Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability of exactly one when events are dependent

  1. Jun 23, 2013 #1
    Probability of "exactly one" when events are dependent

    The question is this:
    "An urn is filled with 8 green balls, 2 red balls, and 6 orange balls. Three balls are selected without replacement."

    What is the probability that exactly one ball is orange?

    I know I could just use the binomial formula if each event were independent (i.e. three balls were selected [with replacement). But I'm not sure how to find the probability in this case because they are dependent events, and the order in which the orange ball is picked affects the probability. I can see three different scenarios of one orange ball being picked:

    P(orange picked)*P(not orange picked)*P(not orange picked)
    or
    P(not orange picked)*P(orange picked)*P(not orange picked)
    or
    P(not orange picked)*P(not orange picked)*P(orange picked)

    The first case corresponds to:
    (6/16)(10/15)(9/14)= 9/56
    The second case corresponds to:
    (10/16)(6/15)(9/14)= 9/56
    The third case corresponds to:
    (10/16)(9/15)(6/14)= 9/56

    Would the probability that exactly one ball is orange just be 3(9/56), or 27/56?

    Would a similar process be done if I wanted to know the probability that at most one ball is red?
     
    Last edited: Jun 23, 2013
  2. jcsd
  3. Jun 24, 2013 #2

    Bacle2

    User Avatar
    Science Advisor

    Looks right. Yes, a similar methd would work to find out P(at most one is red.)
     
  4. Jun 24, 2013 #3
    Thanks! And so for the probability that at most one red ball is picked:

    "at most one" means either 0 red balls are picked or 1 red ball is picked.

    If no red balls are picked, then:
    P(not red)*P(not red)*P(not red)= (14/16)(13/15)(12/14)= 13/20

    and if one red ball is picked it can happen three different ways:
    1) P(red)*P(not red)*P(not red)= (2/16)(14/15)(13/14)= 13/120
    2) P(not red)*P(red)*P(not red)= (14/16)(2/15)(13/14)= 13/120
    3) P(not red)*P(not red)*P(red)= (14/16)(13/15)(2/14)= 13/120
    or P(one red)=3(13/120)= 13/40

    So would the probability of at most one red ball be (13/20)(13/40)= 169/800 ?
     
  5. Jun 25, 2013 #4

    Bacle2

    User Avatar
    Science Advisor

    I don't see why you're multiplying the two probabilities and not adding them, since the events are disjoint --and disjoint events cannot be independent. Sounds counterintuitive, but it's true.

    One way of double-checking your work is by adding P(0 Red)+P(1 Red)+P(2 Reds) and checking that they
    add to 1. So try calculating P( exactly 2 Reds ), and you should get 1/40 .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook