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(C) Let $A$ be the event that face $4$ turns up and $B$ be the event that face $5$ turns up. Given $P(A) = 0.25$ and $P(B) = 0.05$. Since $A$ and $B$

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$\frac{5}{6}$

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(C) Let $A$ be the event that face $4$ turns up and $B$ be the event that face $5$ turns up. Given $P(A) = 0.25$ and $P(B) = 0.05$. Since $A$ and $B$ are mutually exclusive events,the probability that either face $4$ or face $5$ turns up is $P(A \cup B) = P(A) + P(B) = 0.25 + 0.05 = 0.30$. We need to find the conditional probability that the face is $4$ given that either face $4$ or face $5$ has turned up,which is $P(A | A \cup B)$. Using the formula for conditional probability: $P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)} = \frac{0.25}{0.30} = \frac{25}{30} = \frac{5}{6}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

For a biased die,the probabilities for different faces to turn up are

$Face$ $1$ $2$ $3$ $4$ $5$ $6$

$P(F)$ $0.2$ $0.22$ $0.11$ $0.25$ $0.05$ $0.17$

The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is

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For a biased die,the probabilities for different faces to turn up are $Face$ $1$ $2$ $3$ $4$ $5$ $6$ $P(F)$ $0.2$ $0.22$ $0.11$ $0.25$ $0.05$ $0.17$ The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is