## anonymous 4 years ago 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?

1. anonymous

here is what I did

2. anonymous

|dw:1348935331451:dw|

3. 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

4. anonymous

@zarkon

5. 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

6. Zarkon

7. anonymous

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

8. Zarkon

you want P(senior|flu) correct?

9. anonymous

yes, probability of people with flu who are senior

10. Zarkon

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

11. anonymous

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

12. Zarkon

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

13. anonymous

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

14. Zarkon

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

15. Zarkon

use this definition twice to get the formula I have above

16. 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 $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}$

17. anonymous

wow, thank you very much

18. Zarkon

np

19. Zarkon

do you have a new final answer?

20. anonymous

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