## Libniz Group Title 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? one year ago one year ago

1. Libniz Group Title

here is what I did

2. Libniz Group Title

|dw:1348935331451:dw|

3. Libniz Group Title

"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. Libniz Group Title

@zarkon

5. Libniz Group Title

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 Group Title

7. Libniz Group Title

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

8. Zarkon Group Title

you want P(senior|flu) correct?

9. Libniz Group Title

yes, probability of people with flu who are senior

10. Zarkon Group Title

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

11. Libniz Group Title

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

12. Zarkon Group Title

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

13. Libniz Group Title

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

14. Zarkon Group Title

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

15. Zarkon Group Title

use this definition twice to get the formula I have above

16. Zarkon Group Title

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

17. Libniz Group Title

wow, thank you very much

18. Zarkon Group Title

np

19. Zarkon Group Title

do you have a new final answer?

20. Libniz Group Title

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