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A Greek urn contains a red, blue, yellow, and orange ball. A ball is drawn from the urn at random and then replaced. If one does this $4$ times, what is the probability that all $4$ colors were selected?

I approached this questions by doing $(1/4)^4$ because there's always a $1/4$ chance of selected a specific color ball if it's replaced. I also tried doing if not the correct ball was selected; so I did $(3/4)^4$ but that didn't work either. What am I doing wrong?

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    $\begingroup$ "A ball is drawn from the urn at random and then replaced" Replaced by what? by a ball of the same color? By a ball of a random color? $\endgroup$
    – Jorge Leitao
    Dec 15, 2017 at 10:25
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    $\begingroup$ @J.C.Leitão: The ball you drew is re-placed into the urn. $\endgroup$
    – hmakholm left over Monica
    Dec 15, 2017 at 16:05
  • $\begingroup$ Your first approach is basically calculating the odds of getting each ball once in the correct order, but since you don't care what order you get them, that under represents the odds that you care about $\endgroup$
    – Kevin
    Dec 15, 2017 at 18:52

5 Answers 5

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The first ball can be any of the four with probability $\frac{4}{4}$

The second ball must be any of the other three with probability $\frac{3}{4}$

The third ball must be any of the other two with probability $\frac{2}{4}$

The fourth ball must be the ball that hasn't been selected yet with probability $\frac{1}{4}$

All together,

$$\frac{4}{4}\cdot \frac{3}{4}\cdot \frac{2}{4}\cdot\frac{1}{4}=0.09375$$

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  • $\begingroup$ but the balls are being replaced. how is this possible? $\endgroup$
    – minatozaki
    Dec 15, 2017 at 2:39
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    $\begingroup$ The first ball you select can be any of the four colors. Then you replace it and you must select one of the other three balls next, etc. $\endgroup$
    – Remy
    Dec 15, 2017 at 2:40
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    $\begingroup$ @minatozaki If the balls were not replaced, the second ball would be one of the remaining three balls with probability $\frac33,$ etc., and the answer would be $\frac44\cdot\frac33\cdot\frac22\cdot\frac11 = 1$; that is, if you don't replace, then for certain you will get four different colors. $\endgroup$
    – David K
    Dec 15, 2017 at 12:56
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We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates.  There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$

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    $\begingroup$ If there is a -1 is there a reason, anything you'd like explained or improved? $\endgroup$
    – Benji Altman
    Dec 15, 2017 at 2:52
  • $\begingroup$ Someone down-voted my answer too without explanation. $\endgroup$
    – Remy
    Dec 15, 2017 at 2:52
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    $\begingroup$ +1, But one thing to be careful of with this method - it only works when all possible combinations of draws are equally likely. However, in this case, that obviously holds. $\endgroup$
    – Paul Sinclair
    Dec 15, 2017 at 17:45
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The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.

The first draw yields a new ball, guaranteed: $$P(\text{ball 1 new})=1$$

For the second draw, there are $3$ possible new balls and $4$ total balls, so: $$P(\text{ball 2 new})=\frac34$$

For the third, there are $2$ possible new balls and $4$ total balls, so: $$P(\text{ball 3 new})=\frac24=\frac12$$

For the fourth, there is one new ball and there are $4$ total balls, so: $$P(\text{ball 4 new}) = \frac14$$

Thus, the answer is: $$\prod P = 1\cdot\frac34\cdot\frac12\cdot\frac14 = \frac3{32}=0.09375\text{ chance.}$$

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The existing solutions provide the correct probability, but do not directly answer the question "What am I doing wrong?"

$(1/4)^4$ is the probability of a specific sequence of draws such as:

red, blue, yellow, orange

blue, yellow, orange, red

yellow, orange, blue, red

The event that "all 4 colors were selected" would occur if any of these sequences occurred. So we must count the number of such sequences (4! = 24) and add up their probabilities, which yields $\displaystyle\frac{4!}{4^4}$

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The first ball drawn can be any colour. So, the probability is $\frac{4}{4}.$

Since the first ball is replaced, there is a $\frac{1}{4}$ chance that the same ball will be drawn. The chance for a different ball to be drawn is $\frac {3}{4}$.

There is $\frac{2}{4}$ chance that the two drawn balls will duplicate, so the chance for a different ball to be drawn (for the third draw) is $\frac{2}{4}$.

Finally, there is a $\frac {1}{4}$ chance of the final different ball to be chosen as the other three are already drawn (if drawn) and will be duplicated.

$$\frac{4}{4} \times\frac{3}{4}\times\frac {2}{4}\times \frac {1}{4}=\frac{3}{32} =0.09375$$

Conclusion: There is a $\frac{3}{32} $ chance of you getting all four colours.

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