A fair die is tossed until a six is obtained. | vedclass

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Question

(D) The random variable $X$ follows a geometric distribution with probability of success $p = \frac{1}{6}$ and probability of failure $q = \frac{5}{6}

Vedclass · Accepted Answer

$\frac{25}{36}$

Vedclass · Answer

(D) The random variable $X$ follows a geometric distribution with probability of success $p = \frac{1}{6}$ and probability of failure $q = \frac{5}{6}$.By the definition of conditional probability,$P(X \geq 5 \mid X > 2) = \frac{P(X \geq 5 \cap X > 2)}{P(X > 2)}$.Since the event $X \geq 5$ is a subset of $X > 2$,$P(X \geq 5 \cap X > 2) = P(X \geq 5)$.Thus,$P(X \geq 5 \mid X > 2) = \frac{P(X \geq 5)}{P(X > 2)}$.For a geometric distribution,$P(X > k) = q^k = (\frac{5}{6})^k$.Therefore,$P(X \geq 5) = P(X > 4) = (\frac{5}{6})^4$.And $P(X > 2) = (\frac{5}{6})^2$.Substituting these values,we get $P(X \geq 5 \mid X > 2) = \frac{(5/6)^4}{(5/6)^2} = (\frac{5}{6})^2 = \frac{25}{36}$.

$A$ fair die is tossed until a six is obtained. Let $X$ be the number of required tosses,then the conditional probability $P(X \geq 5 \mid X > 2)$ is:

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