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This is a common question I've seen asked but I've searched a little and I'm not sure I've found what I'm looking for.

If I have a bag of 15 marbles where 7 are red/3 Blue and 5 green. If I pick 3 at random what is the probability that I get a marble of each colour? (I.e 1 red/green/blue).

Is there a difference between picking 1 at a time and picking all three at a time? Or can picking all three be thought as picking 1 at a time but 3 times? The total number of marbles decrease in the bag after you pick them so if you were to draw a tree diagram and try to find the probability out that way it would take a long time and it's inefficient

How does this quick way work ?

$\frac{7}{15}$ x $\frac{3}{14}$ x $\frac{5}{13}$ x $3!$

Or any of the numerators re arranged but the denominator kept the same. I do not understand how the anwser can be achieved like this but I understand that there are 6 possible routes which is what 3 factorial is for. Why do you multiply them together even though there are different ways? I.e When you pick there are different choices like this:

$\frac{7}{15}$ , $\frac{5}{14}$ , $\frac{3}{13}$

$\frac{7}{15}$ , $\frac{3}{14}$ , $\frac{5}{13}$

$\frac{5}{15}$ , $\frac{3}{14}$ , $\frac{7}{13}$

$\frac{5}{15}$ , $\frac{7}{14}$ , $\frac{3}{13}$

$\frac{3}{15}$ , $\frac{5}{14}$ , $\frac{7}{13}$

$\frac{3}{15}$ , $\frac{5}{14}$ , $\frac{7}{13}$

How is the quicker way obtained ? I.e what is the method behind it.

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2 Answers 2

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There are many ways to solve the problem. Whether we think of picking the marbles one at a time, or all together, does not alter probabilities, though it will change the way we compute the probabilities.

Imagine the balls are distinct (they all have secret ID numbers). There are $\binom{15}{3}$ equally likely ways to choose $3$ balls from the $15$.

Now we count the number of favourable choices, that is, choices that have $1$ of each colour. There are $\binom{7}{1}\binom{3}{1}\binom{5}{1}$ ways to pick $1$ red, $1$ blue, and $1$ green. Thus our probability is $$\frac{\binom{7}{1}\binom{3}{1}\binom{5}{1}}{\binom{15}{3}}.$$

Or else we calculate the probability by imagining choices are made one at a time. This complicates things somewhat, since the event "we end up with one of each colour" can happen in various ways.

Let us analyze in detail the probability we get GRB (green then red then blue).

The probability the first ball picked is green is $\frac{5}{15}$ (it is best not to simplify). Given that the first ball was green, the probability the second is red is $\frac{7}{14}$. So the probability the first two picks are green, then red, in that order, is $\frac{5}{15}\cdot\frac{7}{14}$. Given that the first two choices were green and red, in that order, the probability the next is blue is $\frac{3}{13}$. We conclude that the probability of GRB is $\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}$.

There is a total of $3$ colour sequences through which we end up with $1$ of each colour. It turns out that each of them has probability $\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}$. So our required probability is $$6\cdot\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}.$$

Why were these six probabilities all equal? One way to think about it is as follows. Suppose that, blindfolded, we use a scoop to pull out $3$ balls, and these happen to represent all the colours. Now we reach into the scoop and pull out the balls one at a time. Of course all $3!$ colour orders will be equally likely.

Another more arithmetical way of seeing it is that the denominators will be, in order, $15$, $15$, and $13$, since we are picking from a diminishing number. Each time we consider a partiular colour, the number of choices will be the number of balls of that colour. So the numbers $7$, $3$, and $5$ must each appear exactly once in the numerators.

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  • $\begingroup$ Been wanting to know what do you mean by the (7,1) is that nCr or nPr $\endgroup$
    – Nubcake
    Commented Apr 25, 2013 at 22:02
  • $\begingroup$ $\binom{n}{r}$ is always nCr (combinations) $\endgroup$
    – Henry Swanson
    Commented Apr 25, 2013 at 22:03
  • $\begingroup$ whats nPr for ? permutations? or does that mean the same thing $\endgroup$
    – Nubcake
    Commented Apr 25, 2013 at 22:03
  • $\begingroup$ ${}_nP_r$ is yes, permutations of $n$ objects, taken $r$ at a time. The notation is not much used except in low level courses. The notation $\binom{n}{r}$ is standard in mathematics for what is called in school $C(n,r)$ or ${}_nC_r$ or $C_r^n$ or ${}^nC_r$ (and there may be other names). $\endgroup$
    – André Nicolas
    Commented Apr 25, 2013 at 22:09
  • $\begingroup$ I think I get it , it doesn't matter what order you multiply them in right? i.e you can do 7 by 5 by 3 or any way . $\endgroup$
    – Nubcake
    Commented Apr 25, 2013 at 22:32
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Here's an attempt to formalize picking 3 marbles at once and checking only their colors.

Treat picking three marbles of some colors as picking one triple from the set of unordered triples of colors. These unordered triples are given the probability the number of ordered triples of marbles with those three colors, out of the number of ordered triples of marbles there are in the set.

e.g. $\{red_1, red_2, red_3\} \rightarrow (red, red, red),$ "all red" should have probability $\frac{{\text{#red} \choose 3}}{\text{#triples}}$ but $(red, blue, red)$ should have probability $\frac{{\text{#red} \choose 2}*\text{#blue}}{\text{#triples}}$. So in all, $\{red, red, blue\}$ (unordered) should have probability $p((red, red, blue))+p((red, blue, red))+p((blue, red, red))=p(\text{onePermutation})*\text{#distinctPermutations}.$ Since the number of triples is ${15 \choose 3}*3!,$ you obtain the original result in this case. Likewise, ${n \choose k}*k!=\frac{n!}{(n-k)!}$, which gives the factor of $\frac{1}{15*14*13}$ you would regularly obtain from the original probability tree.

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