My Math Forum Probability That Three Outcomes Happen Three Times Each

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 May 28th, 2016, 06:45 AM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 638 Thanks: 85 Probability That Three Outcomes Happen Three Times Each There are three equally likely outcomes on each trial, and the trials are independent. After nine trials, what is the probability that the outcomes happened three times each? Thanks from manus
 May 28th, 2016, 07:51 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle \frac{\frac{9!}{3!3!3!}}{3^9} = \frac{560}{6561}$? Thanks from manus
 May 29th, 2016, 12:56 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 638 Thanks: 85 6,561 is 3^8. 3^9 is 19,683. Thanks from manus
 May 29th, 2016, 02:22 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 EvanJ The problem with giving you an answer is that it does not teach you how to find an answer. Let's start with an easier problem. Suppose you had two events and did n trials and wanted to know what was the probability of getting three of event A, when the probability of A in a single trial is 1/3. Well if n = 3, only 1 series of events will help you, namely A, A, and A. So $\dfrac{1}{3} * \dfrac{1}{3} * \dfrac{1}{3} = \dfrac{1}{27}.$ With me on that? If n = 4, 4 different series will help you, AAAB, AABA, ABAA, BAAA. The probability of getting the first is $\left ( \dfrac{1}{3} \right )^3 * \dfrac{2}{3}.$ But there are 4 such series so the probability is $4 * \dfrac{1}{3^3} * \dfrac{2}{3}.$ If n = 5, 10 different series will help you, AAABB, AABAB, ABAAB, BAAAB, AABBA, ABABA, BAABA, ABBAA, BABAA, and BBAAA. The probability of getting the first is $\left ( \dfrac{1}{3} \right )^3 * \left ( \dfrac{2}{3} \right )^2.$ But there are 10 such series so the probability is $10 *\left ( \dfrac{1}{3} \right )^3 * \left ( \dfrac{2}{3} \right )^2 .$ What gives a series like 1, 4, 10? Why $1 = \dbinom{3}{3} = \dfrac{3!}{3! * 0!}\ and\ 4 = \dbinom{4}{3} = \dfrac{4!}{3! * 1!}\ and\ 10 = \dbinom{5}{3} = \dfrac{5!}{3! * 2!}$ So the probability of getting 3 A's in 9 trials is $\dfrac{9!}{3! * 6!} *\left ( \dfrac{1}{3} \right )^3 * \left ( \dfrac{2}{3} \right )^6 .$ Now lets go to your problem with three events, A, B, and C. We can lump B and C together as not-A. So the probability of getting 3 A's and 6 not-A's has already been calculated. Are you with me? Now let's think about conditional probability. What is the probability of getting three B' s in n trials GIVEN that three of the trials were A's? If the first trial is not A, what is the probability that it is B? Well the probability of not A is 2/3, and the probability of B is 1/3 so the probability of B given not A is $\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = \dfrac{1}{2}.$ Now there are six places in the series where the not-A's can go given that there are already 3 A's. We can think about this problem just the way we did about the first problem which gives probability of three B's in the six slots given that they are not A $\dfrac{6!}{3! * 3!} \left ( \dfrac{1}{2} \right )^6$ Have you kept up? So P(X and Y) = P(x | Y) * P(Y) And the answer is $\dfrac{9!}{3! \cancel{6!}} *\left ( \dfrac{1}{3} \right )^3 * \left ( \dfrac{2}{3} \right )^6 * \dfrac{\cancel{6!}}{3! * 3!} \left ( \dfrac{1}{2} \right )^6 = \dfrac{9! * \cancel{2^6}}{3! * 3! * 3! * 3^9 * \cancel{2^6}} = \dfrac{9!}{(3!)^3 * 3^9} \approx 8.5\%.$ If you got lost somewhere, ask more questions. Thanks from 123qwerty, EvanJ and manus
May 29th, 2016, 07:01 PM   #5
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Quote:
 Originally Posted by EvanJ 6,561 is 3^8. 3^9 is 19,683.
It was fraction reduction:
$\displaystyle \frac{\frac{9!}{3!3!3!}}{3^9} = \frac{1680}{19683} = \frac{560}{6561}$

May 31st, 2016, 04:10 PM   #6
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Quote:
 Originally Posted by JeffM1 If you got lost somewhere, ask more questions.
I know the binomial probability formula. If there were only two possible outcomes, I wouldn't need to ask it here. What I don't know is what made you try n = 5. As for the last part, I have memorized that given that B and C are equally likely, the probability of 3 B and 3 C from 6 trials that all must be B or C is 20/64 = 5/16.

