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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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?

Mathematics
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here is what I did
|dw:1348935331451:dw|
"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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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
so what is your answer
well, I thought it was 2/15 but I am having second thoughts
you want P(senior|flu) correct?
yes, probability of people with flu who are senior
\[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]
so , P(F|S) is what I found to be 2/15?
\[P(F|S)=\frac{2}{5}\]
ok, also is this a defination \[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] is the definition of conditional probability
use this definition twice to get the formula I have above
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\] \[P(B|A)=\frac{P(A\cap B)}{P(A)}\Rightarrow P(B|A)P(A)=P(A\cap B) \] then \[P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}\]
wow, thank you very much
np
do you have a new final answer?
(2/5)(1/3) --------- = (1/5) 2/15 ----- = 1/5 2/15 * 5/1=2/3

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