MHB Calculating Conditional Probabilities in Boolean Algebra

[mathmari]

Hey!

I am looking the following concering boolean algebra.

For a certain test group, it is found that $42\%$ of the people have never skied yet, that $58\%$ of them have never flown yet, and that $29\%$ of them have already flown and skied.
Which probability is higher then:
To meet someone, who has already skied, from the group of those who have never flown, or to find someone who has already flown from the group of those who have already skied?

I have done the following:

Let $S$ be the event that someone has already skied and let $F$ be the event that someone has already flown.
Then it is given that $P(\overline{S})=42\%$, $P(\overline{F})=58\%$ and $P(S\land F)=29\%$, or not? (Wondering)

From these probabilities, we get also $P(S)=100\%-P(\overline{S})=100\%-42\%=58\%$ and $P(F)=100\%-P(\overline{F})=100\%-58\%=42\%$.

We are looking for the conditional probabilities $P(S\mid \overline{F})$ and $P(F\mid S)$, or have I understood that wrong? (Wondering)

We have that \begin{align*}&P(S\mid \overline{F})=\frac{P(S\land \overline{F})}{P(\overline{F})} \\ &P(F\mid S)=\frac{P(F\land S)}{P(S)}=\frac{29\%}{58\%}=50\%\end{align*}

How can we calculate $P(S\land \overline{F})$ ? (Wondering)

 

[I like Serena]

Hey mathmari! (Happy)

The easiest way is to make a Venn Diagram. Even if only to understand what we want to achieve.

The following Venn Diagram represents the problem doesn't it? (Thinking)
\begin{tikzpicture}
\begin{scope}[blend group = soft light]
\fill[red!30!white] (-1,0) circle (2);
\fill[blue!30!white] (1,0) circle (2);
\end{scope}
\node at (-1,2) {$S$};
\node at (1,2) {$F$};
\node at (-1.8,0) {$29\%$};
\node at (1.8,0) {$13\%$};
\node at (0,0) {$29\%$};
\end{tikzpicture}
We can immediately see that $P(S\land\bar F)=29\%$.

More formally, we have that $(S\land \bar F)$ and $(S\land F)$ have an empty intersection, and their union is $S$.
Therefore they are mutually exclusive and we can apply the sum rule:
$$P(S)=P\big((S\land F) \lor (S\land\bar F)\big) = P(S\land F) + P(S\land \bar F)
\implies P(S\land\bar F) = P(S)-P(S\land F)$$
This is effectively how the Venn Diagram was constructed in the first place. (Thinking)

 

[mathmari]

The following Venn Diagram represents the problem doesn't it? (Thinking)
\begin{tikzpicture}
\begin{scope}[blend group = soft light]
\fill[red!30!white] (-1,0) circle (2);
\fill[blue!30!white] (1,0) circle (2);
\end{scope}
\node at (-1,2) {$S$};
\node at (1,2) {$F$};
\node at (-1.8,0) {$29\%$};
\node at (1.8,0) {$13\%$};
\node at (0,0) {$29\%$};
\end{tikzpicture}

Ahh yes! (Malthe)

More formally, we have that $(S\land \bar F)$ and $(S\land F)$ have an empty intersection, and their union is $S$.
Therefore they are mutually exclusive and we can apply the sum rule:
$$P(S)=P\big((S\land F) \lor (S\land\bar F)\big) = P(S\land F) + P(S\land\bar F)
\implies P(S\land\bar F) = P(S)-P(S\land\bar F)$$
This is effectively how the Venn Diagram was constructed in the first place. (Thinking)

Ahh ok, I see! So, we have that $$P(S\land\bar F) = P(S)-P(S\land F)=58\%-29\%=29\%$$

Therefore, the probability to meet someone, who has already skied, from the group of those who have never flown is equal to $$P(S\mid \overline{F})=\frac{P(S\land \overline{F})}{P(\overline{F})}=\frac{29\%}{58\%}=50\%$$ and the probability to find someone who has already flown from the group of those who have already skied is equal to $$P(F\mid S)=\frac{P(F\land S)}{P(S)}=\frac{29\%}{58\%}=50\%$$ So, these two probabilities are equal. Is everything correct? (Wondering)

 

[I like Serena]

It looks good to me. (Nod)

 

[mathmari]

It looks good to me. (Nod)

Great! Thanks a lot! (Yes)