Conditional Probability vs Bayes Theorem

Question

Suppose that we have $3$ cards. Both sides of the first card are red, both sides of the second are black, one side of the third card is red and the other side black. $1$ card is selected at random and put down on the ground. If the upper side of the card is red, what's the probability that the other side is black?

I'm struggling with whether to use standard conditional probability or Bayes theorem.

$A= We \ choose \ the \ red \ and \ black \ card$

$B= The \ upper side \ is \ red$

$P(A \mid B)= \frac{P(A \cap B)}{P(B)}= \frac 12*\frac53$~This is clearly wrong

I guess my biggest problem is falsely solving for the union and intersection when not possible. How do I calculate the intersection of $A$ and $B$ in this example?

The correct solution uses Bayes Theorem, which gives the answer of $\frac13$. I can also see intuitively that if we pick one card at random, the probability that it is the one card we are looking for is $\frac 13$.

Why am I not solving for intersection and union correctly. Also, if Bayes theorem is required, why is it so?

Answer 1

I'm struggling with whether to use standard conditional probability or Bayes theorem.

They are not exactly different things.

• Bayes' Rule is: $\mathsf P(A\mid B) = \mathsf P(B\mid A)\,\mathsf P(A)/\mathsf P(B)$
• The definition of conditioning is: $\mathsf P(A\mid B)\,\mathsf P(B) = \mathsf P(A)\,\mathsf P(B\mid A)=\mathsf P(A\cap B)$

So what you use to evaluate $\mathsf P(A\mid B)$ merely depends on what information you have to use.

Suppose that we have $3$ cards. Both sides of the first card are red, both sides of the second are black, one side of the third card is red and the other side black. $1$ card is selected at random and put down on the ground. If the upper side of the card is red, what's the probability that the other side is black?

$A=$ We choose the red and black card

$B=$ The upper side is red

$\mathsf P(A\mid B)~{ = \mathsf P(A\cap B)/\mathsf P(B) \\= \mathsf P(B\mid A)\mathsf P(A)/\mathsf P(B) \\= \tfrac 12\cdot\tfrac 13/\tfrac 12 \\= \tfrac 13}$

Note: $\mathsf P(B)=\tfrac 12$ because each of the faces might show without bias, and half of them are red.   The Law of Total Probability agrees: $\mathsf P(B) ~{= \mathsf P(B\mid A)\mathsf P(A)+\mathsf P(B\mid A^\complement)\mathsf P(A^\complement) \\= \tfrac 12\tfrac 13+\tfrac 12\tfrac 23\\=\tfrac 12}$

Similarly we can say $\mathsf P(A\cap B)=\tfrac 16$ because of the six faces that might show without bias, only one of them is red and belongs to the red-black card.

We can also directly evaluate $\mathsf P(A\mid B)$ by noting that of the three red faces that might be showing without bias, only one belongs to the red-black card.   So thus $\mathsf P(A\mid B)=1/3$ .

Answer 2

If you label the six sides of the cards, "A" through "F," then it should be clear that each letter has an equal chance of appearing on the upper side of the chosen card. So, $P(A \cap B) = 1/6$.

Furthermore, $P(B) = 3/6$ because there are three red sides.

So, your approach (if you computed the two probabilities correctly) yields the same answer as the Bayes's Theorem approach.

---

You should not feel that these are completely different, however, since the numerator and denominator of the complicated side of Bayes's theorem are just a different ways of computing $P(A \cap B)$ and $P(B)$. In this case, it uses the fact that it is easy to compute $P(B \mid A) = 1/2$ and $P(B \mid \text{choose the all black card}) = 0$ and $P(B \mid \text{choose the all red card}) = 1$. In some problems, you must use Bayes's theorem only because you are given certain conditional probabilities in the problem but not others. In this problem however, you can still compute it from elementary principles as above.

Answer 3

Note that there are $6$ equally likely outcomes. Hence: $$P(B_{bottom}|R_{top})=\frac{P(B_{bottom}\cap R_{top})}{P(R_{top})}=\frac{1/6}{3/6}=\frac13.$$