Respuesta :
We observe, that P(V|B)>P(V|A), so the person is more likely to have virus, however is still very small probability (only 15%), so in order to confirm illness, he should make one more test.
The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty.
P(V/A) is not 0.95. It is opposite:
P(A/V)=0.95
From the text we can also conclude, that
P(A/∼V)=0.1
P(B/V)=0.9
P(B/∼V)=0.05
P(V)=0.01
P(∼V)=0.99
P(V/A) is not 0.95. It is opposite:
P(A/V)=0.95
From the text we can also conclude, that
P(A/∼V)=0.1
P(B/V)=0.9
P(B/∼V)=0.05
P(V)=0.01
P(∼V)=0.99
What we need to calculate and compare is P(V/A) and P(V/B)
P(V∩A)=P(A)⋅P(V/A)⇒P(V/A)=P(V∩A)/P(A)
P(V∩A) means, that person has a virus and it is detected, so
P(V∩A)=P(V)⋅P(A/V)
=0.01⋅0.95
=0.0095
P(A) is sum of two options: "Person has virus and it is detected" and "Person has no virus, but it was mistakenly detected", therefore:
P(A)=P(V)⋅P(A/V)+P(∼V)⋅P(A/∼V)=0.01⋅0.95+0.99⋅0.1=0.1085
Dividing those two numbers we obtain
P(V/A)=0.0095/0.1085=0.08755760368663594
Analogically,
P(V/B)=P(V∩B)/P(B)=P(V)⋅P(B/V)/P(V)⋅P(B/V)+P(∼V)⋅P(B/∼V)
=0.01⋅0.9/0.01⋅0.9+0.99⋅0.1
=0.1538461538461539
We see that P(V|B)>P(V|A), meaning the individual is more likely to have a virus, but is still very unlikely (only 15%), therefore person should perform another test to confirm illness.
To learn more about probability visit: brainly.com/question/11234923
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