0
$\begingroup$

If 3 balls are randomly drawn from an urn containing 6 white and 5 black balls, what is the probability that one of the drawn balls is white and the other two black?

1) What is the sample space for this problem?

i.e is it $\{\{w,w,w\},\{b,w,w\},\{b,b,w\},\{b,b,b\}\}$ where ordering does not count or

$\{\{w,w,w\},\{w,w,b\},\{w,b,w\},\{b,w,w\},\{b,b,w\},\{b,w,b\},\{w,b,b\},\{b,b,b\}\}$ where ordering does count.

The solution to the problem counts the number of elements in the sample space as ${11 \choose 3}$ and gives the correct answer eventually as $\frac{4}{11}$.

Can anyone explain what is happening?

$\endgroup$
  • $\begingroup$ I'm not sure what the notation in the answer you propose means. I think, though, that you are incorrectly assuming that all the elements in your "sample space" have the same probability. They do not. $\endgroup$ – lulu Feb 11 at 12:01
  • $\begingroup$ The sample space is simply ${11\choose 3}$ $\endgroup$ – cr001 Feb 11 at 12:06
  • $\begingroup$ why is that? ${11 \choose 3}$ would include {w,w,w} multiple times wouldn't it $\endgroup$ – Jhon Doe Feb 11 at 12:07
  • $\begingroup$ Because the sample space is the space of ALL possible outcomes for drawing three balls. You then find the event space and divide the two for the probability. $\endgroup$ – cr001 Feb 11 at 12:09
  • 2
    $\begingroup$ I think it's clearer to think of it the way I said...with numbered balls. $\endgroup$ – lulu Feb 11 at 12:11
0
$\begingroup$

The sample space can be chosen to be multiple things; both the choices you said would work. However, the most convenient choices are the ones where all the elements of the sample space are equally likely. In this problem, there are $11$ equally likely possibilities for the first ball, and then $10$ equally likely possibilities for the next, and then $9$ for the last, suggesting that a good sample space would consist of all $11\cdot 10\cdot 9$ ordered selections of balls. However, since order does not matter, it is also okay to group the sample space into groups of $3!=6$ corresponding to the six orderings of each unordered selection. Still, every unordered outcome is equally likely.

Note that even though the while balls are indistinguishable, they behave as distinguishable balls when calculating the probability of events.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.