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Let's say there are 12 red 11 green 12 yellow marbles. I pick up 7 marbles randomly. What's the probability of me picking 2 red marbles?

I understand the probability of picking up a red marble is 12/35. How does this relate to picking up 7 marbles and ending up with 2 red ones? I would like an explanation, not just the answer, please.

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    $\begingroup$ Do you mean exactly $2$ or at least $2$ red marbles? $\endgroup$ – Thomas Sep 23 '12 at 18:01
  • $\begingroup$ Exactly 2 marbles. $\endgroup$ – Sai Sep 23 '12 at 18:02
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you need exactly $2$ red marbles out of $7$ you chose from a set of $12$ red, $11$ green and $12$ yellow.

No. of ways to choose $2$ red marbles out of $12$ marbles (red)$={12\choose 2}$

No. of ways to choose remaining $5$ marbles out of $23$ marbles (non-red)= ${23\choose 5}$

Total ways of choosing $7$ marbles out of $35$ marbles =${35\choose 7}$

Therefore, Required Probability = $$\frac{{12\choose 2}{23\choose 5}}{{35\choose 7}}$$

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  • $\begingroup$ You're saying probability of me picking up exactly 2 red marbles out of 7 would be 5.52? Is that 5.52% or .0552%? $\endgroup$ – Sai Sep 23 '12 at 18:22
  • $\begingroup$ @Syedur: How did you come out with $5.52$? There is no $5.52$ in my answer. Did you evaluate the last term? $\endgroup$ – Aang Sep 23 '12 at 18:24
  • $\begingroup$ Am I not seeing your formula properly? I see... ((12/2)(23/5))/(35/7) $\endgroup$ – Sai Sep 23 '12 at 18:26
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    $\begingroup$ oh that's the problem. It is (12 choose 2)(23 choose 5)/(35 choose 7) $\endgroup$ – Aang Sep 23 '12 at 18:28
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    $\begingroup$ oh i forgot. thanks for pointing out. it would be (12 choose 2)(11 choose 4)(12 choose 1)/(35 choose 7) $\endgroup$ – Aang Sep 23 '12 at 18:51

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