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Thread: Conditional Probability with 3 Events: P(A n B)=0.4, P(A n C)=0.2, P(B|A)=0.6 and...

  1. #1

    Conditional Probability with 3 Events: P(A n B)=0.4, P(A n C)=0.2, P(B|A)=0.6 and...

    So I'm having massive difficulty understanding and solving this problem:
    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

    I'm not sure what I should do, if someone could explain what all of this means and how to solve it, that would be greatly appreciated!

  2. #2
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    Hi Math1059

    Are explanations for some of these things not given in your course notes or textbook?

    I can try to help with some conceptual understanding

    Let's say A and B are events that can occur. You can think of them as subsets of an abstract mathematical space of all possible outcomes (what is known as a "set"). In the branch of math known as set theory, the [tex] \cap [/tex] symbol is used to mean the intersection (in a Venn diagram sense) of two sets, i.e. the things (in this case the circumstances) that lie within both sets.

    Therefore the notation [tex] A \cap B [/tex] refers to circumstances in which both A occurs AND B occurs. [tex] P(A \cap B) [/tex] means the probability of event A occurring AND event B also occurring.

    A conditional probability is a probability of some event occurring subject to some other given condition. For example the notation P(A|B) means the probability of event A occurring given that we know it is true that event B has occurred. (So event B is the condition here, and we want to know the probability of A occurring given that condition. That's why it's called the conditional probability of A occurring).

    Similarly P(B|A) is the probability of B occurring given that we know for sure that event A has occurred. (This is the conditional probability of B given A).

    It turns out from probability theory that the probability of A AND B is given by the following product of probabilities

    [tex] P(A \cap B) = P(A|B)P(B) [/tex] [Equation 1]

    This makes sense. Suppose that B has a 10% chance of occurring: P(B) = 0.1. And suppose the conditional probability of A occurring given B has occurred is 50%: P(A|B) = 0.5. So the probability of A occuring AND B occurring, is the probability of B occurring (10% of the time) times the probability of A occurring given B, which happens in half of those cases, i.e. 5% of the time.

    [tex] P(A \cap B) = P(A|B)P(B) = 0.5\times 0.1 = 0.05 [/tex]

    EDIT (added all of this stuff below after initially posting)

    Similarly, we can write the probability of B AND A in terms of a conditional probability:

    [tex] P(B \cap A) = P(B|A)P(A) [/tex] [Equation 2]

    This leads to a very interesting and important result in probability theory. Basically, A AND B occurring is exactly the same outcome as B AND A occurring. In both cases, these two events have both occurred. Since they are the same outcome, their probabilities are equal

    [tex] P(A \cap B) = P(B \cap A) [/tex]

    We can substitute in the expressions for [tex] P(A \cap B) [/tex] and [tex] P(B \cap A) [/tex] that use conditional probabilities, from Equations 1 and 2 above, to obtain:

    [tex] P(A | B)P(B) = P(B | A)P(A) [/tex]

    Or

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

    This result, called Bayes' Theorem (named after the dude who came up with it) has very wide-reaching implications in statistics and probability theory. I think this is all the information you need to solve the problem. The only other notation I see in your original problem statement is [tex] B^\prime[/tex], which means NOT B. So [tex] P(B^\prime)[/tex] is the probability that B does NOT occur. What is the relationship between P(B) and P(B')? Hint: With 100% certainty (P=1) SOMETHING always occurs, either B or NOT B. So the probabilities P(B) and P(B') must sum up to...?
    Last edited by j-astron; 01-27-2018 at 12:25 PM.

  3. #3
    I got 1/2 for P(B'), checked the answers at it was right! 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!

  4. #4
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    Quote Originally Posted by Math1059 View Post
    I got 1/2 for P(B'), checked the answers at it was right! 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!
    You're welcome! Post back here with your attempted steps to solve the other problems if you are stuck.

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