A die is thrown three times. Events A and B a | vedclass

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(A) The sample space $S$ for throwing a die three times has $6 	imes 6 	imes 6 = 216$ outcomes.Event $B$ is defined as $6$ on the first throw and $5

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) The sample space $S$ for throwing a die three times has $6 	imes 6 	imes 6 = 216$ outcomes.Event $B$ is defined as $6$ on the first throw and $5$ on the second throw. Thus,$B = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\}$.The number of outcomes in $B$ is $n(B) = 6$,so $P(B) = \frac{6}{216}$.Event $A$ is defined as $4$ on the third throw.The intersection $A \cap B$ consists of outcomes where $6$ is on the first,$5$ is on the second,and $4$ is on the third throw. Thus,$A \cap B = \{(6, 5, 4)\}$.The number of outcomes in $A \cap B$ is $n(A \cap B) = 1$,so $P(A \cap B) = \frac{1}{216}$.The conditional probability of $A$ given $B$ is $P(A|B) = \frac{P(A \cap B)}{P(B)}$.Substituting the values: $P(A|B) = \frac{1/216}{6/216} = \frac{1}{6}$.

$A$ die is thrown three times. Events $A$ and $B$ are defined as below:
$A$: $4$ on the third throw
$B$: $6$ on the first and $5$ on the second throw
Find the probability of $A$ given that $B$ has already occurred.

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$A$ die is thrown three times. Events $A$ and $B$ are defined as below:$A$: $4$ on the third throw$B$: $6$ on the first and $5$ on the second throwFind the probability of $A$ given that $B$ has already occurred.