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Q&A In general, does $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$?

1 answer  ·  posted 3y ago by DNB‭  ·  last activity 3y ago by whybecause‭

Question probability
#3: Post edited by user avatar DNB‭ · 2021-12-24T06:46:37Z (almost 3 years ago)
  • $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$?
  • In general, does $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$?
  • In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. The author's solution doesn't unfurl how he computed $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I am hankering to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.
  • >25. A crime is committed by one of two suspects, A and B. Initially, there is equal
  • evidence against both of them. In further investigation at the crime scene, it is found
  • that the guilty party had a blood type found in 10% of the population. Suspect A does
  • match this blood type, whereas the blood type of Suspect B is unknown.
  • >
  • >(a) Given this new information, what is the probability that A is the guilty party?
  • >
  • >(b) Given this new information, what is the probability that B's blood type matches
  • that found at the crime scene?
  • >
  • >## Solution:
  • >
  • >(a) Let M be the event that A's blood type matches the guilty party's and for brevity,
  • write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,
  • >
  • >![Image alt text](https://math.codidact.com/uploads/apFEKuigSy4sRBJDyMsGYqH9)
  • Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in the publicly downloadable PDF of curbed solutions.
  • In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. But the author's solution didn't annunciate this, and doesn't unfurl how to compute $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I hanker to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.
  • >25. A crime is committed by one of two suspects, A and B. Initially, there is equal
  • evidence against both of them. In further investigation at the crime scene, it is found
  • that the guilty party had a blood type found in 10% of the population. Suspect A does
  • match this blood type, whereas the blood type of Suspect B is unknown.
  • >
  • >(a) Given this new information, what is the probability that A is the guilty party?
  • >
  • >(b) Given this new information, what is the probability that B's blood type matches
  • that found at the crime scene?
  • >
  • >## Solution:
  • >
  • >(a) Let M be the event that A's blood type matches the guilty party's and for brevity,
  • write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,
  • >
  • >![Image alt text](https://math.codidact.com/uploads/apFEKuigSy4sRBJDyMsGYqH9)
  • Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in the publicly downloadable PDF of curbed solutions.
#2: Post edited by user avatar DNB‭ · 2021-12-23T09:11:46Z (almost 3 years ago)
  • In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. The author's solution doesn't unfurl how he computed $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I am hankering to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.
  • >25. A crime is committed by one of two suspects, A and B. Initially, there is equal
  • evidence against both of them. In further investigation at the crime scene, it is found
  • that the guilty party had a blood type found in 10% of the population. Suspect A does
  • match this blood type, whereas the blood type of Suspect B is unknown.
  • >
  • >(a) Given this new information, what is the probability that A is the guilty party?
  • >
  • >(b) Given this new information, what is the probability that B's blood type matches
  • that found at the crime scene?
  • >
  • >## Solution:
  • >
  • >(a) Let M be the event that A's blood type matches the guilty party's and for brevity,
  • write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,
  • >
  • >![Image alt text](https://math.codidact.com/uploads/apFEKuigSy4sRBJDyMsGYqH9)
  • Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in Student's Solution Manual.
  • In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. The author's solution doesn't unfurl how he computed $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I am hankering to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.
  • >25. A crime is committed by one of two suspects, A and B. Initially, there is equal
  • evidence against both of them. In further investigation at the crime scene, it is found
  • that the guilty party had a blood type found in 10% of the population. Suspect A does
  • match this blood type, whereas the blood type of Suspect B is unknown.
  • >
  • >(a) Given this new information, what is the probability that A is the guilty party?
  • >
  • >(b) Given this new information, what is the probability that B's blood type matches
  • that found at the crime scene?
  • >
  • >## Solution:
  • >
  • >(a) Let M be the event that A's blood type matches the guilty party's and for brevity,
  • write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,
  • >
  • >![Image alt text](https://math.codidact.com/uploads/apFEKuigSy4sRBJDyMsGYqH9)
  • Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in the publicly downloadable PDF of curbed solutions.
#1: Initial revision by user avatar DNB‭ · 2021-12-23T09:08:25Z (almost 3 years ago)
$\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? 
In this question, $\color{forestgreen}{P(A|M)} + \color{red}{P(B|M)} = 1$. The author's solution doesn't unfurl how he computed $\color{red}{P(B|M)}$. Can I simply subtract as $\color{red}{P(B|M)} = 1 - \color{forestgreen}{P(A|M)}$? I am hankering to avoid calculating $\color{red}{P(B|M)}$ by applying Bayes's Rule a second time.  

>25. A crime is committed by one of two suspects, A and B. Initially, there is equal
evidence against both of them. In further investigation at the crime scene, it is found
that the guilty party had a blood type found in 10% of the population. Suspect A does
match this blood type, whereas the blood type of Suspect B is unknown.
>
>(a) Given this new information, what is the probability that A is the guilty party?
>
>(b) Given this new information, what is the probability that B's blood type matches
that found at the crime scene?
>
>## Solution:
>
>(a) Let M be the event that A's blood type matches the guilty party's and for brevity,
write A for "A is guilty" and B for "B is guilty". By Bayes' Rule,
>
>![Image alt text](https://math.codidact.com/uploads/apFEKuigSy4sRBJDyMsGYqH9)

Blitzstein, *Introduction to Probability* (2019 2 edn), Ch 2, Exercise 25, p 87. p 12 in Student's Solution Manual.