Q.3.90 A person tried by a 3-judge pane... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q.3.90 (page 106)

A person tried by a $3$ -judge panel is declared guilty if at least $2$ judges cast votes of guilty. Suppose that when the defendant is in fact guilty, each judge will independently vote guilty with probability . $7$ , whereas when the defendant is, in fact, innocent, this probability drops to . $2$ . If $70$ percent of defendants are guilty, compute the conditional probability that judge number $3$ votes guilty given that
(a) judges $1$ and $2$ votes guilty;
(b) judges $1$ and $2$ casts $1$ guilty and $1$ not guilty vote;
(c) judges $1$ and $2$ both cast not guilty votes.
Let Ei,i= $1, 2, 3$ denote the event that judge i casts a guilty vote. Are these events independent? Are they conditionally independent? Explain

Short Answer

Expert verified

Events Ei are conditionally independent given that the suspect is guilty.

Use Bayes formula in regard to the suspect is, or is not guilty

After calcu膷ating a) and c), b) can be acquired by another Bayes formula, P(E3) being the weighted average of a), b) and c)

$a) \frac{97}{142} =0.683 b) \frac{11}{34} = 0.324 c) \frac{45}{122} =0.369$

Step by step solution

Given Information (part a)

compute the conditional probability that judge number 3 votes guilty given that judges 1 and 2 vote guilty;

Explanation (part a)

Events:

G - the suspect is guility

$E_{i} - judge i = 1, 2, 3 casts a guilty vote$

Probabilities

$P (G) = 0.7 鈫� P (G^{c}) = 0.3$

$P (E_{i} 鈭� G) = 0.7 for i 鈭� {1, 2, 3}$

$P (E_{i} 鈭� G^{c}) = 0.2 for i 鈭� {1, 2, 3}$

$The events E_{1}, E_{2}, E_{3} are independent given G (or G^{c})$

Otherwise, events E1, E2, E3 are dependent. This can be read from the context of the problem or the fact that the problem is not defined if this is not the case (not all needed probabilities are given).

calculate

$a) P (E_{3} 鈭� E_{1} E_{2})$

$b) P [E_{3} 鈭� (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2})]$

$c) P (E_{3} 鈭� E_{1}^{c} E_{2}^{c})$

Explanation (part a)

a) Start with the definition

$P (E_{3} 鈭� E_{1} E_{2}) = \frac{P (E_{1} E_{2} E_{3})}{P (E_{1} E_{2})}$

$Both probabilities on the right hand side can be calculated using Bayes formula with events G and G^{c}$

$P (E_{1} E_{2} E_{3}) = P (E_{1} E_{2} E_{3} 鈭� G) P (G) + P (E_{1} E_{2} E_{3} 鈭� G^{c}) P (G^{c})$

$P (E_{1} E_{2}) = P (E_{1} E_{2} 鈭� G) P (G) + P (E_{1} E_{2} 鈭� G^{c}) P (G^{c})$

When conditional independence is applied to this, we obtain

$P (E_{1} E_{2} E_{3}) = P (E_{1} 鈭� G) P (E_{2} 鈭� G) P (E_{3} 鈭� G) P (G) + P (E_{1} 鈭� G^{c}) P (E_{2} 鈭� G^{c}) P (E_{3} 鈭� G^{c}) P (G^{c})$

$P (E_{1} E_{2}) = P (E_{1} 鈭� G) P (E_{2} 鈭� G) P (G) + P (E_{1} 鈭� G^{c}) P (E_{2} 鈭� G^{c}) P (G^{c})$

substituting the known probabilities here the result is:

$P (E_{1} E_{2} E_{3}) = {0.7}^{3} 鈰� 0.7 + {0.2}^{3} 鈰� 0.3 = 0.2425$

$P (E_{1} E_{2}) = {0.7}^{2} 鈰� 0.7 + {0.2}^{2} 鈰� 0.3 = 0.355$

localid="1648545722244" $鈬� P (E_{3} 鈭� E_{1} E_{2}) = \frac{97}{142}$

$= 0.683$

Final Answer (part a)

Judges 1 and 2 vote guilty is $鈬� P (E_{3} 鈭� E_{1} E_{2}) = \frac{97}{142} = 0.683 .$

Given Information (part c)

The conditional probability that, judges $1$ and $2$ both cast not guilty votes.

