1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook