Probability, one bag or two?

Question

Scenario 1
Say you have two bags, A and B. Bag A contains 1 white ball and 2 black ball. Bag B contains 2 white balls and 3 black balls. You have to chose one of the bags randomly and pick one ball out of the bag you chose.

What is the probability that you'll get a white ball?

P(W) = The probability that the ball you pick is white.

P(A) = The probability that you choose bag A
P(B) = The probability that you choose bag B
P(W|A) = The probability that you pick a white ball given that you choose bag A
P(W|B) = The probability thar you pick a white ball given that you choose bag B

$P(W) = [P(A)\times P(W|A)] + [P(B)\times P(W|B)]$ $P(W) = [\frac{1}{2}\times \frac{1}{3}] + [\frac{1}{2}\times\frac{2}{5}]$
$P(W) = \frac{1}{6} + \frac{1}{5}$
$P(W) = \frac{11}{30}$

Scenario 2
Bag B contains 2 white balls and 1 black ball.$P(W) = [P(A)\times P(W|A)] + [P(W)\times P(W|B)]$
$P(W) = [\frac{1}{2}\times \frac{1}{3}] + [\frac{1}{2}\times \frac{2}{3}]$
$P(W) = \frac{3}{6}$

Notice that for scenario 2 we can consider the two bags as one bag. The total number of balls is 3 (bag A) + 3 (bag B) = 6. The total number of white balls = 1 (bag A) + 2 ( bag B) = 3. $P(W) = \frac{3}{6}$

However this is not true for scenario 1. The total number of balls = 3 (bag A) + 5 (bag B) = 8. The total number of white balls = 1 (bag A) + 2 (bag B) = 3. And ... most importantly ... $P(W) \ne \frac{3}{8}$.

BUT ...

For sceanrio 1 we see that $\frac{3}{8} - \frac{11}{30} = \frac{1}{120} \approx 0.008...$. 0.008... is a really small difference, which lends some legitimacy to treating the two bags as one bag.

Why can we sometimes treat two bags as one bag (scenario 2) and at other times not (scenario 1)?

Thank you.

Addendum

If P(W|A) = $\frac{a}{b}$ and P(W|B) = $\frac{c}{d}$ then ...

when is $P(W) = [P(A)\times \frac{a}{b}] + [P(B)\times \frac{c}{d}] = \frac{a + c}{b + d}$??

EDIT

After all the helpful answers:

The problem boils down to:

Where $x$ and $y$ and $\frac{a}{b}$ and $\frac{c}{d}$ are proper rational fractions

$x(\frac{a}{b}) + y(\frac{c}{d}) = \frac{a + c}{b + d}$ ... equation (i)

AND

$x + y = 1$ ... equation (ii)

Answer 1

I originally misinterpreted the question being asked, and my answer was somewhat misleading and implied incorrect things. true blue anil’s answer is correct, but let me add a proof for this, as well as a generalization.

If $p_A$ and $p_B$ are the probabilities of picking a white ball from bag $A$ and $B$ respectively, (in other words, they are the proportions of white balls to total balls for each respective bag), and $\lambda$ is the fraction of the total balls that are in bag A (so $(1-\lambda)$ is the fraction in bag B), and $\nu$ is the probability of choosing bag A (so $(1-\nu)$ is the probability of picking bag B), then the probability of picking a white ball when the bags are chosen first is $\nu p_A+(1-\nu)p_B$, whereas the probability of picking a white ball from the combined bag is $\lambda p_A + (1-\lambda)p_B$. Therefore the two probabilities will coincide if and only if \begin{align*} \nu p_A+(1-\nu)p_B & = \lambda p_A + (1-\lambda)p_B\\ (\nu-\lambda)p_A + ((1-\nu)-(1-\lambda))p_B &= 0\\ (\nu-\lambda)p_A + (\lambda-\nu)p_B &=0\\ (\nu-\lambda)p_A - (\nu-\lambda)p_B &=0\\ (\nu-\lambda)(p_A-p_B) &=0\text{.} \end{align*} The last equation holds if and only if one of the two factors equals $0$, meaning either

1. $p_A=p_B$, i.e., the proportion of white balls in bag A is the same as the proportion in bag B.

or

2. $\lambda=\nu$, i.e., the probability of picking bag A is the same as the proportion of total balls in bag A. In the special case under consideration ("scenario 2"), this is true with $\lambda=\nu=\frac{1}{2}$.

Remark

Intuitively, you can think of the explanation as being that combining the bags reweighs the probabilities so that the odds of picking each individual ball is the same, which would only have been true in the former case if $\lambda=\nu$. So if $\lambda=\nu$, then you haven't changed the probability of picking any given ball. If you HAVE changed the probabilities of picking individual balls, weighing the balls in one bag more heavily than before, then you need the proportion of whites in each bag to be the same (i.e., $p_A=p_B$) in order for this re-weighing not to matter.

Addendum

Regarding the edit to the original post, which asks for simultaneous solutions to

$$x(\frac{a}{b}) + y(\frac{c}{d}) = \frac{a + c}{b + d}\tag{i}$$ (I have edited (i) from the original post to correct for the intended meaning)

and $$ x+y=1\text{,} \tag{ii}$$

in the above analysis $x=\nu$, $y=1-\nu$, $\frac{a}{b}=p_A$, $\frac{c}{d}=p_B$, $\frac{b}{b+d}=\lambda$, and $\frac{d}{b+d}=(1-\lambda)$. Therefore, if (ii) is satisfied, then (i) becomes equivalent to the condition that either

1. $\frac{a}{b}=\frac{c}{d}$,

or

2. $x = \frac{b}{b+d}$ (or equivalently, $y=\frac{d}{b+d}$).

Answer 2

(You ask "Why can we sometimes treat two bags as one bag (scenario 2) and at other times not (scenario 1)?"

I assume that the bags have equiprobability of being chosen, as is usual for such problems.

Then suppose bag A has $a$ white balls, $b$ black ones, while bag B has $c$ white balls and $d$ black ones,

for us to be able to combine the bags into one,

$P(white) = \frac12\frac{a}{a+b} + \frac12\frac{c}{c+d} \;\;must= \frac{a+c}{a+b+c+d}$

This can only occur if $\frac{a}{a+b}= \frac{c}{c+d}$

ie the probability of drawing white from each bag is equal, OR the total number of balls in each bag is equal.

Yout Scenario $2$ works because of the second clause, ie equal number of balls in both bags.