Probability of a white ball chosen from a bag if there are two bags.

Question

I was reading Bayes' Theorem and I came across a question. The question was:

What is the probability of picking a white ball from a bag, when there are two bags, Bag A and Bag B?
Bag A contains $2$ white and $3$ red balls and Bag B contains $3$ white and $4$ red balls.

So, by Total Probability Theorem,

$$P(W)=P(A)\times P(W \mid A)+P(B)\times P(W \mid B).$$

From the given question, it is very easy to say that $P(W \mid A)$ is $2/5$. As in the Bag A, there are total $5$ balls and $2$ of them is white.

But I am unable to proof this mathematically that $P(W \mid A)$ is $2/5$.

My approach was $P(W \mid A)= P(W \cap A)/P(A).$

I think $P(W)$ and $P(A)$ are dependent events, as $P(W \mid A)$ and $P(W \mid B)$ are not same.

( I think probability of picking a white ball depends on the bag.)

So how can I prove that $P(W \mid A) =2/5$?

Answer 1

Firstly, for probability of drawing white given A,
don't use P(W\A), the correct symbol is $P(W|A)$

And P(W|A) is simply the ratio of white balls to total balls in $A$, so there is nothing to prove here.

So, by Total Probability Theorem,

$P(W)=P(A)*P(W|A)+P(B)*P(W|B) = \Large\frac12\frac25 + \frac12\frac37$

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ADDED

Bayes' Rule would be needed if you has been asked, say, $P(A|W)$

$P(A|W) = \dfrac{P(A)*P(W|A)}{P(A)*P(W|A)+ P(B)*P(W|B)}$

Answer 2

It's easier not to use Bayes thereom to calculate $P(W|A)$. You are being confused by the notation of $|A$. But that just says if your universe of options is pulling marbles from a bag that is $A$ then
$$P(W|A) = P(\text{pulling a white marble from bag A}) =\\ P(\text{pulling a white marble from a bag with 2 whites and 3 reds}) =\\ \frac {\text{favorable outcomes}}{\text{all outcomes}} = \frac 25$$.

If you want to view this in terms of Bayes Theorem then $P(W|A) =\frac {P(W\cap A)}{P(A)}$ which involves calculating $P(W\cap A)$ which... well, we are being circular but...

We have do the experiment $70$ times we have will pick bag $A$ $35$ times and $B$ $35$ times. We will draw each of the marbles of bag $A$ $7$ times and the marbles of Bag $B$ $5$ times. So $P(W\cap A) = \frac {7+7}{70} = \frac 1{5}$.

So $P(W|A) = \frac {P(W\cap A)}{P(A)} =\frac {\frac 1{5}}{\frac 12} =\frac 25$.

This verifies Bayes thereom but it is not the most natural way to set things up.

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It is useful to use Baye's Thereom to calculate $P(M|N)$ when $P(N)$ and $P(M\cap N)$ are easy to calculate, or in cases where Total Probability is not straightforward. In cases were $P(M\cap N)$ or $P(N)$ are not easy to calculate, or where $P(M|N)$ is itself straightforward and especially when Total Probability IS straightforward, then... don't use it.

And in this case $P(W|A)= \frac 25$ and $P(W|B) = \frac 37$ is about as straightforward as it ever gets.

Answer 3

I used a tree diagram.
You draw all the options and their respective probabilities, then you can multiply them to get the probability for the event at the bottom of the tree.
Here, getting White from Bag A would be $2/10$ since you first have a chance of $1/2$ of picking Bag A and then also have to get White which is $2/5$. Since $1/2 \times 2/5$ is $2/10$ that's the result.
Getting White out of Bag B has a different chance because there are different Balls in each bag, even though the chance for each bag is the same. Here getting white out of Bag B would be $3/14$ as shown in the picture below:

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