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Libniz

  • 2 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?

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  1. Libniz
    • 2 years ago
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    here is what I did

  2. Libniz
    • 2 years ago
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    |dw:1348935331451:dw|

  3. Libniz
    • 2 years ago
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    "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
    • 2 years ago
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    @zarkon

  5. Libniz
    • 2 years ago
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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

  6. Zarkon
    • 2 years ago
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    so what is your answer

  7. Libniz
    • 2 years ago
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    well, I thought it was 2/15 but I am having second thoughts

  8. Zarkon
    • 2 years ago
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    you want P(senior|flu) correct?

  9. Libniz
    • 2 years ago
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    yes, probability of people with flu who are senior

  10. Zarkon
    • 2 years ago
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    \[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]

  11. Libniz
    • 2 years ago
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    so , P(F|S) is what I found to be 2/15?

  12. Zarkon
    • 2 years ago
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    \[P(F|S)=\frac{2}{5}\]

  13. Libniz
    • 2 years ago
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    ok, also is this a defination \[P(S|F)=\frac{P(F|S)P(S)}{P(F)}\]

  14. Zarkon
    • 2 years ago
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    \[P(A|B)=\frac{P(A\cap B)}{P(B)}\] is the definition of conditional probability

  15. Zarkon
    • 2 years ago
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    use this definition twice to get the formula I have above

  16. Zarkon
    • 2 years ago
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    \[P(A|B)=\frac{P(A\cap B)}{P(B)}\] \[P(B|A)=\frac{P(A\cap B)}{P(A)}\Rightarrow P(B|A)P(A)=P(A\cap B) \] then \[P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B|A)P(A)}{P(B)}\]

  17. Libniz
    • 2 years ago
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    wow, thank you very much

  18. Zarkon
    • 2 years ago
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    np

  19. Zarkon
    • 2 years ago
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    do you have a new final answer?

  20. Libniz
    • 2 years ago
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    (2/5)(1/3) --------- = (1/5) 2/15 ----- = 1/5 2/15 * 5/1=2/3

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