Combinatorial probability problem
John Uebersax
question:
An urn contains red, blue, and green balls, in equal proportions.
Drawing four balls (with replacement), what is the probability that at
least one color will not be represented among the four.
Thanks in advance. (No, this is not a homework problem ;) )
--
John Uebersax PhD
http://ourworld.compuserve.com/homepages/jsuebersax/agree.htm
iandj...@aol.com
The numbers of red, green and blue balls will have a multi-nomial
distribution.
The only way you will have all 3 colours represented is if there are 2
reds, 1 green and one blue ball or 1 red, 2 greens and one blue ball
or 1 red, 1 green and two blue balls.
The probabality is 3*4!/(2!*1!*1!)*(1/3)^3
The probability of at least one colour not being selected is therefore
5/9.
Ian Smith
Nasser Abbasi
"John Uebersax" <jsueb...@gmail.com> wrote in message
news:1195042122....@o80g2000hse.googlegroups.com...
These are the hard problems I am talking about when it comes to probability.
This is my attempt at it.
at least one color will not be selected means P(exactly one color not
selected) or P(exactly 2 colors not selected)
There are 3^4 ways to draw 4 balls. (wuthout replacement)
There are 2^4 to draw 4 balls with one color missing. But there are 3
colors, hence there are 3*(2^4) ways to draw it.
There are 1 way to draw 4 balls from 2 color missing. But there are 3
colors, hence there are 3 ways to draw it.
Hence the chance at least one color missing is [3*(2^4)+3]/3^4 or 51/81=
0.62963
Almost 60% at least one color is missing? too high
Did I make a mistke? I have a feeling I did :)
Nasser
Jussi Piitulainen
> Would someone kindly tell me what is the *formula* to answer this
> question:
>
> An urn contains red, blue, and green balls, in equal proportions.
> Drawing four balls (with replacement), what is the probability that
> at least one color will not be represented among the four.
Here's an elementary derivation.
Let Rk, Gk, Bk be the statements that there are k Red, Green, Blue
balls among the four drawn. On your information I, there are four
draws, each drawn ball being Red or Green or Blue, each colour equally
likely each time.
Using A+B for disjunction, A,B for conjunction:
P(R0 + G0 + B0|I)
= P(R0|I) + P(G0|I) + P(B0|I) ; repeated use of the sum rule
- P(G0, B0|I) - P(R0, G0|I) - P(R0, B0|I)
+ P(R0, G0, B0|I)
= P(R0|I) + P(G0|I) + P(B0|I) ; use the information I
- P(R4|I) - P(B4|I) - P(G4|I)
+ 0
= 3(P(R0|I) - P(R4|I)) ; colours have the same probabilities
= 3((2/3)^4 - (1/3)^4) ; simple combinatorics
= 5/9.
John Uebersax
Two questions:
I seem, perhaps mistakenly, to enumerate 15 possible combinations, of
which only 3 include all
colors, or P = 3/15. Am I overlooking something obvious:
R B G
-----
4 0 0
3 1 0
3 0 1
2 2 0
2 0 2
2 1 1
1 3 0
1 2 1
1 1 2
1 0 3
0 4 0
0 3 1
0 2 2
0 1 3
0 0 4
2. What is the corresponding probability/formula for all colors being
represented given 5 balls drawn instead of 4?
Thanks,
John Uebersax
John Uebersax
I, perhaps mistakenly, enumerate 15 possible combinations, with 3
meeting the criterion, for P = 3/ 15. Am I overlooking something very
obvious?
R B G
-----
4 0 0
3 1 0
3 0 1
2 2 0
2 0 2
2 1 1
1 3 0
1 2 1
1 1 2
1 0 3
0 4 0
0 3 1
0 2 2
0 1 3
0 0 4
Also, can you tell me how your formula would generalize given five
balls drawn instead of 3.
--
John Uebersax PhD
On Nov 14, 2:06 pm, iandjmsm...@aol.com wrote:
Joe Blow
"Each color is drawn" equivalent to "If the first two balls are the same color, the next two balls must be different from it and each other, and if the first two balls are different colors, the next two must be the same color of one of the remaining two colors."
So,
Prob(each color is drawn) = 1/3*(2/3*1/3) + 2/3(2*1/3*1/3) = 4/9
1-Prob(each color is drawn) = 5/9.
iandj...@aol.com
>
> - Show quoted text -
The probabilities are
R B G
-----
4 0 0 1/81
3 1 0 4/81
3 0 1 4/81
2 2 0 6/81
2 0 2 6/81
2 1 1 12/81
1 3 0 4/81
1 2 1 12/81
1 1 2 12/81
1 0 3 4/81
0 4 0 1/81
0 3 1 4/81
0 2 2 6/81
0 1 3 4/81
0 0 4 1/81
and the sum of the 3 which include all is 36/81 or 4/9.
