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Statistics Problem regarding conditional probabilities

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    A certain delivery service offers both express and standard delivery. Eighty-five percent of parcels are sent by standard delivery and 15% are sent by express. Of those sent standard, 80% arrive the next day, and of those sent express, 95% arrive the next day. A record of parcel delivery is chosen at random from the company’s files.

    Given that the package did not arrive the next day, what is the probability that is was sent standard?

    2. Relevant equations

    Not entirely sure, possibly the statistics probability rule of multiplication or another.

    3. The attempt at a solution

    To begin I made Express Delivery= to the variable "E"
    Standard delivery = "S"
    Next day = "N"

    From here I tried to use the relation P(S|N^c)=(P(N^c|S)P(S))/(P(N^c)) but the P(N^c|S) is one term that I don't have an actual probability for. So I'm going back and forth with one equation and two unknowns. I know the numerical value of the answer already, but I'm trying to follow the steps for the solution to the above listed problem. If someone could show me what the setup is for this solution I'd really appreciate the help.
     
  2. jcsd
  3. Nov 3, 2012 #2

    Ibix

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    They give you P(N|S), so isn't P(Nc|S) just 1-P(N|S)?
     
  4. Nov 3, 2012 #3
    I'm with you now I think. But is it true that:

    P(A^c|B)=1-P(A|B) as well as P(A|B^c)=1-P(A|B)? My book only has the first equation and I wanted to verify that the second was true as well. It works just as you said but I just wanted to make sure that I knew exactly why and that I'm not making up my own rules.

    Thank you.
     
  5. Nov 3, 2012 #4
    I'm with you now I think. But is it true that:

    P(A^c|B)=1-P(A|B) as well as P(A|B^c)=1-P(A|B)? My book only has the first equation and I wanted to verify that the second was true as well. It works just as you said but I just wanted to make sure that I knew exactly why and that I'm not making up my own rules.

    Thank you.
     
  6. Nov 4, 2012 #5

    Ibix

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    Science Advisor

    The first statement is true, but the second is false. Deal one card from a normal pack. It's an ace. What is the probability that the next card dealt is an ace? What is the probability that it is not? That is an example of p(A|B)=1-p(Ac|B). However, draw a card and a second. What is the probability that the second is an ace given that the first was an ace? And given that the first was not an ace? Your relation doesn't hold here.
     
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