There are $20$ balls in an urn: $18$ red, $2$ green.

We have to draw $3$ balls and they can be put back (replaceable), so the sample space is $20^3$, or $8000$.

  1. What is the chance of picking exactly 1 green?

  2. What is the chance of picking exactly $2$ green?

I don't understand how to do these.

Edit: For question 1, I tried doing $18\cdot 18\cdot 2.$

  • $\begingroup$ @platty I edited my question to show what I had tried $\endgroup$ – aryamank Dec 8 '18 at 21:58

Your logic is close; but you need to consider the order the balls are drawn. Since the single green ball can come first, second, or third, you have to choose where to place it when constructing a triple; this gives $3(18)(18)(2)$ as the number of such ways to get exactly one green ball. Dividing this by the total number of ways to draw 3 balls gives $\frac{3(18)(18)(2)}{20^3} = \frac{243}{1000}$.

Now, can you use the same method to figure the chance of getting exactly two green balls?

  • 1
    $\begingroup$ Thanks, that makes sense. Using that logic, the second answer would be 3(18*2*2)/20^3. $\endgroup$ – aryamank Dec 8 '18 at 22:22
  • $\begingroup$ I also have one more question about dealing with a 'without replacement' scenario. Why is the sample space C(20,3) and not 20*19*18? Thanks! $\endgroup$ – aryamank Dec 8 '18 at 22:57
  • $\begingroup$ That depends on how you approach the question — what are the elements of your sample space? It is viable to approach it either way, you just have to count differently (in the second case, order matters, while in the first case, it does not). $\endgroup$ – platty Dec 8 '18 at 22:58
  • $\begingroup$ It is the same scenario wherein the urn has 18 red balls and 2 green balls. $\endgroup$ – aryamank Dec 8 '18 at 23:17

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