# Is this solution for the following Conditional Probability problem correct?

We have a box which contains $3$ white, $4$ red, and $5$ blue balls. We sample $3$ balls without replacement. What is probability that $3$rd ball is not blue given that first ball is not white and second ball is not red.

My solution is $155/252 \approx 0.615078$, but in textbook, solution is $41/67 \approx 0.61194$

Here is my solution:

Let $R_i$, $B_i$, $W_i$ be events that red, blue, or white ball is picked in i-th draw (i.e. $R_1 = \{(red, blue, blue), (red, white, red), ...\}$).

$P(B_3^c | W_1^c, R_2^c) = P(R_1 | W_1^c)\{ P(W_2 | R_1, R_2^c)*P(B_3^c | R_1, W_2) + P(B_2 | R_1, R_2^c)*P(B_3^c | R_1, B_2)\} + P(B_1 | W_1^c)\{P(W_2 | B_1, R_2^c)*P(B_3^3 | B_1, W_2) + P(B_2 | B_1, R_2^c)*P(B_3^c | B_1, B_2)\}$

Expression above is evaluated to: $(4/9)*\{(3/8)*(1/2) + (5/8)*(6/10)\} + (5/9)*\{(3/7)*(6/10) + (4/7)*(7/10)\}$

The idea was to draw a tree which represent all possible outcomes, calculate conditional probabilities, and just traverse and multiply.

This is what i'm talking about:

R_1 corresponds to $R_1$, Bc_2 to $B_2^c$, etc.

This is a complete tree which represents all possible outcomes, but pruned according to conditioning. How i got $P(R_1 | W_1^c) = 4/9$, and $P(B_1 | W_1^c) = 5/9$? Well we know that $W_1$ didn't happend, so all possible outcomes are 4 red + 5 blues which is 9. $P(W_2 | R_1, R_2^c) = 3/8$ because we know that $R_2$ didn't happend, so we only have 5 blues + 3 whites which is 8 (we're not counting that 4-1 red balls). $P(B_2 | B_1, R_2^c) = 4/7$ because our sample space consists only of 3 whites and 5 blues which is 8, but we lost one blue, so we have 7 total balls to draw, and 4 of them are blues.

In which step did i made an error?

• How did you arrive at your solution? Commented Mar 19, 2016 at 21:42
• By drawing tree of all possible outcomes and using law of total probability with conditioning on "first is not white and second is not black" Commented Mar 19, 2016 at 21:47
• If you can show more in your answer how did you get to it, will help. Commented Mar 19, 2016 at 21:48

## 2 Answers

We can tell that the textbook is wrong without checking your work. There are $14!$ equiprobable elementary events (the different orders in which the $14$ balls could be drawn), so the denominator of the result must divide $14!$. The prime number $67\gt14$ doesn't divide $14!$.

(I did check your work, though, and it looks correct.)

Nope. The book's answer is okay. \begin{align}\mathsf P(B_3^\complement\mid W_1^\complement\cap R_2^\complement) ~=&~ \tfrac{\mathsf P(W_1^\complement\cap R_2^\complement\cap B_3^\complement)}{\mathsf P(W_1^\complement\cap R_2^\complement)}\\[1ex]~=&~\tfrac{1-\mathsf P(W_1\cup R_2\cup B_3)}{1-\mathsf P(W_1\cup R_2)}\\[1ex]~=&~\tfrac{1-\mathsf P(W_1)-\mathsf P(R_2)-\mathsf P(B_3)+\mathsf P(W_1, R_2)+\mathsf P(W_1, B_3)+\mathsf P(R_2,B_3)-\mathsf P(W_1,R_2,B_3)}{1-\mathsf P(W_1)-\mathsf P(R_2)+\mathsf P(W_1,R_2)}\\[1ex]~=&~\tfrac{1-\tfrac 3{12}-\tfrac4{12}-\tfrac 5{12}+\tfrac 3{12}\tfrac 4{11}+\tfrac 3{12}\tfrac 5{11}+\tfrac 4{12}\tfrac 5{11}-\tfrac 3{12}\tfrac 4{11}\tfrac 5{10}}{1-\tfrac 3{12}-\tfrac 4{12}+\tfrac 3{12}\tfrac 4{11}}\\[1ex]~=&~\dfrac{41}{67}\end{align}

I think this is what you were trying to accomplish

\begin{align} \mathsf P(B_3^\complement\mid W_1^\complement,R_2^\complement) ~=&~ \tfrac{\mathsf P(B_3^\complement,R_2^\complement,W_1^\complement)}{\mathsf P(R_2^\complement,W_1^\complement)} \\[1ex] ~=&~ \tfrac{ \mathsf P(B_3^\complement,W_2,R_1) + \mathsf P(B_3^\complement,B_2,R_1) + \mathsf P(B_3^\complement,W_2,B_1) + \mathsf P(B_3^\complement,B_2,B_1) }{ \mathsf P(R_2^\complement,W_1^\complement) } \\[1ex] ~=&~ \tfrac{ \mathsf P(B_3^\complement\mid W_2,R_1)~\mathsf P(W_2,R_1) + \mathsf P(B_3^\complement\mid B_2,R_1)~\mathsf P(B_2,R_1) + \mathsf P(B_3^\complement\mid W_2,B_1)~\mathsf P(W_2,B_1) + \mathsf P(B_3^\complement\mid B_2,B_1)~\mathsf P(B_2,B_1) }{ \mathsf P(W_2,R_1) + \mathsf P(B_2,R_1) + \mathsf P(W_2,B_1) + \mathsf P(B_2,B_1) } \\[1ex] ~=&~ \frac{ \tfrac {5}{10}~\tfrac {4}{11}\tfrac 3{12} + \tfrac 6{10}~\tfrac{5}{11}\tfrac{4}{12} + \tfrac 6{10}~\tfrac{3}{11}\tfrac{5}{12} + \tfrac 7{10}~\tfrac{4}{11}\tfrac{5}{12} }{ \tfrac {4}{11}\tfrac 3{12} + \tfrac{5}{11}\tfrac{4}{12} + \tfrac{3}{11}\tfrac{5}{12} + \tfrac{4}{11}\tfrac{5}{12} } \\[1ex] ~=&~ \dfrac{41}{67} \end{align}

• Firstly, thank you for your time you allocated to answer my question. Could you point out errors i made, and why my solution is wrong so i don't repeat it again? Commented Mar 20, 2016 at 8:34
• @StankoKovacevic Difficult. $~$ I can't see where you went wrong because I can't see what you did. $~$ Though I think the above is what you were trying to do. Commented Mar 20, 2016 at 10:46