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

How do you do this? on Probability(without replacement)

  1. Aug 15, 2004 #1


    User Avatar

    A bag contains 4 yellow balls and "n" red balls. Two balls are drawn without replacement. Which expression represents the probability that one ball is yellow and ball is red?

    P.S. the answer is (4/n+4)(n/n+3) + (n/n+4)(4/n+3)
  2. jcsd
  3. Aug 15, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First, what's the probability of drawing a yellow ball? There are four yellow balls, out of (n+4) total balls, so the probability of drawing a yellow ball is:


    Now, after having drawn a yellow ball, there are only three yellow balls and n red balls out of a total of n+3 balls left. Since there are n red balls and n+3 total balls, the probability of drawing a red ball is:


    Now, the probability of the first event occuring (drawing a yellow ball), followed by the second event occuring (drawing a red ball) is the product of the two probabilities. In other words, when you have two events that occur in succession, you multiply their probabilities together. When you multiply them together, you get:

    [tex]\frac{4}{n+4} \cdot \frac{n}{n+3}[/tex]

    That's the probability for drawing a yellow ball, then drawing a red ball.

    Of course, there are two ways to draw one yellow ball and one red ball -- the way we've already described, yellow followed by red, and another way, red followed by yellow. That case, red followed by yellow, is represented by the second half of the expression you listed:


    When there are more than one way to get a result (one red ball and one yellow ball), the probabilities of each way add together. Thus, the probability of getting one red ball and one yellow ball is the sum of the probability of drawing yellow then red and the probability of drawing red then yellow.

    - Warren
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook