r o m s_{\infty} = \frac{G_{1}}{1 - r} s_{\infty} = \frac{1}{1 - r} b u t r = y = \frac{t}{3}$ $s_{\infty} = \frac{1}{1 - r} = \frac{1}{1 - \frac{t}{3}} = \frac{3}{3 - t}$ $E (t^{x}) = \frac{2}{3} (1 + y + y^{2} + y^{3} + . . .) E (t^{x}) = \frac{2}{3} (\frac{3}{3 - t}) = \frac{2}{3 - t}$ $\emptyset_{x} (t) = \frac{2}{3 - t}$ $\emptyset_{x}^{^{'}} (t) = \frac{2}{{(3 - t)}^{2}} \emptyset_{x}^{^{″}} (t) = \frac{2}{{(3 - t)}^{3}} \emptyset_{x}^{^{‴}} (t) = \frac{2}{{(3 - t)}^{3}}$ $E (x) = \emptyset_{x}^{^{'}} (t) = \frac{2}{{(3 - t)}^{2}} t = 1 E (x) = (\frac{2}{{(3 - 1)}^{2}}) = \frac{1}{2}$ $v a r (x) = E (x) - {[E (x)]}^{2}$ $v a r (x) = \emptyset_{x}^{^{″}} (t) + E (x) - {[E (x)]}^{2}$ $v a r (x) = \frac{1}{2} + \frac{1}{2} - {(\frac{1}{2})}^{2} = \frac{3}{4}$
	Answered by Spillover last updated on 12/May/23

	$F r o m p r o b a b i l i t y g e n e r a t i n g f u n c t i o n o f x$ $G_{x} (t) = \frac{2}{3 - t}$ $G_{x}^{^{'}} (t) = \frac{2}{{(3 - t)}^{- 2}} G_{x}^{^{″}} (t) = \frac{2}{{(3 - t)}^{- 3}} G_{x}^{‴} (t) = \frac{2}{{(3 - t)}^{- 4}}$ $G_{x} (t) = G_{x} (0) t^{0} + G_{x} (0) t + G_{x} (0) \frac{t^{2}}{2!} + + G_{x} (0) \frac{t^{3}}{3!} + . . .$ $G_{x} (t) = \frac{2}{3} t^{0} + \frac{2}{9} t + \frac{4}{27} (\frac{t^{2}}{2!}) + \frac{2}{81} (\frac{t^{3}}{3!}) + . . .$ $G_{x} (t) = \frac{2}{3} t^{0} + \frac{2}{9} t + \frac{2}{27} t^{2} + \frac{2}{81} t^{3} + . . .$ $P (X = 0) = \frac{2}{3} P (X = 1) = \frac{2}{9} P (X = 2) = \frac{2}{27} P (X = 3) = \frac{2}{81}$ $P (x) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)$ $P (x) = \frac{2}{3} + \frac{2}{9} + \frac{2}{27} = \frac{36}{81}$
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