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Libniz
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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?
 2 years ago
 2 years ago
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?
 2 years ago
 2 years ago

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Libniz Group TitleBest ResponseYou've already chosen the best response.1
here is what I did
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
dw:1348935331451:dw
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
"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
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
so what is your answer
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
well, I thought it was 2/15 but I am having second thoughts
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
you want P(seniorflu) correct?
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
yes, probability of people with flu who are senior
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[P(SF)=\frac{P(FS)P(S)}{P(F)}\]
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
so , P(FS) is what I found to be 2/15?
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[P(FS)=\frac{2}{5}\]
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
ok, also is this a defination \[P(SF)=\frac{P(FS)P(S)}{P(F)}\]
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[P(AB)=\frac{P(A\cap B)}{P(B)}\] is the definition of conditional probability
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
use this definition twice to get the formula I have above
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[P(AB)=\frac{P(A\cap B)}{P(B)}\] \[P(BA)=\frac{P(A\cap B)}{P(A)}\Rightarrow P(BA)P(A)=P(A\cap B) \] then \[P(AB)=\frac{P(A\cap B)}{P(B)}=\frac{P(BA)P(A)}{P(B)}\]
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
wow, thank you very much
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
do you have a new final answer?
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.1
(2/5)(1/3)  = (1/5) 2/15  = 1/5 2/15 * 5/1=2/3
 2 years ago
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