# Probability of Getting at least rest One Red Ball

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?

• how many balls are you taking out?? – user8795 Feb 12 '17 at 1:21
• How many times do you take a ball out of the urn? Twice? – Owen Sizemore Feb 12 '17 at 1:25
• Assume two balls are grabbed. – darylnak Feb 12 '17 at 1:36

## 1 Answer

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.

• Though I appreciate your answer, it has gone over my head. I'm only in an introductory statistics course; my knowledge is extremely limited. – darylnak Feb 12 '17 at 2:13
• 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. – BruceET Feb 12 '17 at 2:19
• 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. – darylnak Feb 12 '17 at 2:27
• 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 $P(R_1) = 20/100?$ – BruceET Feb 12 '17 at 2:31
• Ah, I just looked at my question and corrected some errors. Please re-read my question to see if it makes sense now. – darylnak Feb 12 '17 at 2:36