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

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:

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.

• 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. – SlipEternal Oct 12 '18 at 14:11
• 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. – saulspatz Oct 12 '18 at 14:27
• 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. – Ned Oct 12 '18 at 17:06

$$P(R) = P(A)P(R|A) + P(B)P(R|B)\\ = .5(1/3) + .5(3/8) = 17/48 = 0.3541667.$$
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,
$$\frac{1/3 + 3/8}{2} = \frac{17}{48}.$$
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.]