A lie detector is known to be 4/5 reliable when the person is guilty and 9/10 reliable when the person is innocent. If a suspect is chosen from a group of suspects of which only 1/100 have ever committed a crime, and the test indicates that the person is guilty, what is the probability that he is guilty?

Answers

Answer 1

Answer:

0.0748 = 7.48% probability that he is guilty

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Tests indicates guilty

Event B: Person is guilty

Probability that the test indicates that the person is guilty:

4/5 = 0.8 of 1/100 = 0.01(person is guilty)

1 - 0.9 = 0.1 of 1 - 0.99(person is not guilty). So

[tex]P(A) = 0.8*0.01 + 0.1*0.99 = 0.107[/tex]

Test indicates guilty and the person is guilty;

0.01 of 0.8. So

[tex]P(A \cap B) = 0.01*0.8 = 0.008[/tex]

What is the probability that he is guilty?

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.008}{0.107} = 0.0748[/tex]

0.0748 = 7.48% probability that he is guilty

Answer 2

The required value is [tex]\simeq0.925 \ or \ 92.5[/tex]%.

Probability:

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it.

Let [tex]T_G[/tex] and [tex]T_I[/tex] denote the test result being guilty or innocent respectively.

And, [tex]G[/tex] and [tex]I[/tex] denote that the person is guilty or innocent respectively.

Given that: [tex]P\left ( T_G\mid I \right )=\frac{4}{5}[/tex]

[tex]P\left ( T_I\mid I \right )=\frac{9}{10}\Rightarrow P\left ( T_G\mid I \right )=1-\frac{9}{10}=\frac{1}{10}[/tex]

[tex]P(G)=\frac{1}{100} \\ \therefore P(I)=1-P(G) \\ =1-\frac{1}{100} \\ =\frac{99}{100}[/tex]

Now, the required probability is,

[tex]P(I\mid T_G)=\frac{P\left ( T_G\mid I \right )\times P(I)}{P(T_G\mid I)\times P(I)+P\left ( T_G\mid G \right )\times P(G)}[/tex]

              [tex]=\frac{\frac{1}{10}\times\frac{99}{100}}{\left ( \frac{1}{10}\times \frac{99}{100}\right )+\left ( \frac{4}{5}\times\frac{1}{100} \right )} \\ =\frac{0.099}{0.099+0.008} \\ =\frac{0.099}{0.107} \\ =0.92523 \\ \simeq 0.925 \\ or \ 92.5%[/tex]

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

 

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Answers

Answer:

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Step-by-step explanation:

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Sorry for getting back to you late.

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Answer:

1) [tex]\frac{\sin x}{1-\cos x} = \csc x + \cot x[/tex]

2) [tex]\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}[/tex]

3) [tex]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}[/tex]

4) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}[/tex]

5) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}[/tex]

6) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}[/tex]

7) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}[/tex]

Step-by-step explanation:

Now we proceed to show all steps needed to demonstrate the trigonometric identity:

1) [tex]\frac{\sin x}{1-\cos x} = \csc x + \cot x[/tex] Given.

2) [tex]\frac{\sin x}{1-\cos x} = \frac{1}{\sin x} + \frac{\cos x}{\sin x}[/tex] Identities for cosecant and cotangent functions.

3) [tex]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}[/tex]             [tex]\frac{a}{b}+\frac{c}{b} = \frac{a+c}{b}[/tex]

4) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x \cdot (1+\cos x)}{\sin^{2}x}[/tex] Existence of additive inverse/Modulative property.

5) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{1-\cos^{2}x}[/tex]    Fundamental trigonometric identity.

6) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x\cdot (1+\cos x)}{(1+\cos x)\cdot (1-\cos x)}[/tex]   Factorization.

7) [tex]\frac{\sin x}{1-\cos x} = \frac{\sin x}{1-\cos x}[/tex] Existence of additive inverse/Modulative property/Result.

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