I have a simple question for you guys but I am studying for exams and I honestly forgot how this was done.

Here is the problem:

"A vase contains 4 white marbles, 3 red and 2 green. You draw marbles 2 without putting the first one back after the second.

What are the chances that you draw a red marble after the first one was green?

What are the chances that you draw 2 white marbles?

What are the chances that you draw a white and a red marble?"

I see I gotta calculate P(A ∩ B) / P(B) but how do I calculate P(A ∩ B) then?

Also the third question from my problem is completely puzzling me even more than the others

I don't have any answers and I have come up with like 100 different answers and I don't even know if any of them are right since I can't compare them to the actual answers and I don't know how to calculate it at all

  • $\begingroup$ An answer to another question very much like this appears at stats.stackexchange.com/a/13449. The methods of that answer will work here. $\endgroup$
    – whuber
    Dec 14, 2011 at 18:12
  • $\begingroup$ I still don't get it. Do I have to do ((9/4) * (9/3))/(9/4) ? $\endgroup$
    – zeta
    Dec 14, 2011 at 18:43
  • $\begingroup$ I assume that "without putting the first one back after the second" is a typographical error for "without putting the first one back before drawing the second". Given that the first marble is green, how many marbles and of what colors remain? What is the probability of drawing a red marble from those that remain after the first green marble is gone? No muss, no fuss, you get $P(\text{second red}\mid \text{first green})$ without having to compute $P(A \cap B)$ etc. $\endgroup$ Dec 14, 2011 at 19:10
  • $\begingroup$ Yes but how do I calculate that then? I honestly am at a loss here I have been staring myself blind at this for 2 hours now. All examples I find on the internet are only involving 2 colors of marbles so I have no idea if what I am doing is right. My answer right now to the first question is 0.1667 and to the second one 1/3 am I right? the third one I don't have yet $\endgroup$
    – zeta
    Dec 14, 2011 at 19:16
  • $\begingroup$ honestly I can't... I am starting myself blindly and stressing a lot here right now and I am starting to blank out completely... All I know is that my book says P(A|B) = P(A∩B) / P(B) This rule should apply to all 3 questions in my OP but I have no idea at all if I am on the right track here or not and what the answers are... $\endgroup$
    – zeta
    Dec 14, 2011 at 19:27

2 Answers 2


One good technique for solving problems that seem to involve arbitrary numbers (like 4, 3, and 2) is to look at a variant involving simpler or smaller values of those numbers. Let's use this to revisit the parts of the question:

  1. What are the chances of drawing a red marble after a green?

    Because one point of possible confusion is that there are three colors, consider a vase having just one of each color, red, white, and green. This makes it feasible to enumerate all the possibilities of two draws. Abbreviating the colors as R, W, and G, and writing the outcome of the first draw at the left, all possible results are

    {RW, RG, WR, WG, GR, GW}.

    Of these, only {GR, GW} consist of drawing a green marble first. They have equal probabilities, because after the green was drawn, there remained just one R and one W in the vase. Therefore the chance of drawing a red after a green is 1/2.

    Let us consider how to generalize this reasoning. It looks like things got simpler once we limited the list only to those results where G was drawn first. After that, there was one less G in the vase and the question became "in a vase with 1-1 = 0 greens, 1 white, and 1 red, what is the chance of drawing a red?" In the present setting, we need to replace these numbers by the arbitrary quantities in the problem statement: "in a vase with 2-1 = 1 greens, 4 white, and 3 red, what is the chance of drawing a red?"

    If this newer problem still seems difficult (because there are three colors), consider that the restatement refers to one color only. The chances therefore must be the same as if we were color blind and could not tell green from white: "in a vase with 3 red marbles and 1+4 = 5 non-red marbles, what is the chance of drawing a red?" The answer is obvious.

  2. What are the chances of drawing two white marbles?

    Once more we can capitalize on the color-blindness argument and rephrase this question as "in a vase with 4 white marbles and 3+2 = 5 non-white marbles, what are the chances of drawing two white marbles?" Again, let's consider a simpler version in which the vase contains only two whites and one non-white (abbreviated "B"). Because the two white marbles are distinct, we must label them to keep them apart: call them W1 and W2, say. The possible results are

    {W1W2, W1B, W2W1, W2B, BW1, BW2}.

    All are equally likely (because at any stage each individual marble has neither greater nor lesser chances of being drawn than any other individual). In two of the six cases both whites are drawn, whence the answer to the simplified question is 2/6.