May 31st, 2016, 06:12 PM   #7
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Quote:
 Originally Posted by EvanJ I know the binomial probability formula. If there were only two possible outcomes, I wouldn't need to ask it here. What I don't know is what made you try n = 5. As for the last part, I have memorized that given that B and C are equally likely, the probability of 3 B and 3 C from 6 trials that all must be B or C is 20/64 = 5/16.
I am not quite sure that I understand what questions you are asking.

First, I did not choose n = 5 to solve your problem. I did not know whether you understood the binomial probability formula or not. So I exemplified (not proved) it by an example using n = 3, n = 4, n = 5 to show the pattern. The example was not meant to address your problem directly. I think I actually said that I would start with a simpler problem than the one you posted. I apologize if that set of examples made you think n = 5 was relevant to your problem.

Second, I solved the POSTED problem in three stages, the first two of which depended on binomial probability. The first stage used n = 9 and treated the events as 3 A and 6 not-A. (You can ALWAYS reduce a problem to two events that way: A and not-A. Of course that is not always useful, but it sometimes is.) The second stage was to calculate 3 B out of only B and C (the not-A in the first stage) in 6 trials (6 = 9 trials total minus 3 resulting in A). The trick there was to figure out that the probability between B and C when A was excluded was (1/2), not (1/3).

Third, the last stage was to use the formula P(X and Y) = P(X | Y) * P(Y).

The rules of probability are simple. It is combining them that lets you solve the difficult problems.

Last edited by JeffM1; May 31st, 2016 at 06:16 PM.

 May 31st, 2016, 06:40 PM #8 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 Oh. Maybe I am being really stupid. $\dbinom{9}{3} * \left ( \dfrac{1}{3} \right )^3 * \left ( \dfrac{2}{3} \right )^6 = \dfrac{9 * 8 * 7}{3 * 2} * \dfrac{64}{19683} = \dfrac{84 * 64}{19683}= \dfrac{5376}{19683} = \dfrac{1792}{6561}.$ $\dbinom{6}{3} * \left ( \dfrac{1}{2} \right )^3 * \left ( \dfrac{1}{2} \right )^3 = \dfrac{6 * 5 * 4}{3 * 2} * \dfrac{1}{64} = \dfrac{20}{64} = \dfrac{5}{16}.$ $\dfrac{1792}{6561} * \dfrac{5}{16} = \dfrac{8960}{104976} \approx 8.5\%.$ Which is exactly the same answer I got before using powers and factorials. Thanks from EvanJ and manus
June 3rd, 2016, 06:57 PM   #9
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Quote:
 Originally Posted by JeffM1 First, I did not choose n = 5 to solve your problem. I did not know whether you understood the binomial probability formula or not. So I exemplified (not proved) it by an example using n = 3, n = 4, n = 5 to show the pattern. The example was not meant to address your problem directly. I think I actually said that I would start with a simpler problem than the one you posted. I apologize if that set of examples made you think n = 5 was relevant to your problem. Second, I solved the POSTED problem in three stages, the first two of which depended on binomial probability. The first stage used n = 9 and treated the events as 3 A and 6 not-A. (You can ALWAYS reduce a problem to two events that way: A and not-A. Of course that is not always useful, but it sometimes is.) The second stage was to calculate 3 B out of only B and C (the not-A in the first stage) in 6 trials (6 = 9 trials total minus 3 resulting in A). The trick there was to figure out that the probability between B and C when A was excluded was (1/2), not (1/3).
The first paragraph answers my question. What I put in bold is useful. I understand your last post. Thank you.

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