What is the probability that a red ball will be selected?

Suppose there are two jars, $A,B$

$A$ has $2$ red, $4$ green

$B$ has $3$ red, $5$ green

An urn is selected at random, giving each of the urns a probability of $1/2$

A random urn is selected, and one ball is selected from that urn. What is the probability that a $G$ ball is selected.

I can use a tree diagram, in which my answer is very clear:

enter image description here

We can see that the probability that a $R$ ball is selected is $(1/2)(2/6)+(1/2)(3/8)=17/48$

The way my textbook does it is using the theorem of total probability (conditional version).

That is, $P(R)=P(R|A)P(A)+P(R|B)P(B)$

Thinking on pure intution, $P(R|A)$ is asking me "what is the probability that you selected a red ball, knowing that you already selected A". Well thats just $2/6$. I can basically just look at everything after the $A$ in the diagram.

So our equation ends up becoming the same as the thereom of probability.

Here is where my question is:

Using the formula, $P(R|A)=\dfrac{P(R\cap A)}{P(A)}$

But What is $P(R\cap A)$? If $P(A) = 1/2$, then surely the numerator must be $1/6$, since our intuitive approach told us that this conditional probability is $2/6$.

But I don't understand where this $1/6$ comes from. I can see that this is probably just $(1/2)(2/6)$ (basically we just multiply the entire branch), but why does this work? I know that you can multiple the probabilities of two independent events, but how is this independent? Selecting box A effected the number of red balls we had.

  • 1
    $\begingroup$ Select a random ball from A and a random ball from B. There is a $\dfrac{2}{6}$ chance the ball from A is red. There is a $\dfrac{3}{8}$ chance the ball from B is red. Now choose one of those two balls at random. What is the probability that the ball you selected is a red ball from A? That is what $P(R\cap A)$ represents. $\endgroup$ – SlipEternal Oct 12 '18 at 14:11
  • $\begingroup$ One problem is that $P(A\cap B)=P(A)P(B)$ is the definition of independence. Still I would say that choosing the urn and choosing a ball from the urn are independent events. Imagine that in one room, Alice randomly chooses a ball from urn A and in another, Ben randomly chooses a ball from urn B. Meanwhile, in another room, Chuck chooses the urn at random. $R\cap A$ is the event that Chuck chooses urn A, and Alice chooses a red ball. $\endgroup$ – saulspatz Oct 12 '18 at 14:27
  • $\begingroup$ The right hand side of your tree consists of four pairwise exclusive, exhaustive events, each of which is an intersection. The 1/6 is the probability of following the path that includes $A$ then $R$, that is, of picking urn A and also choosing a Red ball. $\endgroup$ – Ned Oct 12 '18 at 17:06

Law of total probability:

$$P(R) = P(A)P(R|A) + P(B)P(R|B)\\ = .5(1/3) + .5(3/8) = 17/48 = 0.3541667.$$

If you want an intuitive approach: Because urns $A$ and $B$ are equally likely, the answer is the average of the respective probabilities, $1/3$ and $3/8,$ of getting a red ball from each urn individually,

$$\frac{1/3 + 3/8}{2} = \frac{17}{48}.$$

Your diagram is an easy and correct approach, but it amounts to a graphical representation of the Law of Total Probability.

Note: The formula $P(B|A) = P(B \cap A)/P(A)$ is one definition of conditional probability, provided $P(A) > 0.$ In the form $P(B \cap A) = P(A)P(B|A),$ it is sometimes called the General Multiplication Rule. In this particular problem its use is natural because we know both $P(A)$ and $P(B|A).$ [If events $A$ and $B$ were independent, we would have $P(B \cap A) = P(B)P(A),$ but in this problem $A$ and $B$ aren't independent events.]


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