The efficacy of mumps vaccine is about 80% that is 80% of those recieving the mumps vaccine will not contract the disease when exposed.

Question

The efficacy of mumps vaccine is about 80% that is 80% of those recieving the mumps vaccine will not contract the disease when exposed. Assume that each person's response to the mumps is independent of another person's response. Find the probability that at least one exposed person will get the mumps if n are exposed where: n = 2 and when n = 4

Let C be the event of contracting the disease M the event of the person getting mumps

What I have said is that $0.8 = P (\bar C \mid M) $

Now, what they want is $P (M) $ when n equals the different values.

How can I get it?

I found a solution like this,

My question is how did they get that 0.2? For me 0.2 is just the value of $P (C \mid M) $

What's wrong?

Answer 1

The following figure shows the populations that might be considered in this problem.

Let $M = \left\{ {{\rm{PeopleContractingMumps}}} \right\}$

As has been suggested, this is not actually a conditional probability problem. You can simply observe that the population of interest in the problem is the set of people who have been vaccinated and exposed. When you focus on that "sub"population, it is clear that if the probability of not getting mumps is 0.8, then the probability of getting the mumps must be 0.2 (in answer to the question).

If you consider the given population to be all vaccinated people, then you might consider the given probability to be conditional: $P\left( {\bar M|V \cap E} \right) = 0.8.$ Then, a beginner might question whether or not $P\left( {M|V \cap E} \right) = 0.2.$ That seems to have led to this question.

Notice that

[\begin{array}{l} P\left( {M|V \cap E} \right) + P\left( {\bar M|V \cap E} \right)\\ = \frac{{P\left( {M \cap V \cap E} \right)}}{{P\left( {V \cap E} \right)}} + \frac{{P\left( {\bar M \cap V \cap E} \right)}}{{P\left( {V \cap E} \right)}}\\ = \frac{{P\left[ {\left( {M \cap V \cap E} \right) \cup \left( {\bar M \cap V \cap E} \right)} \right]}}{{P\left( {V \cap E} \right)}} = 1. \end{array}]

Thus, $P\left( {M|V \cap E} \right) = 0.2$ and the answer to the problem is found using the binomial distribution as observed.