Explanation (part c)

The same method as for a)

$P (E_{3} 鈭� E_{1}^{c} E_{2}^{c}) = \frac{P (E_{1}^{c} E_{2}^{c} E_{3})}{P (E_{1}^{c} E_{2}^{c})}$

Bayes formula

$P (E_{1}^{c} E_{2}^{c} E_{3}) = P (E_{1}^{c} E_{2}^{c} E_{3} 鈭� G) P (G) + P (E_{1}^{c} E_{2}^{c} E_{3} 鈭� G^{c}) P (G^{c})$

$P (E_{1}^{c} E_{2}^{c}) = P (E_{1}^{c} E_{2}^{c} 鈭� G) P (G) + P (E_{1}^{c} E_{2}^{c} 鈭� G^{c}) P (G^{c})$

$Conditional independence given G (and G^{c}) is applied to this, then, the probabilities are substituted to obtain$

$P (E_{1}^{c} E_{2}^{c} E_{3}) = (1 鈭� 0.7)^{2} 鈰� 0.7 鈰� 0.7 + (1 鈭� 0.2)^{2} 鈰� 0.2 鈰� 0.3 = 0.0825$

$P (E_{1}^{c} E_{2}^{c}) = (1 鈭� 0.7)^{2} 鈰� 0.7 + (1 鈭� 0.2)^{2} 鈰� 0.3 = 0.255$

鈬�P(E3鈭�E1E2)=1134=0.324localid="1648545962939" $鈬� P (E_{3} 鈭� E_{1} E_{2}) = \frac{11}{34} = 0.3235$

Final Answer (part c)

The conditional probability that, judges $1$ and $2$ both cast not guilty votes.
localid="1648545996450" $鈬� P (E_{3} 鈭� E_{1} E_{2}) = \frac{11}{34} = 0.3235$

Given Information (part b)

Judges $1$ and $2$ cast $1$ guilty and $1$ not guilty vote;

Explanation (part b)

$There are many Bayes formulae for calculating P (E_{3}), two of them are:$

$P (E_{3}) = P (E_{3} 鈭� G) P (G) + P (E_{3} 鈭� G^{c}) P (G^{c}) = {0.7}^{2} + 0.2 鈰� 0.3 = 0.55$

$P (E_{3}) = P (E_{3} 鈭� E_{1} E_{2}) P (E_{1} E_{2}) + P [E_{3} 鈭� (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2})] P (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2}) + P (E_{3} 鈭� E_{1}^{c} E_{2}^{c})$

$Conclusion is to equate right hand side in the second equation with P (E_{3}) = 0.55$

$If we first note that E_{1} E_{2}, E_{1}^{c} E_{2}^{c} and E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2} are mutually exclusive, and their union is the whole space, from axiom 3$

$P (E_{1} E_{2}) + P (E_{1}^{c} E_{2}^{c}) + P (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2}) = 1 \overset{Substitution}{鈬�} P (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2}) = 0.61$

And:

$P (E_{3} 鈭� E_{1} E_{2}) P (E_{1} E_{2}) = P (E_{3} E_{1} E_{2}) = 0.2425$

$P (E_{3} 鈭� E_{1}^{c} E_{2}^{c}) P (E_{1}^{c} E_{2}^{c}) = P (E_{3} E_{1}^{c} E_{2}^{c}) = 0.0825$

$Now there is only one unknown in equation (2)$

$0.55 = 0.2425 + P [E_{3} 鈭� (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2})] 鈰� 0.61 + 0.0825$

From this ,

localid="1648546032593" $P [E_{3} 鈭� (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2})] = \frac{45}{122} 鈮� 0.369$

Final Answer (part b)

The conditional probability that, judges $1$ and $2$ cast $1$ guilty and $1$ not guilty vote is

localid="1648546061166" $P [E_{3} 鈭� (E_{1} E_{2}^{c} 鈭� E_{1}^{c} E_{2})] = \frac{45}{122} = 0.369$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Day	P(mail/accepted)	P(mail/rejected)
Monday	$. 15$	$. 05$
Tuesday	$. 20$	$. 10$
Wednesday	$. 25$	$. 10$
Thursday	$. 15$	$. 15$
Friday	$. 10$	$. 20$

91影视

Short Answer

Step by step solution

Given Information (part a)

Explanation (part a)

Explanation (part a)

Final Answer (part a)

Given Information (part c)

Explanation (part c)

Final Answer (part c)

Given Information (part b)

Explanation (part b)

Final Answer (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Applied Mathematics

Discrete Mathematics

Pure Maths

Geometry

Study anywhere. Anytime. Across all devices.