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An urn contains $29$ red, $18$ green, and $12$ yellow balls. Draw two balls without replacement What is the probability that the number of red balls in the sample is exactly $1$ or the number of yellow balls in the sample is exactly $1$ (or both)? What about with replacement?

I can't seem to figure this out. My closest attempt

${\frac{29 \choose 1}{59 \choose 2}} + {\frac{12 \choose 1}{59 \choose 2}} + {\frac{29 \choose 1 }{59 \choose 2}} * {\frac{29 \choose 1}{59 \choose 2}}$

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    $\begingroup$ You seem to have made multiple errors, but they're all conceptual errors in how you think about creating formulas for the relevant probabilities. Since you just presented a formula without explanation, it's hard to say exactly what your mistakes were. If you explain in detail where each of the terms such as $\binom{29}{1}$ means to you and what each of the intermediate results means to you (after division, after multiplication), you might get more useful advice. $\endgroup$
    – David K
    Sep 18, 2018 at 22:37
  • $\begingroup$ @DavidK: I second that! $\endgroup$
    – String
    Sep 18, 2018 at 22:38

2 Answers 2

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I'm pretty much a rookie in combinatorics but I'd like to leave my attempt (it will also count as an exercise for me so why not).

The number of ways you can fail at picking only one yellow, one red, or one yellow and one red, is by picking either two green, two yellow or two red balls. This can be done:

  • Picking two greens among the $18$ possible and $0$ among the $41$ remaining, which can be done in ${18 \choose 2}{41 \choose 0}$ ways.
  • Picking two yellow among the $12$ possible and $0$ among the $47$ remaining, which can be done in ${12 \choose 2}{47 \choose 0}$ ways.
  • Picking two red among the $29$ possible and $0$ among the $30$ remaining, which can be done in ${29\choose 2}{30 \choose 0}$ ways.

So the probability that you fail is $\frac{{18 \choose 2}{41 \choose 0} + {12 \choose 2}{47 \choose 0} + {29\choose 2}{30 \choose 0}}{ {59\choose 2}} = \frac{625}{1711}\approx 0.365 $

The probability of success is $ 1 - \frac{625}{1711} = \frac{1086}{1711} \approx 0.635$

In case of replacement inside the urn, the fact that you extract for instance two yellow becomes such that there's no influence of knowing that you already picked up a yellow one in the probability to pick it up again, so the probability of picking two yellow balls will be $(\frac{12}{59})^2$ for independence. Just reproduce keeping this in mind the same reasoning, cfr. this can be a good reading.

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Let $r$ denote the event of having exactly $1$ red. Similarly, let $y$ denote the event of having exactly $1$ yellow. Then we have: $$ |r\cup y|=|r|+|y|-|r\cap y| $$ Furthermore, we have the following event-counts: $$ \begin{align} |r| &= 29\cdot 30 \\ |y| &= 12\cdot 47 \\ |r\cap y| &= 29\cdot 12 \end{align} $$ and so the number of succesful outcomes must be: $$ |r\cup y|=29\cdot 30+12\cdot 47-29\cdot 12 $$ In case of no replacement, the number of possible outcomes will be $\binom{59}2$ whereas with replacement it becomes $59^2$. The rest is just arithmetic.

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    $\begingroup$ For some reason I like this answer more than my own one. $\endgroup$ Sep 18, 2018 at 22:40
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    $\begingroup$ @Bafforasta: Well, thanks! There will be plenty of ways to break this problem down. You tried out two different approaches (I saw your edits;), and I gave yet another approach. Good on you for both putting in the effort as a kind of self-training! And thanks again for the kind remark! $\endgroup$
    – String
    Sep 18, 2018 at 22:48

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