# Why doesn't the general formula for conditional probability produce the same answer as the tree diagram in this case?

I am solving the following question:

There are two urns, the first urn contains 2 black balls and 3 white balls. The second urn contains 4 black balls and 3 white balls. An urn is chosen at random, and a ball is chosen at random from that urn.

So given that a black ball that was chosen the probability that it is from the first urn can be solved using the formula for conditional probability like this: $$P(first urn| black ball) = \frac{P(first urnAND blackball)}{P(blackball)}$$

Similarly if I am asked for the probability that given the first urn is chosen, a black ball is chosen, I would do the following

$$P(black ball| first urn) = \frac{P(first urnAND blackball)}{P(firsturn)}$$

But, this incorrectly evaluates to

$$P(black ball| first urn) = \frac{P(first urnAND blackball)}{P(firsturn)} = \frac{2/5}{1/2} = \frac{4}{5}$$

whereas by the tree diagram below it is $\frac{2}{5}$ (I believe by the Multiplication Rule for Independent Events $P(AB) = P(A)P(B)$)

Tree diagram of two urns

Why are the events "first urn" and "black ball" treated as independent events in one case but not in the other?

• This probability of the desired event cannot directly be determined by the probabilities in the tree diagram. Note that $P(B|A)\ne P(A|B)$ in general. – Peter Oct 4 '17 at 20:35
• The tree diagram contains implicitely the probability of a black ball given that the first urn was chosen. So, we only need to multiply the probabilities (for one path) , but nevertheless the events are not independent. To get the probability that a black ball is drawn, you must add the probabilities of two paths (and in each part you multiply the probabilities) – Peter Oct 4 '17 at 20:39
• @Peter I understand how the tree diagram is used, but if I did not make a tree diagram and I am just looking at the two probabilities, what is to say that $P(firsturn | blackball)$ can be solved by the formula and $P(blackball| firsturn)$ cannot? – EvaD Oct 4 '17 at 20:45
• Vice versa, the probability $P("blabkball"|"firstturn")$ appears directly in the diagram : If we have chosen the first urn, one path leads to a black ball and the probability appearing there is the conditional probability of $P("blackball"|"firsturn")$ , the other conditional probability has to be calculated. – Peter Oct 4 '17 at 20:50
• Ignoring the tree diagram : The probability to chose the first urn is trivial, but the probability that a black ball is drawn must be calculated using the law of total probability. This is the reason that one of the two conditional probabilities can easily be determined, the other more complicated. – Peter Oct 4 '17 at 20:55

The probability to choose the first urn and a black ball is $$\frac{1}{2}\cdot \frac{2}{5}=\frac{1}{5}$$
The probability to choose the second urn and a black ball is $$\frac{1}{2}\cdot \frac{4}{7}=\frac{2}{7}$$
Since not both events can occur, we can sum up the probabilities, so we draw a black ball with probability $\frac{1}{5}+\frac{2}{7}=\frac{17}{35}$
So, the probability that the first urn was choosen given that we draw a black ball is $$\frac{\frac{1}{5}}{\frac{17}{35}}=\frac{7}{17}$$
The probability to draw a black ball given that the first urn was chosen is $$\frac{\frac{1}{5}}{\frac{1}{2}}=\frac{2}{5}$$