Drawing balls from an urn

[daon]

This isn't tricky at all, should be straight forward. I'm sure many here are better versed in counting problems than I

I thought I had solved this two ways, but both are probably wrong (they certainly both aren't right).

The Problem: An urn contains 7 blue balls, 5 red balls. You randomly remove 6 balls and place them into a second empty urn. You then remove one ball from the second urn and see it is blue. What is the probability the balls left in the second urn consist of 3 red and 2 blue?

My first solution: Here I used "intuition" which normally gets me in trouble. Since we know that one ball is blue, we may disregard it and simplify the problem to counting the number of arrangements of 3 Red balls and 2 Blue balls, taking care to remove one blue from the sample space.

\(\displaystyle Answer \,\, 1: \,\, \,\, \frac{{7 \choose 3} {4 \choose 2}}{{11 \choose 5}}\)


Second solution: I attempted to use standard conditional probability.
Let E(C) be the event of choosing a ball from the second urn of color C.
We are looking for:

\(\displaystyle P(3R2B | E(B)) = \frac{P(E(B) | 3R2B) \cdot P(3R2B)}{P(E(B))}\)

But this got extremely hectic; The numerator is fine to calculate. But, what does event E(B) even mean if we don't know which balls have been placed in the 2nd urn?

Then I thought to break it up... for any event K we have:

\(\displaystyle P(E(B)) = P(E(B)|K)P(K) + P(E(B)|K^c)P(K^c)\)

I tried letting K=3R2B, but then, K^c is not fun to think about.

Looking forward to someone's insight

 

[Denis]

I got this kwicky thought:
answer is probability of ending up with 3 of each in 2nd urn :idea:

SO:
7C3 * 5C3 / 12C6 = 35 * 10 / 924 = 350 / 924 = 25 / 66 = .378787878....

 

[soroban]

Hello, daon!

This is a long and tedious problem . . . I don't see any way around it.

An urn contains 7 blue balls, 5 red balls.
You randomly remove 6 balls and place them into a second empty urn.
You then remove one ball from the second urn and see it is blue.
What is the probability the balls left in the second urn consist of 3 red and 2 blue?

\(\displaystyle \text{Just before the ball was removed from the second urn, it contained 3 Blue and R red.}\)

\(\displaystyle \text{We have: }\;P(\text{3B,3R}\,|\,\text{B drawn}) \;=\;\frac{P(\text{3B,3R}\,\wedge\,\text{B drawn})}{P(\text{B drawn})}\)


The denominator requires a lot of work.
. . There are six scenarios . . .


[1] 6 Blue, 0 Red are moved.
. . \(\displaystyle P(\text{6B,0R}) \:=\:\frac{{7\choose6}{5\choose0}}{{12\choose6}} \;=\;\frac{7\cdot1}{924} \:=\:\frac{1}{132}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{6}{6} \:=\:1\)
. . \(\displaystyle \text{Hence: }\;P(\text{6B,0R}\,\wedge\,\text{B drawn}) \;=\;\frac{1}{132}\cdot1 \:=\:\frac{1}{132}\)


[2] 5 Blue, 1 Red are moved.
. . \(\displaystyle P(\text{5B,1R}) \:=\:\frac{{7\choose5}{5\choose1}}{{12\choose6}} \:=\:\frac{21\cdot5}{924} \:=\:\frac{5}{44}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{5}{6}\)
. . \(\displaystyle \text{Hence: }\;P(\text{5B,1R}\,\wedge\,\text{B drawn}) \:=\:\frac{5}{44}\cdot\frac{5}{6} \:=\:\frac{25}{264}\)


[3] 4 Blue, 2 Red are moved.
. . \(\displaystyle P(\text{4B,2R}) \:=\;\frac{{7\choose4}{5\choose2}}{{12\choose6}} \:=\:\frac{35\cdot10}{924} \:=\:\frac{25}{66}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{4}{6}\:=\:\tfrac{2}{3}\)
. . \(\displaystyle \text{Hence: }\;P(\text{4B,2R}\,\wedge\,\text{B drawn}) \:=\:\frac{25}{66}\cdot\frac{2}{3}\:=\:\frac{25}{99}\)


[4] 3 Blue, 3 Red are moved.
. . \(\displaystyle P(\text{3B,3R}) \:=\:\frac{{7\choose3}{5\choose3}}{{12\choose6}} \:=\:\frac{35\cdot10}{924} \:=\:\frac{25}{66}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{3}{6}\:=\:\tfrac{1}{2}\)
. . \(\displaystyle \text{Hence: }\;P(\text{3B,3R}\,\wedge\,\text{B drawn}) \:=\:\frac{25}{66}\cdot\frac{1}{2} \:=\:\frac{25}{132}\)


[5] 2 Blue, 4 Red are moved.
. . \(\displaystyle P(\text{2B,4R}) \:=\:\frac{{7\choose2}{5\choose4}}{{12\choose6}} \:=\:\frac{21\cdot5}{924} \:=\:\frac{5}{44}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{4}{6} \:=\:\tfrac{2}{3}\)
. . \(\displaystyle \text{Hence: }\;P(\text{2B,4R}\,\wedge\,\text{B drawn}) \:=\:\frac{5}{44}\cdot\frac{2}{3} \:=\:\frac{5}{66}\)


[6] 1 Blue, 5 Red are moved.
. . \(\displaystyle P(1B,5R}) \:=\:\frac{{7\choose1}{5\choose5}}{{12\choose6}} \:=\:\frac{7\cdot1}{924} \:=\:\frac{1}{132}\)
. . \(\displaystyle P(\text{B drawn}) \:=\:\tfrac{1}{6}\)
. . \(\displaystyle \text{Hence: }\;P(\text{1B,5R}\,\wedge\,\text{B drawn}) \:=\:\frac{1}{132}\cdot\frac{1}{6} \:=\:\frac{1}{792}\)



\(\displaystyle \text{The denominator is the }s\:\!\!um\text{ of these six probabilities:}\)

. . \(\displaystyle \frac{1}{132} + \frac{25}{264} + \frac{25}{99} + \frac{25}{132} + \frac{5}{66} + \frac{1}{792} \;=\; \frac{492}{792} \:=\:\frac{41}{66}\)


\(\displaystyle \text{The numerator is [4] on our list: }\;P(\text{3B,3R}\,\wedge\,\text{B drawn}) \:=\:\frac{25}{132}\)



\(\displaystyle \text{Therefore: }\;P(\text{3B,3R}\,|\,\text{B drawn}) \;=\;\frac{\frac{25}{132}}{\frac{41}{66}} \;=\;\frac{25}{82}\)


But check my work . . . please!
.

 

[daon]

Yeah, I'm pretty sure that is the way to go about it. I didn't think it would have to be that long. Thanks.