Suppose that the probability of exposure to the flu during an epidemic
is 0.6.  Experience has shown that a serum is 80% successful in
preventing an innoculated person from acquiring the flu, if exposed to
it.  A person not inoculated faces the probability of .90 of acquiring
the flu if exposed to it.  Two persons, one innoculated and one not,
perform a highly specialized task in business.  Assume that they are
not at the same location, not in contact with the same people, and
cannot expose eachother to the flu.  What is the problem that at least
one will get the flu?

Please describe each step as you go though it, I have no clue what to do!
Thanks

Hello and thank you for your question.

The .6 means that there's a 60% chance of being exposed to the flu,
and a 40% chance of not.

But the problem goes on to say that if you are exposed, then if you've
been inoculated there is still a .2 (20%) chance of getting sick
(because .8 of the time you'll be protected), while if you haven't
been inoculated then there is a 90% chance of getting sick.

So for an inoculated person, the chance of getting flu is .6 x .2 = .12 or 12%

For a not-inoculated person, the chance of getting flu is .6 x .9 = .54 or 54%

Finally, they want to know the chance either (or both) of these two
people will get the flu.

One way to calculate that is to figure the chance that neither will
get flu, and subtract that from 1.0 (100%).

The chance of two independent (that's why there's all that talk about
their not interacting, encountering the same people, etc.) events both
happening is each probability times the other (the product).

So the inocuted person will be flu-free 1 - .12 = 88% of the time
and the not-inoculated person will be flu-free 1 - .54 = 46% of the time

The chance both will be flu-free is .88 x .46 = .4048  (40.48%)

The chance one, or the other, or both will catch flu is 1 - .4048 = .5952 (59.52%).

[if you want to only work from the chance each will get sick, you have
to subtract for the double-counting of the case where both of them get
sick:
  .12 + .54 - (.12 x .54) = .5952  
This second way is easier, if you understand why you're subtracting the .12 x .54 ]

That's it!

Take a look at question 5 of
http://statweb.calpoly.edu/chance/stat221/quizzes/exam1sols.html
to see how it works out.

Again, thanks for bringing us your question.

-Richard-ga

thanks!

I think if a serum is 80% successful, that means that 80% of the times
a subject WOULD have gotten the flu, he doesn't.  Would you say a
serum was 50% successful if half of innoculated people get sick, and
half of non-innoculated people get sick?

So, I think the probability that the innoculated person gets the flu is 

0.6 * 0.2 * 0.9 = 0.108    (instead of 0.12).

This gives a probablity of 0.58968 that at least one of the two gets the flu.

Racecar-ga

Thank you for your comment.  It seems to me that to evaluate a serum,
you would give your subject two needles--a shot of flu germs and a
shot of serum.  And the observation in that case would be that the
subject gets sick 20% of the time, while absent the inoculation the
subject gets sick 100% of the time.

But I see your point.
  
Richard-ga