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Q: An urn contains 5 red, 2 blue, and 9 green balls. Six balls are drawn. Assuming the drawing is WITH replacement, what is the probability of getting 1 red, 2 blue, and 3 green balls?

This is an exam question I got wrong. My answer was:

$\frac{{5 \choose 1}{2 \choose 2}{9 \choose 3}}{{16 \choose 6}} $

I checked other questions, such as this one, and they approached it the same way. What am I missing?

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  • $\begingroup$ You miss that in your problem drawing is with replacement. For example, with your calculation you would get zero probability to draw $3$ green balls, while it's clearly possible. $\endgroup$ – mihaild Apr 30 '19 at 17:44
  • $\begingroup$ Why? I included "9 choose 3" in the fraction. $\endgroup$ – juliodesa Apr 30 '19 at 17:46
  • $\begingroup$ Oops, a typo from my part: it should be $3$ blue balls. $\endgroup$ – mihaild Apr 30 '19 at 17:46
  • $\begingroup$ But there are only 2 blue balls in the urn. $\endgroup$ – juliodesa Apr 30 '19 at 17:49
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    $\begingroup$ And yet, since each time a ball is pulled out, immediately after looking at its color it is put back in it has the ability of being drawn again immediately after. $\endgroup$ – JMoravitz Apr 30 '19 at 17:50
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Hint: the question you linked is looking at drawing balls without replacement. The way to approach the problem changes if you put each ball back into the urn after you draw it (i.e. with replacement), and the formula you used isn't correct for this situation.

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  • $\begingroup$ So this would be the answer to the question if it was asking without replacement? What's the answer for with replacement then? $\endgroup$ – juliodesa Apr 30 '19 at 17:47
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    $\begingroup$ @juliodesa You've heard of the binomial distribution I hope, yes? How about the multinomial distribution? $\endgroup$ – JMoravitz Apr 30 '19 at 17:48
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Since there is replacement there is no conditional probability, this means that the answer is simply:

$$\frac{6!}{3!2!1!}p( \color{red}\bullet \cap \color{blue}\bullet \cap \color{blue} \bullet \cap\color{green}\bullet \cap\color{green}\bullet\cap\color{green}\bullet)=\frac{6!}{3!2!1!}p(\color{red}\bullet)p^2(\color{blue}\bullet )p^3(\color{green}\bullet \color{green})=60\left(\frac{5}{16}\right)\left(\frac{2}{16}\right)^2 \left(\frac{9}{16}\right)^3=\frac{54675}{1048576}\approx 5.21 \% $$

Where the factor $\frac{6!}{3!2!1!}$ indicates the number of permutations of the sequence of drawings.

:)

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    $\begingroup$ My graphic is so lovely xd :) $\endgroup$ – Eureka Apr 30 '19 at 17:56
  • $\begingroup$ @callculus I was too concentrated on the graphics lol $\endgroup$ – Eureka Apr 30 '19 at 17:58

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