# Conditional Probability - Marbles Example

I have two bowls, Bowl A and Bowl B. Bowl A has 1 blue marble, and 4 yellow marbles. Bowl B has 3 blue and 2 yellow marbles.

You randomly chose a bowl, and randomly pick a marble. This marble is blue. What is the probability you picked from bowl A, given that you have picked a blue marble?

Initially I used the conditional probability formula as follows:

$$P(Bowl \, A| Picking\,Blue\, Marble) = \frac{P(Bowl\,A \cap Picking\,Blue\, Marble)} {P(Picking\,Blue\, Marble)} = \frac{\frac{1}{5}}{\frac{4}{10}} = \frac{1}{2}$$

My reasoning was that picking Bowl A and Picking a Blue marble is 1/2*4/10 = 1/5. And the probability of picking a blue marble from either bowl is 1/4.

However, the answer is a bit more intuitive. Without the formula it is simply 1/4. What have I done wrong?

$$P(\text{Bowl A} \cap \text{Picking blue marble})=P(\text{Bowl A})\cdot P(\text{Picking blue marble|Bowl A})$$
$$P(\text{Picking blue marble|Bowl A})$$: Probability of picking a blue marble, given it is drawn from Bowl A.
Your first problem is that you assume the probability of picking from bowl A and the probability that the marble is blue are independent, when this is not the case. So, your assumption that $P(A \cap Blue)=P(A)P(Blue)$ is not correct.
Second, you are not applying Bayes' theorem correctly, what you want is $P(A|B)P(B)=P(B|A)P(A)$, in your case, this simplifies to $P(A|Blue)=\frac{P(Blue|A)P(A)}{P(Blue)}$ Applying this, you should readily find your answer of $\frac{1}{4}$