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(C) Let $A$ be the event that an even face turns up,so $A = \{2, 4, 6\}$.The probability of event $A$ is $P(A) = P(2) + P(4) + P(6) = 0.24 + 0.18 + 0.

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$0.75$

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(C) Let $A$ be the event that an even face turns up,so $A = \{2, 4, 6\}$.The probability of event $A$ is $P(A) = P(2) + P(4) + P(6) = 0.24 + 0.18 + 0.14 = 0.56$.Let $B$ be the event that the face is $2$ or $4$,so $B = \{2, 4\}$.The intersection $B \cap A$ is the event that the face is $2$ or $4$ $AND$ it is even,which is simply $B = \{2, 4\}$.The probability of $B \cap A$ is $P(B \cap A) = P(2) + P(4) = 0.24 + 0.18 = 0.42$.Using the formula for conditional probability,$P(B|A) = \frac{P(B \cap A)}{P(A)}$.Substituting the values,$P(B|A) = \frac{0.42}{0.56} = \frac{42}{56} = \frac{3}{4} = 0.75$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

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

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

$P(F)$ $0.1$ $0.24$ $0.19$ $0.18$ $0.15$ $0.14$

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: $Face$ $1$ $2$ $3$ $4$ $5$ $6$ $P(F)$ $0.1$ $0.24$ $0.19$ $0.18$ $0.15$ $0.14$ If an even face has turned up,then the probability that it is face $2$ or face $4$ is: