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

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