A biased die is tossed if an even face has turned up then the probability that it is face $2$ or face $4$ is

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

The correct option is C

$0.75$

The explanation for correct option:

Finding the probability:

Let the event that the even face turns up be $A$ and the event that the face is $2\mathrm{or}4\mathrm{be}\mathrm{B}$

$P\left(A\right)=P\left(2\right)+P\left(4\right)+P\left(6\right)$

$\Rightarrow P\left(A\right)=0.24+0.18+0.14$

$\Rightarrow P\left(A\right)=0.56$

$P\left(B\right)=P\left(2\right)+P\left(4\right)$

$\Rightarrow P\left(B\right)=0.24+0.18$

$\Rightarrow P\left(B\right)=0.42$

Hence, the required probability is

$$\begin{gathered} P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)} \\ =\frac{P\left(B\right)}{P\left(A\right)} \\ =\frac{0.42}{0.56} \\ =0.75. \end{gathered}$$

Hence, option(C) is correct answer