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(C) Let $A$ be the event that an even face turns up,and $B$ be the event that the face is $2$ or $4$.The even faces are $2, 4,$ and $6$.The probabilit

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$0.75$

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(C) Let $A$ be the event that an even face turns up,and $B$ be the event that the face is $2$ or $4$.The even faces are $2, 4,$ and $6$.The probability of event $A$ is $P(A) = P(2) + P(4) + P(6) = 0.24 + 0.18 + 0.14 = 0.56$.The event $B$ is that the face is $2$ or $4$. Since $B$ is a subset of $A$,$P(B \cap A) = P(B) = P(2) + P(4) = 0.24 + 0.18 = 0.42$.Using the conditional probability formula,$P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{0.42}{0.56}$.Simplifying the fraction,$P(B|A) = \frac{42}{56} = \frac{3}{4} = 0.75$.

$Face$	$1, 2, 3, 4, 5, 6$
$Probability$	$0.1, 0.24, 0.19, 0.18, 0.15, 0.14$

$A$ biased die is tossed and the respective probabilities for various faces to turn up are given below:

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

$Probability$ $0.1, 0.24, 0.19, 0.18, 0.15, 0.14$

If an even face has turned up,then the probability that it is face $2$ or face $4$ is:

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$A$ biased die is tossed and the respective probabilities for various faces to turn up are given below: $Face$ $1, 2, 3, 4, 5, 6$ $Probability$ $0.1, 0.24, 0.19, 0.18, 0.15, 0.14$ If an even face has turned up,then the probability that it is face $2$ or face $4$ is: