# Difference between Conditional Probability and Bayes Theorem

To ask my question I will present an example. Suppose we have $2$ boxes with yellow and blue balls. Box 1 has $5$ yellow and $4$ blue balls and Box 2 has $7$ yellow and $5$ blue balls. We draw a ball at random from one of the boxes ($50-50$ chance) and the ball is yellow. What is the probability of the ball originating from Box 1?

To solve this, I used conditional probability which I understand is wrong and the correct answer would be to use Bayes theorem, which I didn't even know about. However because the chance is $50-50$ for either box the results were very similar.

My question is, what is the difference between the two formulas?

• Apr 27, 2018 at 12:02

First I explain why there is no real difference between both, then I tackle your example.

There is no fundamental difference between conditional probability and Bayes' theorem. Actually, Bayes' theorem is a simple rewrite of the definition of conditional probability.

Let $A$ and $B$ be two events.

The conditional probability of event $A$ knowing $B$ occurred is:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$

Basically this formula restricts the sample space to what happens within event B, and normalize the new probability measure by dividing by $P(B)$ to ensure the sum is $1$ as it should be.

If you manipulate this expression, you find:

$$P(A \cap B) = P(A\mid B)\,P(B)$$

Of course, by symmetry, replace $A$ by $B$ and you get:

$$P(B \cap A) = P(B\mid A)\,P(A)$$

Now, Bayes' theorem simply merges these two expressions in a different form. In a way, it's not really a "theorem" because its proof is trivial. But its implication for statistics are so wide that it's called "theorem"

Bayes theorem:

$$P(A \mid B) = \frac{P(B \cap A)}{P(B)} = \frac{P(B \mid A)\,P(A)}{P(B)}$$

Asking the difference between Bayes' theorem and conditional probability is like asking the difference between these two equations: $$x = \frac{a}{b} \quad \text{ and } \quad b\times x = a$$

Hope this helps

Edit:

Let $B_1$ the event "the ball is drawn from box 1" and $B_2$ the event "the ball is drawn from box 2". We know that $P(B_1) = P(B_2) = 0.5$.

In box 1, there are 5 yellow and 4 blue balls so if $Y$ is the event "we draw a yellow ball", then $P(Y \mid B_1) = 5 / (4+5) = 5/9$.

Likewise for box 2, $P(Y \mid B_2) = 7 / (7 + 5) = 7 / 12$

With this setup, we have:

$$P(Y) = P(Y \mid B_1)\,P(B_1) + P(Y \mid B_2)\,P(B_2)$$ $$P(Y) = 5/9 * 0.5 + 7/12 * 0.5 \approx 0.56944444$$

This formula I used is called "law of total probability". It can be used when two (or more) events partition the sample space $\Omega$. That is: $B_1 \cap B_2 = \emptyset$ and $B_1 \cup B_2 = \Omega$.

Now, you can compute $P(B_1 \mid Y)$ in a similar fashion:

$$P(B_1 \mid Y) = \frac{P(B_1 \cap Y)}{P(Y)} = \frac{P(Y \mid B_1)\,P(B_1)}{P(Y)} = \frac{5/9 * 0.5}{5/9 * 0.5 + 7/12 * 0.5} = 0.4878...$$

• +1, if only for the recapitulation at the end.
– Did
Apr 27, 2018 at 10:07
• @Did You wrote "+1", but didn't upvote. What does your comment mean then? Apr 27, 2018 at 10:12
• Thank you for this answer. In the case of my example, the solutions are not exactly the same when using conditional probability or bayes theorem. For C.P. I get 0.486 but for Bayes I get 0.49 Apr 27, 2018 at 10:18
• Done now.  
– Did
Apr 27, 2018 at 10:49
• @System, the answer should be the same whatever means you use if your calculation is right. I find 0.4878... as shown in my edit. Maybe you did a small mistake or your calculator is too approximate? In particular, you should use the exact value of $P(Y)$ to compute the last fraction, not the approximate value. Otherwise, you accumulate approximations. In any case, remember to mark this answer as accepted if it answers your question. Best Apr 27, 2018 at 11:58