Answer
a)\[P({{X}^{C}}\cap Y)=\frac{1}{8}\]
b)\[P(X\cap {{Y}^{C}})=\frac{1}{4}\]
Work Step by Step
(a)
A coin is to be tossed four times, so there are 16 possible outcomes in sample space S:
\[S=\left\{ \begin{array}{*{35}{l}}
HHHH,\text{ }HHHT,\text{ }HHTH,\text{ }HTHH, \\
THHH,\text{ }HHTT,\text{ }HTHT,\text{ }THHT, \\
THTH,\text{ }TTHH,\text{ }HTTT,\text{ }THTT, \\
TTHT,TTTH,\text{ }HTTH,\text{ }TTTT \\
\end{array} \right\}\]
Define events X and Y such that
X: first and last coins have opposite faces
Y: exactly two heads appear
Hence, the possible outcomes of events X and Y are:
$X=\{HHHT,\text{ }THHH,\text{ }HHTT,\text{ }HTHT,\text{ }THTH,\text{ }TTHH,\text{ }HTTT,\text{ }TTTH\}$
$Y=\{HHTT,\text{ }HTHT,\text{ }THTH,\text{ }THHT,\text{ }TTHH,\text{ }HTTH\}$
Assume that each of the sixteen head or tail sequences has the same probability.
Since, there are 8 possible outcomes in event X and 6 in the event Y, their probabilities would be
\[\begin{align}
& P(X)=\frac{8}{16} \\
& P(Y)=\frac{6}{16} \\
\end{align}\]
We know that
\[P({{X}^{C}}\cap Y)=P(Y)-P(X\cap Y)\]
For this we want to find \[P(X\cap Y)\]as:
\[\begin{align}
& (X\cap Y)=\{HHTT,\text{ }HTHT,\text{ }THTH,\text{ }TTHH\} \\
& P(X\cap Y)=\frac{4}{16} \\
\end{align}\]
Therefore,
\[\begin{align}
& P({{X}^{C}}\cap Y)=P(Y)-P(X\cap Y) \\
& P({{X}^{C}}\cap Y)=\frac{6}{16}-\frac{4}{16} \\
& P({{X}^{C}}\cap Y)=\frac{2}{16} \\
& P({{X}^{C}}\cap Y)=\frac{1}{8} \\
\end{align}\]
The required probability is \[\frac{1}{8}\].
(b)
Define events X and Y such that
X: first and last coins have opposite faces
Y: exactly two heads appear
And their probabilities are,
\[\begin{align}
& P(X)=\frac{8}{16} \\
& P(Y)=\frac{6}{16} \\
\end{align}\]
We know that
\[P(X\cap {{Y}^{C}})=P(X)-P(X\cap Y)\]
For this we want to find \[P(X\cap Y)\]as:
\[\begin{align}
& (X\cap Y)=\{HHTT,\text{ }HTHT,\text{ }THTH,\text{ }TTHH\} \\
& P(X\cap Y)=\frac{4}{16} \\
\end{align}\]
Therefore,
\[\begin{align}
& P(X\cap {{Y}^{C}})=P(X)-P(X\cap Y) \\
& P(X\cap {{Y}^{C}})=\frac{8}{16}-\frac{4}{16} \\
& P(X\cap {{Y}^{C}})=\frac{4}{16} \\
& P(X\cap {{Y}^{C}})=\frac{1}{4} \\
\end{align}\]
The required probability is \[\frac{1}{4}\].
