Question

On a normal standard die, one of the $21$ dots from any one of the six faces is removed at random with each dot equally likely to be chosen. The die is then rolled. If the probability that the top face has an odd number of dots is $\frac{p}{q}$ where $p$ and $q$ are in their lowest form, then the value of $\frac{(p+q)}{4}$ is

Answer 1

$E_1:$ Dot removed from odd face.
$P\left(E_1\right)=\frac{9}{21}$

$E_2$: Dot removed from even face.
$P\left(E_2\right)=\frac{12}{21}$

$E:$ Die shows odd numbers of dots.
$P\left(E\right)=P(E\cap E_1)+P(E\cap E_2)=P\left(E_1\right)P\left(\frac{E}{E_1}\right)+P\left(E_2\right)P\left(\frac{E}{E_2}\right)=\frac{9}{21}\times \frac{2}{6}+\frac{12}{21}\times \frac{4}{6}=\frac{11}{21}=\frac{p}{q}$
Now, $\frac{11+21}{4}=8$