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?

Question

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?

anonymous · Answer

here is what I did

anonymous · Answer

|dw:1348935331451:dw|

anonymous · Answer

"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

anonymous · Answer

@zarkon

anonymous · Answer

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 · Answer

so what is your answer

anonymous · Answer

well, I thought it was 2/15 but I am having second thoughts

zarkon · Answer

you want P(senior|flu) correct?

anonymous · Answer

yes, probability of people with flu who are senior

zarkon · Answer

$$P(S|F)=\frac{P(F|S)P(S)}{P(F)}$$

anonymous · Answer

so , P(F|S) is what I found to be 2/15?

zarkon · Answer

$$P(F|S)=\frac{2}{5}$$

anonymous · Answer

ok,

also

is this a defination
$$P(S|F)=\frac{P(F|S)P(S)}{P(F)}$$

zarkon · Answer

$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

is the definition of conditional probability

zarkon · Answer

use this definition twice to get the formula I have above

zarkon · Answer

$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

$$P(B|A)=\frac{P(A\cap B)}{P(A)}\Rightarrow P(B|A)P(A)=P(A\cap B) $$

then 

$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}$$

anonymous · Answer

wow, thank you very much

zarkon · Answer

np

zarkon · Answer

do you have a new final answer?

anonymous · Answer

(2/5)(1/3)
---------  =
    (1/5)

2/15
----- =
1/5

2/15 * 5/1=2/3