    Turning to the original question with four whites and 5 non-whites, let's distinguish them as W1, W2, W3, W4 and B1, B2, B3, B4, B5. The larger quantities create a challenge, because there are many possibilities to enumerate ($\binom{9}{2} = 36$) so let's see whether the problem can be analyzed in stages. Go back to the case of two whites and one non-white. For the first draw, the chance of a white obviously is 2/3. This leaves a vase with one less white and the same number of non-whites: one of each. Evidently the chance of drawing a white, given that a white has just be drawn (i.e., removed from the vase), is 1/2.

    This reasoning is reflected in our earlier enumeration: of the six cases listed, 2/3 of them (that is, 2/3 * 6 = 4) consist of an initial white draw: {W1W2, W1B, W2W1, W2B}. Of these, 1/2 (that is 1/2 * 4 = 2) are followed by another white draw, {W1W2, W2W1}. It appears that the correct mathematical operations to follow are these:

    a. Find the total number of possibilities, 6.

    b. Multiply by the chances of drawing a white, 2/3.

    c. Multiply that by the chances of drawing a white after a white has been removed, 1/2.

    d. Finally, as always, divide the result by the total number of possibilities.

    The answer is 6 * 2/3 * 1/2 / 6. But the sixes represent the same thing--the total number of possibilities--and they cancel in the calculation, leaving just 2/3 * 1/2 = 1/3, exactly the answer we obtained directly.

    The beauty of this is that we do not need to know how many total possibilities there are. We just multiply.

    The general rule is, that when one outcome follows another, to find the chance of both outcomes in succession, you multiply the probabilities. In math notation this can be written like

    $$\Pr[WW] = \Pr[W] \times \Pr[W\ \vert\ W].$$

    To solve the original problem, compute the chances represented by the right hand side and multiply.

  3. What are the chances you draw a white and a red marble?

    One way to solve this is to compute the chance of drawing a white and then a red and also compute the chance of drawing a red and then a white. The first result represents a collection of possible outcomes; so does the second; and no outcomes are common to both results.

    Look at the situation with, say, a vase with two whites (W1 and W2), a red, and a green. Here are all the possible results:

    {W1W2, W1R, W1G, W2W1, W2R, W2G, RW1, RW2, RG, GW1, GW2, GR}.

    There are 12 of them. The cases where a white and then a red are drawn are

    {W1R, W2R}

    and the cases where a red and then a white are drawn are

    {RW1, RW2}.

    Therefore the total number of cases is 2 + 2 = 4, because no case appears in both lists. (If it did, it would be wrong to sum the counts because the common case(s) would be doubly counted.) Consequently, reasoning as before, the chance of obtaining a white and a red equals 4/12 = 1/3.

    This reasoning suggests that when two events have no outcomes in common, the chances add. Consequently we would compute

    $$\Pr[\text{W and R}] = \Pr[WR] + \Pr[RW] = 1/6 + 1/6 = 1/3$$

    using the rule from the second question. It should be clear now how to proceed with the original question and the numbers in it.

To summarize, by looking at simplified versions of the questions, we were able to develop two simple rules:

  1. When one outcome follows another, to find the chance of both outcomes in succession, you multiply the probabilities.

  2. When two events have no outcomes in common, you add the probabilities.

With these rules, most elementary probability problems can be solved easily.

I advocate not memorizing mathematical formulas like those shown here. Doing so is unenlightening. Of far greater value is remembering the reasoning that led up to them. The fundamental methods used here included:

  1. Analyze simpler versions of a problem, using a complete enumeration of all results if need be. Then generalize.

  2. When necessary, label distinct objects that are indistinguishable.

  3. Simplify the problem by removing distinguishing elements of the objects when it does not matter (the "color blind" argument).

  4. Reduce the evaluation of probabilities to situations where you are drawing one of $n$ equally likely objects out of a vase and want to know the chance that it is one of $k$ such objects, because then the chance is obviously $k/n$.


Look up the hypergeometric distribution. It's useful in explaining why experimenters randomize subjects into treatment groups.

  • 1
    $\begingroup$ Using the hypergeometric distribution is way too complicated an approach to this fairly elementary problem. $\endgroup$ Dec 14, 2011 at 18:57
  • $\begingroup$ I don't know the answers to this exercise though so I have no idea if I am calculating it right at all $\endgroup$
    – zeta
    Dec 14, 2011 at 18:57
  • $\begingroup$ see last line in my OP, I edited it to what my book says I need to do but I don't know how to do it $\endgroup$
    – zeta
    Dec 14, 2011 at 19:03

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