2
$\begingroup$

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.

$\endgroup$
3
  • 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
2
$\begingroup$

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.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.