GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A terrible disease sweeps around the world, luckily only affecting 1
[#permalink]
16 Aug 2021, 08:03
16 Aug 2021, 08:03
15
Bookmarks
00:00
Question Stats:
5% (02:03) correct
94% (01:56) wrong based on 34 sessions
Hide Show timer Statistics
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
Attachments
Probability (2).jpg [ 16.57 KiB | Viewed 2604 times ]
Probability.jpg [ 12.96 KiB | Viewed 2611 times ]
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
