Problem source

\begin{questionparts}
\item The probability that a hobbit smokes a pipe is 0.7 and the probability that a hobbit wears a hat is 0.4. The probability that a hobbit smokes a pipe but does not wear a hat is $p$. Determine the range of values of $p$ consistent with this information.
\item  The probability that a wizard wears a hat is 0.7; the probability that a wizard wears a cloak is 0.8; and the probability that a wizard wears a ring is 0.4. The probability that a wizard does not wear a hat, does not wear a cloak and does not wear a ring is 0.05. The probability that a wizard wears a hat, a cloak and also a ring is 0.1.  Determine the probability that a wizard wears exactly two of a hat, a cloak, and a ring. 
The probability that a wizard wears a hat but not a ring, \textbf{given} that he wears a cloak, is $q$. Determine the range of values of $q$ consistent with this information.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{center}
    \begin{tikzpicture}
    
        % \node at (1,2) {Number of total episodes of case};
    
        \draw  (-2.5,-1.5)  rectangle (4.5,1.5);
    
        \draw[fill=blue!70,fill opacity=0.1] (0,0) ellipse (2cm and 1cm);
        \draw[fill=red!70,fill opacity=0.1] (2,0) ellipse (2cm and 1cm);
        
        % \node at (-1,0) {$9$};
        % \node at (1,0) {$242$};
        % \node at (3,0) {$180$};
        \node at (-1,1.2) {pipe};
        \node at (3,1.2) {hat};
        % \node at (-2.3,-1.3) {$\emptyset$};
        
    \end{tikzpicture}
\end{center}

The overlap can be at most 0.4, which would mean $p =0.7-0.4 = 0.3$

It must be at least 0.1, which would mean $p =0.7-0.1 = 0.6$ so $0.3 \leq p \leq 0.6$

\item 
\begin{center}
    \begin{tikzpicture}
    
        % \node at (1,2) {Number of total episodes of case};

        \draw[fill=blue!70,fill opacity=0.1] (-1, 0) circle (1.5cm);
        \draw[fill=red!70,fill opacity=0.1] (1, 0) circle (1.5cm);
        \draw[fill=yellow!70,fill opacity=0.1] (0, 1.73) circle (1.5cm);
    
        \draw  (-3,-2.25)  rectangle (3,3.75);
    
        % \draw[fill=blue!70,fill opacity=0.5] (0,0) ellipse (2cm and 1cm);
        % \draw[fill=red!70,fill opacity=0.5] (2,0) ellipse (2cm and 1cm);
        
        % \node at (-1,0) {$9$};
        % \node at (1,0) {$242$};
        % \node at (3,0) {$180$};
        \node at (1,3.2) {hat};
        \node at (-2,-1.5) {cloak};
        \node at (2,-1.5) {ring};

        \node at (-2, 3.2) {$0.05$};
        \node at (0, 0.5) {$0.1$};
        \node at (0, -0.4) {$cr$};
        \node at (.7, 0.8) {$hr$};
        \node at (-.7, 0.8) {$hc$};
        % \node at (-2.3,-1.3) {$\emptyset$};
    \end{tikzpicture}
\end{center}

Notice that:

\begin{align*}
    && 1 &= 0.05 + 0.7 -(hc+hr+0.1) + \\
    &&&\quad\quad 0.8 - (hc+cr + 0.1) + \\
    &&&\quad \quad \quad 0.4 - (hr+cr+0.1) +\\
    &&&\quad \quad \quad \quad hc+hr+cr+0.1  \\
    && &= 0.05 +0.7+0.8+0.4 - (hc+hr+cr)-2\cdot 0.1 \\
    \Rightarrow && hc+hr+cr &=0.05 +0.7 + 0.8 + 0.4 - 0.2-1 \\
    \Rightarrow && \mathbb{P}(\text{exactly 2}) &= 0.75
\end{align*}

Notice $q = \frac{hc}{0.8}$

Notice that we must have: $hc, hr cr \geq 0$ as well as $hc+hr+cr = 0.75$

\begin{align*}
&& \mathbb{P}(\text{only hat}) &=  0.7 -(hc+hr+0.1) \geq 0 \\
\Rightarrow && hc+hr & \leq 0.6 \\
&& \mathbb{P}(\text{only cloak}) &=  0.8 - (hc+cr + 0.1)\geq 0 \\
\Rightarrow &&hc+cr & \leq 0.7 \\
&& \mathbb{P}(\text{only ring}) &=  0.4 - (hr+cr+0.1) \geq 0 \\
\Rightarrow && hc+hr & \leq 0.3 \\
\end{align*}

To find the minimum for $hc$ we want to maximise $hr+cr = 0.3$, so $hc = 0.75 - 0.3 = 0.45$.

To find the maximum for $hc$ we want to minimise $hr$ and $cr$ $cr \leq 0.7 - hc$ and $hr \leq 0.6 - hc$ so $0.75 \leq hc + (0.6 - hc) + (0.7 - hc) = 1.3-hc$ so $hc \leq 1.3 - 0.75 = 0.55$

Therefore the range for $q$ is $\frac{.45}{.8}$ to $\frac{.55}{.8}$ or $\frac9{16} \leq q \leq \frac{11}{16}$

\end{questionparts}