Answer
\[P(A\cap {{B}^{C}})=\frac{1}{12}\]
Work Step by Step
Two dice are tossed, so there are 36 possible outcomes in sample space S:
\[S=\{(1,1),\text{ (1,2),}.................\text{,(6,6)}\}\]
Let A be the event that the sum of the faces showing is 6,and let B be the event that the face showing on one die is twice the face showing on the other. Hence, the possible outcomes of events A and B are:
$A=\left\{ \left( 1,5 \right),\text{ }\left( 5,1 \right),\text{ }\left( 2,4 \right),\text{ }\left( 4,2 \right),\text{ }\left( 3,3 \right) \right\}$
$B=\left\{ \left( 1,2 \right),\text{ }\left( 2,1 \right),\text{ }\left( 2,4 \right),\text{ }\left( 4,2 \right),\text{ }\left( 3,6 \right),\left( 6,3 \right) \right\}$
Assume that each possible outcome has a probability \[{}^{1}/{}_{36}\] Since, there are 5 possible outcomes for A and 6 for B, their probabilities would be
\[\begin{align}
& P(A)=\frac{5}{36} \\
& P(B)=\frac{6}{36} \\
\end{align}\]
We know that
\[P(A\cap {{B}^{C}})=P(A)-P(A\cap B)\]
For this we want to find \[P(A\cap B)\]as:
\[\begin{align}
& (A\cap B)=\{\left( 2,4 \right),\text{ }\left( 4,2 \right)\} \\
& P(A\cap B)=\frac{2}{36} \\
\end{align}\]
Therefore,
\[\begin{align}
& P(A\cap {{B}^{C}})=P(A)-P(A\cap B) \\
& P(A\cap {{B}^{C}})=\frac{5}{36}-\frac{2}{36} \\
& P(A\cap {{B}^{C}})=\frac{3}{36} \\
& P(A\cap {{B}^{C}})=\frac{1}{12} \\
\end{align}\]
The required probability is \[\frac{1}{12}\].
