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(D) The sequence $'1001'$ can occur in two patterns based on the starting position:Case $1$: The sequence starts at an odd place.Odd place $(1)$: $P(1

Vedclass · Accepted Answer

$\frac{1}{9}$

Vedclass · Answer

(D) The sequence $'1001'$ can occur in two patterns based on the starting position:Case $1$: The sequence starts at an odd place.Odd place $(1)$: $P(1) = 1 - \frac{1}{3} = \frac{2}{3}$Even place $(0)$: $P(0) = \frac{1}{2}$Odd place $(0)$: $P(0) = \frac{1}{3}$Even place $(1)$: $P(1) = 1 - \frac{1}{2} = \frac{1}{2}$Probability $= \frac{2}{3} 	imes \frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{2} = \frac{2}{36} = \frac{1}{18}$Case $2$: The sequence starts at an even place.Even place $(1)$: $P(1) = 1 - \frac{1}{2} = \frac{1}{2}$Odd place $(0)$: $P(0) = \frac{1}{3}$Even place $(0)$: $P(0) = \frac{1}{2}$Odd place $(1)$: $P(1) = 1 - \frac{1}{3} = \frac{2}{3}$Probability $= \frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{2} 	imes \frac{2}{3} = \frac{2}{36} = \frac{1}{18}$Total Probability $= \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9}$

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