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A terrible disease sweeps around the world, luckily only affecting 1

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KarunMendiratta
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A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   16 Aug 2021, 08:03
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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.

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
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C
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A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   19 Aug 2021, 07:23
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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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A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   16 Aug 2021, 10:39
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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

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.

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
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A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   19 Aug 2021, 04:46
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Explanation:

If a person has a positive report then, there is 99% of chance that he is actually positive
i.e. He will be the 1 out of those 10000 tested.

So, Probability = \(\frac{1}{10000}(0.99) = \frac{99}{1000000}\)

There is also a 1% chance that he was tested fake positive
i.e. He will be the 9999 out of those 10000

So, Probability = \(\frac{9999}{10000}(0.01) = \frac{9999}{1000000}\)

Thus required Probability = \(\frac{0.000099}{(0.000099 + 0.009999)} = \frac{0.000099}{0.010098} = 0.009803\)

Col. A: 0.009803
Col. B: 0.009803

Hence, option C
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Re: A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   19 Aug 2021, 05:16
KarunMendiratta wrote:
Explanation:

If a person has a positive report then, there is 99% of chance that he is actually positive
i.e. He will be the 1 out of those 10000 tested.

So, Probability = \(\frac{1}{10000}(0.99) = \frac{99}{1000000}\)

There is also a 1% chance that he was tested fake positive
i.e. He will be the 9999 out of those 10000

So, Probability = \(\frac{9999}{10000}(0.01) = \frac{9999}{1000000}\)

Thus required Probability = \(\frac{0.000099}{(0.000099 + 0.009999)} = \frac{0.000099}{0.010098} = 0.009803\)

Col. A: 0.009803
Col. B: 0.009803

Hence, option C

Could you please help me understand if we already considered true positive then why do we need to consider false positive?
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Re: A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   19 Aug 2021, 05:51
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koala wrote:
KarunMendiratta wrote:
Explanation:

If a person has a positive report then, there is 99% of chance that he is actually positive
i.e. He will be the 1 out of those 10000 tested.

So, Probability = \(\frac{1}{10000}(0.99) = \frac{99}{1000000}\)

There is also a 1% chance that he was tested fake positive
i.e. He will be the 9999 out of those 10000

So, Probability = \(\frac{9999}{10000}(0.01) = \frac{9999}{1000000}\)

Thus required Probability = \(\frac{0.000099}{(0.000099 + 0.009999)} = \frac{0.000099}{0.010098} = 0.009803\)

Col. A: 0.009803
Col. B: 0.009803

Hence, option C

Could you please help me understand if we already considered true positive then why do we need to consider false positive?


Because the total outcomes would be all possibilities of getting a positive result i.e. 99% positive (Actual) + 1% positive (Fake)
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Re: A terrible disease sweeps around the world, luckily only affecting 1 [#permalink] New post   14 Nov 2024, 17:36
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Re: A terrible disease sweeps around the world, luckily only affecting 1 [#permalink]
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