Conditional Probability with 3 Events

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SUMMARY

This discussion focuses on solving a conditional probability problem involving three events: A, B, and C. Key values provided include P(A ∩ B) = 0.4, P(A ∩ C) = 0.2, P(B|A) = 0.6, and P(B) = 0.5. The solutions derived include P(A|B) = 0.6, P(B') = 1/3, P(A) = 2/3, P(C|A) = 0.2, and the odds in favor of B as 2 out of 5. The discussion emphasizes the importance of understanding the relationships between these probabilities for effective problem-solving.

PREREQUISITES
  • Understanding of conditional probability
  • Familiarity with probability notation (e.g., P(A ∩ B))
  • Knowledge of Bayes' theorem
  • Basic skills in solving algebraic equations
NEXT STEPS
  • Study the derivation of Bayes' theorem in depth
  • Practice solving problems involving multiple events in probability
  • Learn about the law of total probability
  • Explore advanced topics in probability such as Markov chains
USEFUL FOR

This discussion is beneficial for students preparing for exams in statistics or probability, educators teaching these concepts, and anyone looking to strengthen their understanding of conditional probability with multiple events.

Math1015
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I'm currently stuck on a question that involves conditional probability with 3 events. This is a concept that I'm having the most trouble grasping and trying to solve in this subject. I am not sure how to start this problem.

The Question:
Given that P(A n B) = 0.4, P(A n C) = 0.2, P(B|A)=0.6 and P(B)=0.5, find the following.
a) P(A|B)
b) P (B')
c) P(A)
d) P (C|A)
e) the odds in favor of B

If someone could provide an explanation on how to solve this and guide me through it, that would be greatly appreciated! I really want to understand how to do problems like these.
 
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$P(A|B)= \frac{P(A\cap B)}{P(B)}$. You are given that $P(A|B)= 0.6$ and that $P(A\cap B)= 0.4$ so $0.6= \frac{.4}{P(B)}$. $P(B)= \frac{0.4}{0.6}= \frac{2}{3}$. "(e) the odds in favor of B" are "2 out of 5" and "(b) P(B')= 1- 2/3= 1/3".
 
Thanks for the fast reply! We have an exam on monday and I've been stuck on this all day and understand it better now!
 

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