Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
1 answer:
Answer:
A)The probability that someone who tests positive has the disease is 0.9995
B)The probability that someone who tests negative does not have the disease is 0.99999
Step-by-step explanation:
Let D be the event that a person has a disease
Let be the event that a person don't have a disease
Let A be the event that a person is tested positive for that disease.
P(D|A) = Probability that someone has a disease given that he tests positive.
We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive
So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988
We are also given that one person in 10,000 people has a rare genetic disease.
So,
Only 0.4% of the people who don't have it test positive.
= probability that a person is tested positive given he don't have a disease = 0.004
Formula:
P(D|A)==0.9995
P(D|A)=
A)The probability that someone who tests positive has the disease is 0.9995
(B)
=probability that someone does not have disease given that he tests negative
=probability that a person tests negative given that he does not have disease =1-0.004
=0.996
=probability that a person tests negative given that he has a disease =1-0.988=0.012
Formula:
B)The probability that someone who tests negative does not have the disease is 0.99999
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Answer:
(a) 0.343
(b) 0.657
(c) 0.189
(d) 0.216
(e) 0.353
Step-by-step explanation:
Let P(a vehicle passing the test) = p
Let P(a vehicle not passing the test) = q
q = 1 - p
q = 1 - 0.7 = 0.3
(a) P(all of the next three vehicles inspected pass) = P(ppp)
= 0.7 × 0.7 × 0.7
= 0.343
(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)
= (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)
= 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027
= 0.657
(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)
= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)
= 0.063 + 0.063 + 0.063
= 0.189
(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)
= (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)
= 0.063 + 0.063 + 0.063 + 0.027
= 0.216
(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?
P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)
= (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)
= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063
= 0.973
With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,
= 0.353