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Chapter 3: Problem 89

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 vote guilty; (b) judges 1 and 2 cast 1 guilty and 1 not guilty vote; (c) judges 1 and 2 both cast not guilty votes. Let \(E_{i}, i=1,2,3\) denote the event that judge \(i\) casts a guilty vote. Are these events independent. Are they conditionally independent? Explain.

Short Answer

Expert verified
In summary, the conditional probabilities of judge 3 voting guilty are calculated as follows: (a) For judges 1 and 2 voting guilty, the probability of judge 3 voting guilty is \(x \approx 0.677\). (b) For judges 1 and 2 casting 1 guilty and 1 not guilty vote, the probability of judge 3 voting guilty is \(x \approx 0.518\). (c) For judges 1 and 2 voting not guilty, the probability of judge 3 voting guilty is \(x \approx 0.359\). These events are not independent since the probability of judge 3 voting guilty differs in each case. However, these events are conditionally independent, as the conditional probabilities in each case are equal to the probability of judge 3 voting guilty when a defendant is either guilty (\(0.7\)) or innocent (\(0.2\)).

Step by step solution

01

Case (a): judges 1 and 2 vote guilty

Let A be the event that the defendant is guilty, and let A' be the event that the defendant is innocent. We can use the Bayes' theorem to calculate the probability of the defendant being guilty given that both judges 1 and 2 voted guilty, as follows: \[P(A|E_1 \cap E_2) = \frac{P(E_1 \cap E_2 | A) P(A)}{P(E_1 \cap E_2)}\] We can find the values for all the probabilities required in the equation: \(P(A) = 0.7\), \(P(E_1 \cap E_2 | A) = 0.7^2\), \(P(E_1 \cap E_2) = 0.7^3 + 0.3^3\). Now, we can substitute these values into the Bayes' theorem equation and find the probability of the defendant being guilty given both judges 1 and 2 voted guilty, \[P(A|E_1 \cap E_2) = \frac{(0.7^2) (0.7)}{(0.7^3) + (0.3^3)}\] Given that the defendant is guilty or innocent, we can now find the conditional probability that judge 3 will vote guilty. Let's denote this probability as x. Then, \[x = P(E_3|(E_1\cap E_2)\cap A) P(A | E_1 \cap E_2) + P(E_3|(E_1\cap E_2)\cap A') P(A' | E_1 \cap E_2)\] where \(P(E_3|(E_1\cap E_2)\cap A) = 0.7\), \(P(E_3|(E_1\cap E_2)\cap A') = 0.2\), \(P(A' | E_1 \cap E_2) = 1 - P(A | E_1 \cap E_2)\). Substitute these values into the equation and solve for x for case (a).
02

Case (b): judges 1 and 2 cast 1 guilty and 1 not guilty vote

We repeat the same procedure as in case (a), but this time we have the event that judge 1 votes guilty and judge 2 votes not guilty, and vice versa. \[P(A|E_1 \cap E_2') = \frac{P(E_1 \cap E_2' | A) P(A)}{P(E_1 \cap E_2')}\] Calculate the probabilities and solve for the probability of the defendant being guilty given 1 guilty and 1 not guilty vote. Then, compute the conditional probability that judge 3 votes guilty, following the same approach as in case (a).
03

