# Urn problem - conditional probability

• Sep 10th 2008, 05:32 PM
ban26ana
Urn problem - conditional probability
Consider three urns, A, B, and C. Urn A contains 6 white balls and 4 black balls, urn B contains 2 white balls and 6 black balls, and urn C is empty. Two balls are drawn at random from each one of the urn A and urn B, and are placed into urn C. (4 balls are now in urn C.) Then a ball is drawn at random from urn C. What is the probability the ball drawn from urn C is black?

I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. At least I think so. But now I don't know what to do with that. (My backslashes are for vertical lines I don't know how to type that.)
• Sep 10th 2008, 09:24 PM
Soroban
Hello, ban26ana!

I wouldn't call this "conditional probability" . . .

Quote:

There are three urns, A, B, and C.
Urn A contains 6 white balls and 4 black balls,
urn B contains 2 white balls and 6 black balls,
and urn C is empty.

Two balls are drawn at random from each of urn A and urn B,
and are placed into urn C.
Then a ball is drawn at random from urn C.
What is the probability the ball drawn from urn C is black?

This takes a lot of preliminary work . . . bear with me.

Urn A
There are: .$\displaystyle {10\choose2} = 45$ outcomes.

$\displaystyle \begin{array}{cccccccc} WW\!: & {6\choose2} = 15\text{ ways} & P_A(WW) &=& \frac{15}{45} &=& \frac{5}{15} \\ \\[-3mm] WB\!: & 6\cdot 4 = 24\text{ ways} & P_A(WB) &= & \frac{24}{45} &=& \frac{8}{15} \\ \\[-3mm] BB\!: & {4\choose2} = 6\text{ ways} & P_A(BB) & = & \frac{6}{45} &=& \frac{2}{15}\end{array}$

Urn B
There are: .$\displaystyle {8\choose2} = 28$ outcomes.

$\displaystyle \begin{array}{cccccccc} WW\!: & {2\choose2} = 1\text{ way} & P_B(WW) &=& \frac{1}{28} \\ \\[-3mm] WB\!: & 2\cdot6 = 12\text{ ways} & P_B(WB) &=& \frac{12}{28} \\ \\[-3mm] BB\!: & {6\choose2} = 15\text{ ways} & P_B(BB) &=& \frac{15}{28} \end{array}$

Urn C

$\displaystyle WWWW\!:\;\;\begin{Bmatrix} P_A(WW)\cdot P_B(WW) &=&\frac{5}{15}\cdot\frac{1}{28} &=&\frac{5}{420} \end{Bmatrix} \qquad P(W,W,W,W) \:=\:\frac{5}{420}$

$\displaystyle WWWB\!:\;\;\begin{Bmatrix} P_A(WW)\cdot P_B(WB) & =&\frac{5}{15}\cdot\frac{12}{28} &=& \frac{60}{420} \\ \\[-3mm] P_A(WB)\cdot P_B(WW) &=& \frac{8}{15}\cdot\frac{1}{28} &=& \frac{8}{420} \end{Bmatrix} \qquad P(W,W,W,B) \:=\:\frac{68}{420}$

$\displaystyle WWBB\!:\;\;\begin{Bmatrix} P_A(WW)\cdot P_B(BB) &=& \frac{5}{15}\cdot \frac{15}{28} &=& \frac{75}{420} \\ \\[-3mm] P_A(BB)\cdot P_B(WW) &=& \frac{2}{15}\cdot\frac{1}{28} &=& \frac{2}{420} \\ \\[-3mm] P_A(WB)\cdot P_B(WB) &=& \frac{8}{15}\cdot\frac{12}{28} &=& \frac{96}{420}\end{Bmatrix} \qquad P(W,W,B,B) \:=\:\frac{173}{420}$

$\displaystyle WBBB\!:\;\;\begin{Bmatrix} P_A(WB)\cdot P_B(BB) &=& \frac{8}{15}\cdot\frac{15}{28} &=& \frac{120}{420} \\ \\[-3mm] P_A(BB)\cdot P_B(WB) &=& \frac{2}{15}\cdot\frac{12}{28} &=&\frac{24}{420} \end{Bmatrix} \qquad P(W,B,B,B) \:=\:\frac{144}{420}$

$\displaystyle BBBB\!:\;\;\begin{Bmatrix} P_A(BB)\cdot P_B(BB) &=& \frac{2}{15}\cdot\frac{15}{28} &=& \frac{30}{420}\end{Bmatrix} \qquad P(B,B,B,B) \:=\:\frac{30}{420}$

Now in each of these five cases, what is the probability of drawing a black ball?

$\displaystyle \begin{array}{cccccccc} P(WWWW) \:=\: \frac{5}{420}, & P(B) \:=\: 0 && P(WWWW \wedge B) &=& \frac{5}{420}\cdot0 &=& 0 \\ \\[-3mm] P(WWWB) \:=\:\frac{68}{420}, &P(B) \:=\:\frac{1}{4} && P(WWWB \wedge B) &=& \frac{68}{420}\cdot\frac{1}{4} &=& \frac{68}{1680} \\ \\[-3mm] P(WWBB) \:=\:\frac{173}{420}, & P(B) \:=\:\frac{2}{4} && P(WWBB \wedge B) &=&\frac{173}{420}\cdot\frac{2}{4} &=& \frac{346}{1680} \\ \\[-3mm]\end{array}$
$\displaystyle \begin{array}{cccccccc} P(WBBB) \:=\:\frac{144}{420}, & \;\;\;P(B) \:=\:\frac{3}{4} && P(WBBB \wedge B) &\;\;=&\frac{144}{420}\cdot\frac{3}{4} &=& \frac{432}{1680} \\ \\[-3mm] P(BBBB) \:=\:\frac{30}{420}, & \;\;\;P(B) \:=\:\frac{4}{4} && P(BBBB \wedge B) &\;\;=&\frac{30}{420}\cdot\frac{4}{4} &=& \frac{120}{1680}\end{array}$

Therefore: . $\displaystyle P(B) \;=\;\frac{68}{1680} + \frac{346}{1680} + \frac{432}{1680} + \frac{120}{1680} \;=\;\frac{966}{1680} \;=\;\boxed{\frac{161}{280}}$

I need a nap . . .
.
• Sep 10th 2008, 10:45 PM
ban26ana
Thank you very much. Really, thank you. When solving a problem like this, what if it is impossible to list all the possible outcomes? What if there were 30 outcomes for each step. You obviously wouldn't list them all. Is there a way to solve it for that, or are you stuck doing this huge amount of work for every problem.