Exercise


ABy Admin
Jun 26'24

Answer

Solution: A

We have

[[math]]P(T\gt6|T\gt3) = \frac{P(T\gt6)}{P(T\gt3)}[[/math]]

. However, [math]T[/math] has distribution [math]P(T=k) = (5/6)^{k-1}(1/6) [/math] which means that

[[math]] \frac{P(T\gt6)}{P(T\gt3)} = \frac{\sum_{k\geq 7} (5/6)^{k-1}}{\sum_{k \geq 4} (5/6)^{k-1}} = \frac{(5/6)^6}{(5/6)^3} = (5/6)^3 = 0.5787. [[/math]]

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