Do we have to calculate these probabilities?

Question

In a small town there are $5600$ people, who are older than $60$ years. Of these, $3200$ were vaccinated against flu last winter. Of those over $60$, $1800$ had the flu. Of those not vaccinated over $60$, $950$ remained flu-free.

With what probability

a) Got a vaccinated person the flu.

b) Was a flu patient vaccinated.

c) Did a non-vaccinated person leave flu-free.

d) has a healthy person not been vaccinated.

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Here we need the conditional probability, right?

I have done the following:

Let V the event that a perosn got vaccinated and F that a person is a flu-patient.

Do we have to calculate the following probabilities?

a) $P(F\mid V)$

b) $P(V\mid F)$

c) $P(\overline{F}\mid \overline{V})$

d) $P(\overline{V}\mid \overline{F})$

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EDIT1:

So we have that a) $P(F\mid V)=\frac{P(F\cap V)}{P(V)}$

b) $P(V\mid F)=\frac{P(V\cap F)}{P(F)}$

c) $P(\overline{F}\mid \overline{V})=\frac{P(\overline{F}\cap \overline{V})}{P(\overline{V})}$

d) $P(\overline{V}\mid \overline{F})=\frac{P(\overline{V}\cap\overline{F})}{P(\overline{F})}$ or not?

Do we have that $P(F)=\frac{1800}{5600}$, $P(V)=\frac{3200}{5600}$, $P(\overline{V}\cap\overline{F})=\frac{950}{5600}$ ?

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EDIT2:

Does it hold that \begin{align*}P(F\cap V)&=1-P(\overline{F}\cup \overline{V})\\ & =1-\left (P(\overline{F})+P(\overline{V})-P(\overline{F}\cap \overline{V})\right ) \\ & =1-\left (1-\frac{1800}{5600}+1-\frac{3200}{5600}-\frac{950}{5600}\right )\end{align*} ?

Answer 1

No, $$P(F\cap V)=1-P(\overline{F\cap V})= 1 - P(\overline{F}\cup\overline{V})$$ by De Morgan's law.

I suggest you draw a Venn diagram.

P.s. Once again, you should add the additional question to the body not the comments. As it stands, someone reading only your question and my answer wouldn't understand what I'm talking about.

Answer 2

In situations like this, it is often best to start with a $2 \times 2$ table.

You are given the following information:

Flu\Vac     Yes    No      Total
--------------------------------
Yes                         1800
No                950           
--------------------------------
Total      3200             5600

From that information, simple subtraction allows you to complete all entries:

Flu\Vac     Yes    No      Total
--------------------------------
Yes         350  1450       1800
No         2850   950       3800
--------------------------------
Total      3200  2400       5600

Then you are ready to reinforce your understanding of conditional probability and to compute all of the required answers.

(a) For the probability that a vaccinated person got the flu: To condition on 'Vaccinated', focus on the 'Vaccinated=Yes' column: $P(F|V) = 350/3200.$

(b) For the probability that a flu patient was vaccinated: To condition on 'Flu' focus on the 'Flu = Yes' row: $P(V|F) = 350/1800.$

And so on. Once the table is complete, there is no question about 'Flu' and 'Vaccinated' that you can't answer.

For example, possibly a bit later in your course:

Are events $V$ = 'was vaccinated' and $F$ = 'caught flu' independent?

If so, then we must have $P(F \cap V) = P(F)P(V).$ From the table we have $P(F \cap V) = 360/5600=0.0643,\,$ $P(F) = 1800/5600= 0.3214,$ and $P(V) = 3200/5600= 0.5714.$ But $0.3412 \times 0.5714 = 0.1950 \ne 0.0643,$ so the events are not independent (to the delight of flu vaccine makers). [Also note: Unconditional probabilities all have the grand total 5600 in the denominator.]

Then you can explore various rules of logic (such as De Morgan's laws) and various probability results such as $P(F \cap V) = P(F)P(V|F).$

Sometimes, Venn Diagrams can also be helpful, but it is not easy to put all of the counts in a $2 \times 2$ table into a Venn Diagram of two events.