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A pot contains $25$ balls: $5$ red, $5$ blue, $5$ green, $5$ yellow and $5$ white. Extracting $3$ balls without reintroduction and without order.

What's the probability that:

1) The $3$ balls are all red.

2) The $3$ balls have all the same color.

3) The $3$ balls are all different.

I was reasoning in the following way:

The set of all possibilities $\Omega$ is defined by $$\Omega=\frac{C_{25,3}}{u}\ \ \ \ or \ \ \ \ \ \Omega=C_{25,3}-u $$

With $C_{25,3}$ the combinations of $25$ balls in $3 $ position and $u$ (unknown) is all combination I should remove (e.g. $\{R,R,R\}$ which in Combinations is counted more than one time) but I don't really know how to do it and what's the number of terms I have to remove.

Can you help me to find $u$, please? And can anyone confirm if these solutions are right?

1) I need to take $3$ out of $5$ balls of the same color, so: $$\frac{\binom{5}{3}}{\Omega}$$

2) I need to take 3 our of 5 balls of the same color and this color can be one of the $5$ colors, so: $$\frac{\binom{5}{3}+\binom{5}{3}+\binom{5}{3}+\binom{5}{3}+\binom{5}{3}}{\Omega}$$

3) I need to take all different balls, so $1$ ball for each group but only $3$ groups can be extracted, so: $$\frac{C_{5,3}}{\Omega}=\frac{\binom{5}{3}}{\Omega}$$

Thank you.

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2 Answers 2

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We temporarily assume that all balls are labeled and distinct. This allows us to use the sample space: $\{\text{all ways to pull 3 balls out of 25 distinct balls}\}$ with size $\binom{25}{3}$, noting that each of these combinations are equally likely to occur. There is no reason to introduce a variable "$u$." Note that otherwise, we may run into problems with the outcomes not being equally likely to occur, for example the outcome "three reds were drawn" is much less likely than the outcome "one of each color, red green and blue, were drawn."

We have then answers to $1$ and $2$ being $\binom{5}{3}/\binom{25}{3}$ and $5\binom{5}{3}/\binom{25}{3}$, just as you expected.

Now... for part three however, I suggested that you treat all balls as labeled and distinct. The $\binom{5}{3}$ here for part 3) is just for choosing which colors are present. Now, you still have to choose which specific balls were used for each of the chosen colors.

$\binom{5}{3}\cdot 5^3/\binom{25}{3}$

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  • $\begingroup$ But dude, if I take $3$ red balls, I have only one case ($\{R,R,R\}$) and not many cases ($\{R_1,R_2,R_3\}$ or $\{R_1,R_2,R_4\}$ or $\{R_1,R_2,R_5\}$ and so on). So I have to remove something. $\endgroup$
    – Spaggy
    Commented Jul 28, 2017 at 19:19
  • $\begingroup$ That's just it though, by introducing labelings, we can look at an equiprobable sample space. Any arrangement from the original sample space which counts as "three reds" would also count as "three reds" in the improved sample space and vice versa, and probability arguments are much easier to craft in equiprobable sample spaces. See again the other comment thread. This is in the same vein as why for a bag with $10$ balls, one of which red and the other nine blue, the probability of drawing the red ball is $\frac{1}{10}$ not $\frac{1}{2}$ despite only two outcomes (until labeled balls). $\endgroup$
    – JMoravitz
    Commented Jul 28, 2017 at 19:31
  • $\begingroup$ Okay, so.. are you saying that the probability to extract $3$ red balls in a sample space $\Omega_1$ (the sample space I was trying to caclulate with $u$ term) is the same probability to extract $3$ red balls enumerated in an equiprobable sample space $\Omega_2=\binom{25}{3}$? $\endgroup$
    – Spaggy
    Commented Jul 28, 2017 at 20:18
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    $\begingroup$ Yes... though in your sample space, finding an appropriate $ u $ will be very difficult and even after finding it, calculating probabilities will be even worse because we can't use counting techniques... again because the sample space isn't equiprobable. You may only use probability calculations like we are here where we count number of favorable outcomes and divide by sample space size if each outcome were equally likely to have occurred. $\endgroup$
    – JMoravitz
    Commented Jul 28, 2017 at 20:18
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The total number of draws is just $T = {25 \choose 3}$. No removals.. There are ${5 \choose 3}$ ways to pick $3$ red balls.

All the same colour: We have 5 similar probabilities, one for each colour. These are all mutually incompatible so we can add them, and get:

$$5 \cdot \frac{5 \choose 3}{T}$$

All different is a bit more tricky: we have to pick 1 ball each of 3 different colours. We have $5$ colours so we can pick the colours to pick one from in $5 \choose 3$ ways, and so we get

$$ \frac{\binom{5}{3}\binom{5}{1}\binom{5}{1}\binom{5}{1}}{T} $$

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    $\begingroup$ The total number of outcomes to the problem where the balls are all considered identical except for color does not have sample space size $\binom{25}{3}$ but rather has sample space size $\binom{3+5-1}{5-1}=\binom{7}{4}$ (via stars and bars). The downside to this sample space however is that the outcomes are not equally likely and so we can't use counting techniques to find the probability. Until you make the change to the problem that the balls are distinctly labeled, the number $\binom{25}{3}$ does not make sense in this context. $\endgroup$
    – JMoravitz
    Commented Jul 28, 2017 at 19:06
  • $\begingroup$ @JMoravitz This is common practice in these vase problems. It doesn't matter if we label them, because we ignore the labels and just note the result. (e.g. 3 red, instead of $r_1, r_3, r_4$ e.g. But labelling makes everything equiprobable, so much more convenient as you wrote in your answer. $\endgroup$ Commented Jul 28, 2017 at 19:10
  • $\begingroup$ You're right, @JMoravitz, balls are not labeled, so if I take $3$ red balls, I have $\{R,R,R\}$ which is different from $\{R_1,R_2,R_3\}$ or $\{R_1,R_2,R_4\}$ or $\{R_1,R_2,R_5\}$ and so on. But I can't understand what you're saying in the first part of your comment: what does stars and bars mean? Where does the binom $\binom{7}{4}$ come from? $\endgroup$
    – Spaggy
    Commented Jul 28, 2017 at 19:15
  • $\begingroup$ @ChuckTheYellow stars and bars is a commonly applied technique for counting how many arrangements there are of objects where order doesn't matter but object type does. For example, "How many ways can you give $3$ identical cookies to $5$ distinct children?" The name stars and bars is in reference to replacing such arrangements with stars representing the cookies, and bars representing where one person's cookie pile ends and the next person's begins. We then count how many sequences of stars and bars to count original problem. $\endgroup$
    – JMoravitz
    Commented Jul 28, 2017 at 19:20
  • $\begingroup$ @ChuckTheYellow that the probabilities aren't equally likely can be seen with a smaller example of two red balls and two blue balls, taking two balls at a time. The outcome of $\{r,r\}$ where balls aren't labeled only corresponds to the outcome $\{r_1,r_2\}$ where the balls are labeled, but the outcome $\{b,r\}$ corresponds to $\{b_1,r_1\},\{b_1,r_2\},\{b_2,r_1\}$ and $\{b_2,r_2\}$ and is four times as likely to occur as two-reds was, despite each only accounting for a single outcome in the "unlabeled balls" sample space. $\endgroup$
    – JMoravitz
    Commented Jul 28, 2017 at 19:22

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