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, find $( p + q )$.

Answer 1

$ E_1$ : event that the dot is removed from an odd face; $P\left(E_1\right)=\frac{1+3+5}{21}=\frac{9}{21}$
$E _2$ : dot is removed from the even face; $P \left( E _2\right)=\frac{2+4+6}{21}=\frac{12}{21}$
$E$ : die thrown has an odd number of dots on its top face
$P(E)=P\left(E \cap E_1\right)+P\left(E \cap E_2\right)$

$=P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) $
$=\left(\frac{9}{21}\right) \cdot \frac{2}{6}+\left(\frac{12}{21}\right) \cdot \frac{4}{6} $
$=\frac{9}{21} \cdot \frac{1}{3}+\frac{12}{21} \cdot \frac{2}{3}=\frac{3}{21}+\frac{8}{21}=\frac{11}{21}=\frac{p}{q} \Rightarrow p+q=32$