A Warning - Yellow Fever
A century ago, yellow fever was a source of terror—decimating populations and destroying economies. Mass vaccination campaigns led to a dramatic drop in cases worldwide; but the early 2000s witnessed a resurgence of the acute viral hemorrhagic disease in Africa and the Americas, where 40 countries are considered at highest risk.
Suppose a healthy mosquito of the genre Aedes or Haemogogus has a chance of of getting infected upon bitting an infected person. If the chance of transmitting the virus upon a bite to a healthy person is and the proportion of the infected human population is , what is the chance that a healthy mosquito will bite an infected person and just next infect a second (uninfected) person?
For more information about yellow fever, please visit CDC's page or WHO's page . If you live in a high-risk area, get vaccinated. We won't lose this fight.
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1 solution
The chance that a healthy mosquito is bitting someone who is infected is exactly the proportion of infected human population f . We want the mosquito to get infected, so the probability of that happening is f P g . The probability that this mosquito is bitting someone who is not infected and infecting is ( 1 − f ) P i . Therefore the probability of a healthy mosquito bitting an infected person and just next infect (and therefore he was infected) a second (uninfected) person is just the product of these values:
P g P i f ( 1 − f ) = P g P i ( f − f 2 ) .
Notice that this chance will look like this
where we can nicely see that the maximum probability of spreading the infection happens when exactly half of the population is sick ( f = 1 / 2 ) . We can also see that by maximizing the value of P ( f ) = P g P i ( f − f 2 ) where P g P i is a constant:
P ′ ( f ) = P g P i ( 1 − 2 f ) = 0 ⇒ f = 1 / 2 .