Problem source

Explain why, if $\mathrm{A, B}$ and  $\mathrm{C}$ are three events, 
\[
\mathrm{P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) +P(A \cap B \cap C)},
\]
where $\mathrm{P(X)}$ denotes the probability of event $\mathrm{X}$.
A cook makes three plum puddings for Christmas. He stirs $r$ silver sixpences thoroughly into the pudding mixture before dividing it into three equal portions. Find an expression for the probability that each pudding contains at least one sixpence. Show that the cook must stir 6 or more sixpences into the mixture if there is to be less than ${1 \over 3}$ chance that at least one of the puddings contains no sixpence. Given that the cook stirs 6 sixpences into the mixture and that each pudding contains at least one sixpence, find the probability that there are two sixpences in each pudding.

Solution source

\begin{center}
    \begin{tikzpicture}
    
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        \draw (1, 0) circle (1.5cm);
        \draw (0, 1.73) circle (1.5cm);
    
        \draw  (-3,-2.25)  rectangle (3,3.75);

        \filldraw (-1, 0) circle (1.5pt);
        \filldraw (1, 0) circle (1.5pt);
        \filldraw (0, 1.75) circle (1.5pt);

        \filldraw (0, 0.05) circle (1.5pt);
        \filldraw (0, -0.15) circle (1.5pt);
        \filldraw (.6, 0.875) circle (1.5pt);
        \filldraw (-.6, 0.875) circle (1.5pt);
        \filldraw (.4, 0.875) circle (1.5pt);
        \filldraw (-.4, 0.875) circle (1.5pt);

        \filldraw (0, .6) circle (1.5pt);
        \filldraw (0.1, .7) circle (1.5pt);
        \filldraw (-0.1, .7) circle (1.5pt);
    
        % \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) {$A$};
        \node at (-2,-1.5) {$B$};
        \node at (2,-1.5) {$C$};
        % \node at (-2.3,-1.3) {$\emptyset$};
        
    \end{tikzpicture}
\end{center}

When we add everything in $A$,$B$, $C$ we overcount the overlaps. When we remove the overlaps we remove the centre section too many times, so we have to add it back on in the end.

Let $X_i$ be the probability that the $i$th pudding contains a sixpence.

\begin{align*}
&& \mathbb{P}(X_1^c \cup X_2^c \cup X_3^c) &=\mathbb{P}(X_1^c \cap X_2^c \cap X_3^c) + \mathbb{P}(X_1^c)+\mathbb{P}(X_2^c)-\mathbb{P}(X_3^c)+\\
&&&\quad\quad-\mathbb{P}(X_1^c \cap X_2^c )-\mathbb{P}( X_2^c \cap X_3^c)-\mathbb{P}(X_1^c  \cap X_3^c) \\
&&&= 0 + (\tfrac23)^r+ (\tfrac23)^r+ (\tfrac23)^r + \\
&&&\quad\quad - (\tfrac13)^r- (\tfrac13)^r- (\tfrac13)^r \\
&&&= \frac{3\cdot2^r-3}{3^{r}} \\
\Rightarrow && \mathbb{P}(\text{all contain a sixpence}) &= 1 - \frac{3\cdot2^r-3}{3^{r}} \\
&&&= \frac{3^r-3\cdot2^r+3}{3^r}
\end{align*}

When $r = 5$ we have $\frac{3 \cdot 32-3}{3^5} = \frac{31}{81} > \frac13$
When $r = 6$  we have $\frac{3 \cdot 64-3}{3^6} = \frac{7}{27} < \frac13$

Therefore, the chef must stir in at least $6$.

\begin{align*}
&& \mathbb{P}(\text{two in each}|\text{at least 1 in each}) &= \frac{ \mathbb{P}(\text{two in each} \cap \text{at least 1 in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\
&&&= \frac{ \mathbb{P}(\text{two in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\
&&&= \frac{90/3^6}{20/27} \\
&&&= \frac{1/3}{2} = \frac16
\end{align*}