Dear co-listers. Here you have a summary of all the responses about my past
query.
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Hello,
have you other information concerning the probability to be inoculate? My
idea is to use the Bayes theorem and the conditionnal probability. But
whithout the probabiltiy of the exposure P(E)... it's difficult.
Regards,
E.L
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I think you're right if the independence assumption holds.
Abderrahim
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Hi, my answer is different from both:
The probability that the person not inoculated being infected is 0.6. If he
is not sick (with probability 0.4), then the probability the other person is
sick is 0.6 * 0.2 = 0.12. Therefore the probability that at least one person
is sick is 0.6 + 0.4 * 0.12 = 0.648.
That's assuming I've understood the question correctly. If the probability
that an inoculated person is sick is 0.8, overall, regardless of whether
he's infected, then the overall probability would be 0.6 + 0.4 * 0.2 = 0.68.
Tim
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Whatever the probability, I don't see how it can be less than 0.6, since
this is the probability that the non-inoculated one gets flu regardless of
what happens to the other.
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Rodrigo,
I have still another result adding to yours. I'd say there is at least one
error in your reasoning (but there might still be one in mine as well):
The probabilities of the inoculated are not correct. The probability for
this one to NOT get sick is 0.4 + 0.6*0.8 = 0.88, as he has a 40% chance of
not being affected by the flue at all (in that case it doesn't matter
whether or not he is inoculated, he will not get sick at all) and a 60%
chance to be infected, but with 80% chance he is still protected due to the
inoculation.
On the other hand, the prob for him to get sick is 0 + 0.6*0.2 = 0.12, as
there is no chance in case he is not infected at all and a 20% chance in
case he is infected, which occurs with prob = 0.6.
So your three individual probabilties would be 0.528 + 0.048 + 0.072 =
0.648, which is my solution. I had the same result with two different
approaches over which I will not go into detail for brevity and not to cause
any confusion, but this is different from both your result and the result
given in the textbook.
Please let me know what other ALLSTATERS wrote and whether they gave a
similar answer or found a way to reproduce the textbook's result.
HTH,
Michael
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It's the same reasoning that you used:
you wrote:
Non-inoculated sick, inoculated non-sick: 0.6 X 0.8 = 0.48
Non-Inoculated non-sick, inoculated sick: 0.4 X 0.2 = 0.08
Both Sick: 0.6 X 0.2 = 0.12
Including my new probabilties for the inoculated, you get
Non-inoculated sick, inoculated non-sick: 0.6 X 0.88 = 0.528
Non-Inoculated non-sick, inoculated sick: 0.4 X 0.12 = 0.048
Both Sick: 0.6 X 0.12 = 0.072
Regards, Michael
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I think this is a bad problem.
It is not clearly stated.
The solution is wrong.
You (in my view correctly) read the first sentence of the problem to mean
that for a not inoculated person the probability of getting flu is
0.6 . It immediately follows that the probability of this person and/or any
other person getting flu is AT LEAST 0.6 .
So the answer 0.5952 is wrong.
The problem is not clearly stated. I would have liked to know what is the
probability of being exposed during an epidemic.
If I assume it is 100%, like you did, I agree with the outcome of your
calculation as a normal common-sense interpretation of the text. Even then I
would have preferred a clearer problem text, something like "without
inoculation, 60% fall ill, of the inoculated, 20% fall ill"
where the denominator of the percent is not in doubt. Because the present
text leaves me guessing ... perhaps they meant that there was 80%
effectiveness among the 60% that would have fallen ill?
But if the exposure rate among the whole population is 100% anyway, then
what about the whole condition "Suppose that they aren=92t in the same
place, they are not in contact with the same people and they can=92t be
infected between each other" . Then this is superfluous (superFLUous ;-) ).
Regards,
Peter Das
Netherlands
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With Respect
I thought the catch was that the inoculation reduces the chance of
getting flu by 80%, or 0.48, so that the inoculated have a chance of getting
sick of
0.12 not 0.2
-but this gives me 0.528 + 0.048 + 0.072 = 0. 648
Note the textbook answer gives a probability of boh employees not being
sick of 0.4048, more than, the 0.4 both Mr Briceno and I took as the chance
of the non-inoculated employee not being sick.
Yours Sincerely,
Alan E. Dunne
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Actually, that should be "doubt about the given answer."
I concur with the other two observed responders but would like to emphasize
the issue that Mr. (Dr.?) Dunne raises, which is the inherent illogic of the
textbook answer. Although I am too far removed from such exercises to be
trustworthy, it seems to me that, if the probability of non-illness of the
noninoculated person is .4, the probability of non-illness in both _cannot_
be greater than .4. Some law of nature that I seem to recall states: p(a and
b) <= p(a). I shall be much obliged to the person who can show this to be
wrong. Or inapplicable to the present exercise.
Respectfully,
Michael
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Regards
Rodrigo Briceņo
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