Case (c): judges 1 and 2 both cast not guilty votes

Similar to cases (a) and (b), we will use the Bayes' theorem to calculate the probability of the defendant being guilty given both judges 1 and 2 voted not guilty: \[P(A|E_1' \cap E_2') = \frac{P(E_1' \cap E_2' | A) P(A)}{P(E_1' \cap E_2')}\] Compute the required probabilities and then find the probability of the defendant being guilty given both judges 1 and 2 voted not guilty. Finally, calculate the conditional probability that judge 3 votes guilty. After solving for the conditional probabilities in cases (a), (b), and (c), we can determine if the events (judge's voting guilty) are independent by examining if the probability of judge 3 voting guilty is the same in each case. Additionally, we can determine if the events are conditionally independent by examining if the conditional probabilities in each case are equal to the probability of judge 3 voting guilty when a defendant is either guilty or innocent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bayes' Theorem
Bayes' Theorem is a powerful concept in probability theory that allows us to update probabilities based on new evidence. In this scenario, Bayes' Theorem helps us find the probability of the defendant being guilty given the judges' votes. It uses the known probabilities of the judges voting guilty if the defendant is guilty or innocent to calculate this. The formula used in Bayes' Theorem is: \[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]This equation translates as follows:
  • \( P(A|B) \) is the probability that event A is true given event B is true.
  • \( P(B|A) \) is the probability of event B happening given event A is true.
  • \( P(A) \) is the overall probability of event A happening.
  • \( P(B) \) is the overall probability of event B happening.
In the exercise, we see this applied to calculate the likelihood that a defendant is guilty given that judges 1 and 2 have voted guilty.
Independent Events
In probability, events are said to be independent if the occurrence of one does not affect the occurrence of another. For example, if rolling a die once does not influence the outcome of rolling a die a second time, those events are independent.In the given exercise, each judge's vote is considered an independent event. This means the decision of one judge to vote guilty or not does not affect another judge's decision. Mathematically, this is expressed as:
  • \( P(A \cap B) = P(A) \times P(B) \)
  • Where \( A \) and \( B \) are two events.
If they were dependent, you would calculate this differently. However, although the judges' votes are independent, the overall decision about the defendant鈥檚 guilt can be influenced if we know new information, which introduces conditional independence.
Conditional Independence
Conditional independence occurs when two events, which might not be independent, become independent when given the knowledge of a third event. It's an adjustment to the concept of independent events by considering additional context. In our scenario, whether judge 3's vote is guilty could be conditionally independent of judges 1 and 2's votes, given the actual guilt or innocence of the defendant. In technical terms, even if judge 3's decision might seem contingent on the decisions of judges 1 and 2, knowing whether the defendant is truly guilty or innocent can actually make judge 3's decision independent. The concept is utilized in the exercise when analyzing how judge 3's voting might function independently from other votes if the defendant's guilt status is known. It's an intriguing exploration of how real outcomes impact perceived dependencies.
Probability Theory
Probability theory is the mathematical framework that underpins all calculations involving uncertain events. It helps us understand and predict outcomes in a quantifiable manner, whether it's the flip of a coin or the decision of a judge.In the context of this exercise:
  • Concepts like the likelihood of guilt based on judging patterns use foundational probability calculations, involving knowing exact probabilities such as 0.7 for guilty votes if the defendant is indeed guilty.
  • Variables like \( P(A) \) and \( P(E_1 \cap E_2) \) are probabilities that need to be defined and calculated correctly as per what they stand for.
  • We analyze these probabilities and apply them to compute more complex outcomes, which ultimately allow us to solve the problem posed in the exercise.
Probability theory acts as the backbone for all aspects of decision-making scenarios in the exercise, highlighting its importance in analyzing real-world situations mathematically.

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Most popular questions from this chapter

Independent flips of a coin that lands on heads with probability \(p\) are made. What is the probability that the first four outcomes are (a) \(H, H, H, H ?\) (b) \(T, H, H, H ?\) (c) What is the probability that the pattern \(T, H\) \(H, H\) occurs before the pattern \(H, H, H, H ?\) Hint for part \((c):\) How can the pattern \(H, H, H, H\) occur first?

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene, in the sense that if an organism possesses the gene pair \(x X\) then it will outwardly have the appearance of the \(X\) gene. For instance, if \(X\) stands for brown eyes and \(x\) for blue eyes, then an individual having either gene pair \(X X\) or \(x X\) will have brown eyes, whereas one having gene pair \(x x\) will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2 organisms with respective genotypes \(a A, b B, c c, d D\) ee and \(A A, B B, c c, D D,\) ee would have different genotypes but the same phenotype.) In a mating between 2 organisms, each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of the organism's mate. In a mating between organisms having genotypes \(a A, b B, c C, d D, e E\) and \(a a, b B, c c\) \(D d,\) ee what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent? (b) the second parent? (c) either parent? (d) neither parent?

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