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In probabilityland, one third of population is senior. Every winter 20% of population gets the flu. on average, two out of five seniors get the flu. what is the probability that a person with the flu is senior?
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here is what I did
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|dw:1348935331451:dw|
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"that's probability of senior with flu"
whereas they are asking for
"probability of person with a flu being a senior"
I am not sure if they are same
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@zarkon
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or
let event A be senior
event B be people with flu
A=1/3 S
B=.2 S
2/5B=(A ∩ B)
probability of people with flu being a senior
(A ∩ B)
------------
B
Zarkon
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so what is your answer
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well, I thought it was 2/15 but I am having second thoughts
Zarkon
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you want P(senior|flu) correct?
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yes, probability of people with flu who are senior
Zarkon
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\[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]
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so , P(F|S) is what I found to be 2/15?
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\[P(F|S)=\frac{2}{5}\]
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ok,
also
is this a defination
\[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]
Zarkon
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\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
is the definition of conditional probability
Zarkon
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use this definition twice to get the formula I have above
Zarkon
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\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
\[P(B|A)=\frac{P(A\cap B)}{P(A)}\Rightarrow P(B|A)P(A)=P(A\cap B) \]
then
\[P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}\]
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wow, thank you very much
Zarkon
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np
Zarkon
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do you have a new final answer?
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(2/5)(1/3)
--------- =
(1/5)
2/15
----- =
1/5
2/15 * 5/1=2/3