MHB Probability of White Ball Drawn Twice From Urn I & II

[Jameson]

Basic problem:
Urn I has 4 white and 4 black balls. Urn II has 2 white and 6 black balls. Flip a fair coin. If the outcome is heads, then a ball from urn I is selected, whereas if the outcome is tails, then a ball from urn II is selected. Suppose a white ball is selected and the replaced. Denote this event by $W_1$. Now another ball is withdrawn at random from the same urn.

(a) What is the probability that the first ball is from urn I (given that it is white)?

My solution to part (a):
This can be solved by Bayes' Theorem in a pretty straightforward way. Let the notation $W_1$ mean that the first ball is white and let $U_1$ mean that it was chosen from urn I.

$$P \left( U_1|W_1 \right) = \frac{ P \left( W_1 \cap U_1 \right)}{P \left(W_1 \right)} = \frac{P \left( W_1|U_1 \right)P(U_1)}{P \left( W_1|U_1 \right)P(U_1)+P \left( W_1|U_2 \right)P(U_2)}$$.

Now plugging in the information I have into the last expression I get:

$$P \left( U_1|W_1 \right)=\frac{\frac{1}{2}\frac{1}{2}}{\frac{1}{2} \frac{1}{2}+\frac{1}{4}\frac{1}{2}}=\frac{2}{3}$$

So my question is how does that look?

(b) What is the probability that the second ball is also white?

My solution to part (b):
This one I'm not 100% sure on. I think it should be similar to part (a) except the values of $$P(U_1)$$ and $$P(U_2)$$ will be different. Instead of $\frac{1}{2}$ and $\frac{1}{2}$ they will be $\frac{2}{3}$ and $\frac{1}{3}$ respectively. So the solution should be found using the same method just replacing those values, correct?

 

[Jameson]

I think I solved it now. My above idea was wrong.

What I need to find is $$P(W_2|W_1)=\frac{P(W_2 \cap W_1)}{P(W_1)}$$. Both the numerator and denominator should be broken up into two cases. Once I do that I get the following:

$$P(W_2|W_1)=\frac{\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{4} \frac{1}{4}\frac{1}{2}}{\frac{1}{2}\frac{1}{2}+ \frac{1}{4}\frac{1}{2}}=\frac{5}{12}$$

 

[I like Serena]

Basic problem:
Urn I has 4 white and 4 black balls. Urn II has 2 white and 6 black balls. Flip a fair coin. If the outcome is heads, then a ball from urn I is selected, whereas if the outcome is tails, then a ball from urn II is selected. Suppose a white ball is selected and the replaced. Denote this event by $W_1$. Now another ball is withdrawn at random from the same urn.
My solution to part (a):
This can be solved by Bayes' Theorem in a pretty straightforward way. Let the notation $W_1$ mean that the first ball is white and let $U_1$ mean that it was chosen from urn I.

$$P \left( U_1|W_1 \right) = \frac{ P \left( W_1 \cap U_1 \right)}{P \left(W_1 \right)} = \frac{P \left( W_1|U_1 \right)P(U_1)}{P \left( W_1|U_1 \right)P(U_1)+P \left( W_1|U_2 \right)P(U_2)}$$.

Now plugging in the information I have into the last expression I get:

$$P \left( U_1|W_1 \right)=\frac{\frac{1}{2}\frac{1}{2}}{\frac{1}{2} \frac{1}{2}+\frac{1}{4}\frac{1}{2}}=\frac{2}{3}$$

So my question is how does that look?

Looks fine!

Since the probability on urn I and urn II is equally likely, so any ball is equally likely, you can shorten it to:
$$P(U_1|W_1) = \frac{P(U_1 \wedge W_1)}{P(W_1)} = \frac{\frac 4 {16}}{\frac 6 {16}} = \frac 2 3$$

My solution to part (b):
This one I'm not 100% sure on. I think it should be similar to part (a) except the values of $$P(U_1)$$ and $$P(U_2)$$ will be different. Instead of $\frac{1}{2}$ and $\frac{1}{2}$ they will be $\frac{2}{3}$ and $\frac{1}{3}$ respectively. So the solution should be found using the same method just replacing those values, correct?

I think I solved it now. My above idea was wrong.

What I need to find is $$P(W_2|W_1)=\frac{P(W_2 \cap W_1)}{P(W_1)}$$. Both the numerator and denominator should be broken up into two cases. Once I do that I get the following:

$$P(W_2|W_1)=\frac{\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{4} \frac{1}{4}\frac{1}{2}}{\frac{1}{2}\frac{1}{2}+ \frac{1}{4}\frac{1}{2}}=\frac{5}{12}$$

Working it out mathematically:

$$\begin{aligned}
P(W_2|W_1)&=\frac {P(W_1 \wedge W_2)}{P(W_1)} \\
&= \frac {P((U_1 \wedge W_1 \wedge W_2) \vee (U_1 \wedge W_1 \wedge W_2))}{P(W_1)} \\
&= \frac {P(U_1 \wedge W_1 \wedge W_2) + P(U_1 \wedge W_1 \wedge W_2)}{P(W_1)} && \text{disjoint sum rule}\\
&= \frac {P(W_2|U_1 \wedge W_1)P(U_1 \wedge W_1) + P(W_2|U_2 \wedge W_1)P(U_2 \wedge W_1)}{P(W_1)} && \text{product rule} \\
&= \frac {\frac 4 8 \cdot \frac 4 {16} + \frac 2 8 \cdot \frac 2 {16}}{\frac 6 {16}} && \text{rule for equally likely outcomes}\\
&= \frac 5 {12} \\
\end{aligned}$$

 

[Jameson]

Thank you for your reply I like Serena! :)

I am leaving now for a study session so will have to reply later but I feel like my method is correct. Also, I think you might be interpreting the problem as without replacement but it's with replacement. Again, my apologies that I can't answer in more detail now but trust me I will later tonight and really appreciate the help!

I broke down the numerator as follows: $$P(W_1 W_2)=P(W_1 W_2|U_1)P(U_1)+P(W_1 W_2|U_2)P(U_2)$$ and the numbers I wrote follow. Do you see any flaw in this reasoning?

 

[I like Serena]

Thank you for your reply I like Serena! :)

I am leaving now for a study session so will have to reply later but I feel like my method is correct. Also, I think you might be interpreting the problem as without replacement but it's with replacement. Again, my apologies that I can't answer in more detail now but trust me I will later tonight and really appreciate the help!

I broke down the numerator as follows: $$P(W_1 W_2)=P(W_1 W_2|U_2)P(U_2)+P(W_1 W_2|U_2)P(U_2)$$ and the numbers I wrote follow. Do you see any flaw in this reasoning?

Yes. Sorry. I had just realized my mistake and corrected it.
And your reasoning is flawless (although you made a typo with $U_1$ and $U_2$ ;)).

$$P(W_1 W_2)=P(W_1 W_2|U_1)P(U_1)+P(W_1 W_2|U_2)P(U_2)$$

 

[Jameson]

Typo fixed and we now get the same answer, $\frac{5}{12}$, so I am marking the thread solved. :)

I feel confident about my midterm tomorrow. :D