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More Probability

  1. Mar 12, 2007 #1
    1. The problem statement, all variables and given/known data
    An experiment consists of drawing two balls out of a bag containing 3 red balls, 2 green, and 1 blue at random and without replacement. Figure out the probability of getting exactly one red ball.

    3. The attempt at a solution

    I wrote down all the possible scenarios I could think of and I came out with:
    RG, RB, RR, GR, GB, GG, BR, BG
    So I thought the answer was 4/8 or .50%

    Then I thought but BR=RB and BG=GB and GR=RG so then i eliminated those and thought the answer was 2/5.

    The answer given is 3/5 or 60%

    Any help is appreciated. Thank you.
  2. jcsd
  3. Mar 12, 2007 #2


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    Why? Here you have two choices BR and GR, but above you also have RR. That's 3/5...
  4. Mar 12, 2007 #3
    You have three red and three non-red balls. It doesn't matter what the other colors are.
  5. Mar 12, 2007 #4
    (N= non red)

    After one ball is drawn

    50% (R)RRNNN, 50% RRR(N)NN

    You should be able to get it from here. Consider the 4 possibilities for two balls being drawn.
  6. Mar 12, 2007 #5
    It asks for the probability of exactly one red ball, so RR wouldn't be an option, right?

    The four possibilities i come up with is;
    3 Red 3 Non red


    isn't that 2/4 or 1/2

    I'm missing something here....
  7. Mar 12, 2007 #6
    OK, another hint.

    After one ball is drawn
    50% (R)RRNNN, 50% RRR(N)NN

    In the first case you're left with RRNNN. That means there's only a 40% chance to draw a second red (2 red out of 5 balls).

    where ()=first ball []=second ball
    (The 50% comes from the prob. of drawing the first red).

    There are three other possibilities to consider.

    (Sorry for my weird notation)
  8. Mar 12, 2007 #7
    okay so now we have 1/5 for the first possibility.
    Are the other possibilities:
    1. if you pull a non-red ball first and then another non-red ball second
    which will yield another 1/5 just like above.
    2. if you pull a non-red ball first and then a red ball second, which will also yield 1/5

    do you then add those up to get the 3/5?

    It's not entirely making sense to me because you stated there were 3 other possibilities. Also, by doing it this way it seems like we are factoring the possibility of getting two red balls and not just one.
  9. Mar 12, 2007 #8
    Here's another one of the 4 branches...

    The initial distribution is RRRNNN
    50 % probability of N first
    This leaves RRRNN
    60 % probability (once N is picked first) of picking R second.

    60% of 50 % = 30%

    Out of the 4 branches, 2 of them correspond to one R out of two picks.

    (I'm trying hard not to do the whole problem for you. The book answer is correct. The total probability is 60%)
  10. Mar 12, 2007 #9
    Answer: P(Red and not Red)+P(not Red and Red) = 3/6 x 3/5 + 3/6 x 3/5 = 3/5

  11. Mar 13, 2007 #10
    Just a note to help you with your intuition:

    Suppose there were 500 red balls and 2 blue balls.
    Would you suggest that since the possibilities are RR, RB, BR, or BB that the probability of getting exactly one red ball is 50%?
    Remember, each of those possibilities (and in your first post) don't have the same probability of occurring.
  12. Mar 13, 2007 #11


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    Dearly Missed

    Try to think as follows:

    P(exactly one red in two moves)=P(red in first, no red in second)+P(no red in first, red in second)

    Now, try to determine P(red in first, no red in second)and P(no red in first, red in second):

    P(red in first, no red in second)=P(red in first)*P(no red in second, given that you drew red in first)=(3/6)*(3/5)=3/10

    Make a similar calculation for the other probability.
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