With similar logic, 3*(prob of selecting 3,1,1 + prob of selecting
2,2,1), the probability that at least one color will not be
represented among the five balls drawn is 31/81.
Ian Smith
Karl Ove Hufthammer
> The probabality is 3*4!/(2!*1!*1!)*(1/3)^3
The last factor should be (1/3)^4 (which you have correctly used in the
final answer).
> The probability of at least one colour not being selected is therefore
> 5/9.
--
Karl Ove Hufthammer
Karl Ove Hufthammer
> I seem, perhaps mistakenly, to enumerate 15 possible combinations, of
> which only 3 include all
> colors, or P = 3/15. Am I overlooking something obvious:
>
> R B G
> -----
> 4 0 0
> 3 1 0
> 3 0 1
> 2 2 0
They don't all have the same probability. For instance, you can get
400 in only one way, by picking RRRR (probability: (1/3)^4), but you
can get 220 in several (6) ways: RRBB RBRB BRBR BBRR BRRB RBBR
(Probability: 6 × (1/3)^4.)
The shortcut 'favourable outcomes divided by possible outcomes' only works
when each outcome has the same probability (i.e., uniform distribution).
Here 220 is six times as likely as 400 to occur, so you can't use this
shortcut.
--
Karl Ove Hufthammer
Nasser Abbasi
"Nasser Abbasi" <n...@12000.org> wrote in message
news:PQC_i.11741$4k....@newsfe11.phx...
>
> "John Uebersax" <jsueb...@gmail.com> wrote in message
> news:1195042122....@o80g2000hse.googlegroups.com...
>> Would someone kindly tell me what is the *formula* to answer this
>> question:
>>
>> An urn contains red, blue, and green balls, in equal proportions.
>> Drawing four balls (with replacement), what is the probability that at
>> least one color will not be represented among the four.
>>
>> Thanks in advance. (No, this is not a homework problem ;) )
>> --
>> John Uebersax PhD
>> http://ourworld.compuserve.com/homepages/jsuebersax/agree.htm
>>
>
>
> at least one color will not be selected means P(exactly one color not
> selected) or P(exactly 2 colors not selected)
>
> There are 3^4 ways to draw 4 balls. (wuthout replacement)
>
> There are 2^4 to draw 4 balls with one color missing. But there are 3
> colors, hence there are 3*(2^4) ways to draw it.
> There are 1 way to draw 4 balls from 2 color missing. But there are 3
> colors, hence there are 3 ways to draw it.
> Hence the chance at least one color missing is [3*(2^4)+3]/3^4 or 51/81=
> 0.62963
>
Opps, I overcounted again! I should have subracted P(exactly 2 colors not
selected) not added it (becuase it is counted allready)
Replace [3*(2^4)+3]/3^4 by [3*(2^4)-3]/3^4 and the answer will come out
5/9
Nasser
John Uebersax
Okay, I see now.
Thanks,
John
Ray Koopman
Let k = the # of colors, let n = the # of draws, and let [x1,...,xk]
denote the number of times each color appeared. Sum xj = n.
Then the usual multinomial probability formula, with all pj = 1/k,
gives P[x1,...,xk] = n!/(k^n Prod xj!).
The probability that at least one color is not represented is the
sum of the probabilities of all possible outcome vectors that have
at least one zero; or, alternatively, one minus the sum of the
probabilities of all possible outcome vectors that have no zeros.
That's about as close as you're likely to get to a general *formula*.
Here's an example for k = 4, n = 6. The "patterns" are all possible
nonincreasing sequences of k non-negative integers that sum to n.
pattern # of distinguishable permutations
6 0 0 0 4
5 1 0 0 4*3
4 2 0 0 4*3
4 1 1 0 4*3
3 3 0 0 4C2
3 2 1 0 4!
3 1 1 1 4
2 2 2 0 4
2 2 1 1 4C2
6! 4 4C2 195
P[no zeros] = ---*(----------- + -----------) = ---
4^6 3!*1!*1!*1! 2!*2!*1!*1! 512
jos jansen
was about drawing with replacement?
Ray Koopman
No, those answers (and mine) are for sampling *with* replacement,
which gives a multinomial distribution.