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Suppose we have an urn containing 100 balls. 20 are red, 10 are green, and the rest are neither red or green. What is the probability of grabbing at least one red ball, assuming the balls are not placed back in the urn? Assume two balls are always grabbed.

  • Let A denote the event of getting at least one red ball.
  • Let R1 denote the event of the first ball being red.
  • Let R2 denote the event of the second ball being red.

Attempt at question:

P(A) = P(R1∩R2̅) + P(R1̅∩R2) + P(R1∩R2)

Is this the correct equation?

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    how many balls are you taking out?? – 2017-02-12
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    How many times do you take a ball out of the urn? Twice? – 2017-02-12
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    Assume two balls are grabbed. – 2017-02-12

1 Answers 1

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Hypergeometric distribution. Let $X$ be the number of red balls drawn (in two draws without replacement). You seek $$P(X \ge 1) - 1 - P(X=0) = 1 - \frac{{80 \choose 2}}{{100 \choose 2}} = 1 - \frac{80}{100}\times\frac{79}{99} = 1-P(R_1^c)P(R_2^c|R_1^c).$$

In R statistical software:

1 - dhyper(0, 20, 80, 2)
## 0.3616162
1 - choose(80,2)/choose(100,2)
## 0.3616162

Also, $P(R_1) = 20/100,$ and by symmetry $P(R_2) = P(R_1).$ The latter can be shown rigorously as $P(R_2) = P(R_1 \cap R_2) + P(R_1^c \cap R_2),$ where each term can be found using a product, as at the end of the displayed equation above. Or maybe make a tree diagram.

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    Though I appreciate your answer, it has gone over my head. I'm only in an introductory statistics course; my knowledge is extremely limited. – 2017-02-12
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    If not already covered, everything here should be in an elementary statistics course. Now is a good time for your knowledge not to be so limited. Do you know $P(A\cap B) = P(A)P(B|A),$ which is equivalent to the definition $P(B|A) = P(A\cap B)/P(A).$ Might start there. – 2017-02-12
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    Yes, I must use those equations in a separate part of my assignment, which is why I wanted to make sure my value for P(A) was correct. – 2017-02-12
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    Sorry, but your expression for $P(A)$ makes no sense to me. In words, what is $A?$ Does $P(R_1^c)P(R_2^c|R_1^c)$ make any sense to you? By $R_1^c$ I mean that the first ball is not red. Do you understand why $P(R_1) = 20/100?$ – 2017-02-12
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    Ah, I just looked at my question and corrected some errors. Please re-read my question to see if it makes sense now. – 2017-02-12
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    Still not making sense of what you wrote. Try simpler notation: Let R be red, N not red. $P(A) = P(RR) + P(RN) + P(NR)$ $=\frac{20}{100}\cdot\frac{19}{99}+\frac{20}{100}\cdot\frac{80}{99}+\frac{80}{100}\cdot\frac{20}{99};$ $P(R_1) = 20/100;$ $P(R_2) = P(RR) + P(NR),$ for which the two terms are evaluated above. Turns out, after some arithmetic, that $P(R_1) = P(R_2).$ Hope this does it. – 2017-03-02