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A terrible disease sweeps around the world, luckily only affecting 1
[#permalink]
16 Aug 2021, 08:03
16 Aug 2021, 08:03
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Question Stats:
3% (01:28) correct
96% (01:58) wrong based on 32 sessions
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A terrible disease sweeps around the world, affecting 1 in 10,000. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease or not. In other words, the test yields the correct positive or correct negative result 99% of the time.
A person took the test and a week later, he received a positive test report.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
A person took the test and a week later, he received a positive test report.
Quantity A |
Quantity B |
Probability that he actually has the disease |
\(\frac{1}{102}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
A terrible disease sweeps around the world, luckily only affecting 1
[#permalink]
19 Aug 2021, 07:23
19 Aug 2021, 07:23
1
revised
Bayes' Theorem: \(P(A|B)=\frac{P(B|A)*P(A)}{P(B)}\), where P(A) is the probability of event A, P(B) is the probability of event B, P(A|B) is the probability of observing event A if B is true, P(B|A) is the probability of observing event B if A is true.
P(Positive|Sick)=0.99
P(Sick)=1/10,000
P(Positive)=101/10,000
\(P(Sick|Positive)=\frac{P(Positive|Sick)*P(Sick)}{P(Positive)}\)
\(P(Sick|Positive)=\frac{0.99*0.0001}{0.0101}=0.0098\)
Quantity A: 0.0098
Quantity B: 0.0098
Correction: answer is C
Bayes' Theorem: \(P(A|B)=\frac{P(B|A)*P(A)}{P(B)}\), where P(A) is the probability of event A, P(B) is the probability of event B, P(A|B) is the probability of observing event A if B is true, P(B|A) is the probability of observing event B if A is true.
P(Positive|Sick)=0.99
P(Sick)=1/10,000
P(Positive)=101/10,000
\(P(Sick|Positive)=\frac{P(Positive|Sick)*P(Sick)}{P(Positive)}\)
\(P(Sick|Positive)=\frac{0.99*0.0001}{0.0101}=0.0098\)
Quantity A: 0.0098
Quantity B: 0.0098
Correction: answer is C
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General Discussion
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
A terrible disease sweeps around the world, luckily only affecting 1
[#permalink]
16 Aug 2021, 10:39
16 Aug 2021, 10:39
1
1
Bookmarks
Quantity A:\(\frac{1}{10,000} * \frac{9,999}{10,000} : \frac{99+99}{10,000} = 0.005\)
Quantity B: \(\frac{1}{102}=0.009\)
Quantity A < Quantity B
Answer must be B
Quantity B: \(\frac{1}{102}=0.009\)
Quantity A < Quantity B
Answer must be B
KarunMendiratta wrote:
A terrible disease sweeps around the world, affecting 1 in 10,000. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease or not. In other words, the test yields the correct positive or correct negative result 99% of the time.
A person took the test and a week later, he received a positive test report.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
A person took the test and a week later, he received a positive test report.
Quantity A |
Quantity B |
Probability that he actually has the disease |
\(\frac{1}{102}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
