$P( \text{ Launch a project }|\text{ Stock increases })=0.75 * 0.6 = 0.45$, is wrong? Here is given an example of conditional probability. If you look at this part:

If they want to find the probabilities they have written out after the graph, they're supposed to do the following calculations:
Let events $L = \text{ Launch a Project }$, $S = \text{ Stock price increases }$
Then $P(L|S)=\frac{P(S|L)P(L)}{P(S)}$
$P(S|L)= 0.75$
$P(L)= 0.6$
$P(S)=P(S|L)P(L)+P(S|L^c)P(L^c)=0.75*0.6+0.4*0.3=0.45+0.12=0.57$
So $P(L|S) = \frac{0.75*0.6}{0.57}=0.45/0.57= 0.78$
So the way they write  $P(  \text{ Launch a project }|\text{ Stock increases })=0.75 * 0.6 = 0.45$, is wrong.
Am I correct?
 A: Also according to me it is wrong, even if I am not used to these tree representation (but sounds intuitive).
One can make also a calculation without Bayes:
$p(L|S)=\frac{p(L,S)}{p(S)}=\frac{0.75*0.6}{0.75*0.6+0.4*0.3}=\frac{0.45}{0.57}\sim 0.79$
that is what you get. The calculation they did considers only the numerator for some reason ?
A: The intended calculation should have been $$\Pr[S] = \Pr[L \cap S] + \Pr[\bar L \cap S], \tag{correct}$$ rather than what they actually wrote, $$\Pr[S] = \Pr[L \mid S] + \Pr[\bar L \mid S]. \tag{incorrect}$$  In turn, what was written should have been
$$\begin{align}
\Pr[L \cap S] &= \Pr[L] \Pr[S \mid L] = (0.6) (0.75) = 0.45, \\
\Pr[\bar L \cap S] &= \Pr[\bar L] \Pr[S \mid \bar L] = (0.4) (0.3) = 0.12, \end{align} \tag{correct}$$ rather than
$$\begin{align}
\Pr[L \mid S] &= \Pr[L] \Pr[S \mid L] = (0.6) (0.75) = 0.45, \\
\Pr[\bar L \mid S] &= \Pr[\bar L] \Pr[S \mid \bar L] = (0.4) (0.3) = 0.12. \end{align} \tag{incorrect}$$
The final answer, $\Pr[S] = 0.45 + 0.12 = 0.57$, is correct, but they made typographical errors, confusing joint probability $\cap$ with conditional probability $\mid$.
To calculate $\Pr[L \mid S]$ for instance, we would need to use Bayes' rule, as you noted:  $$\Pr[L \mid S] = \frac{\Pr[S \mid L]\Pr[L]}{\Pr[S]} = \frac{0.45}{0.57}.$$

The point here is that if you read the actual web page, it is clear that what they really meant to write was $\cap$ instead of $|$, and "Using the decision tree, we can calculate the following joint probabilities."  There is no contextual interest in the posterior probability of having launched given that the stock price increased--indeed, there is no reason to be interested in such a probability because the temporal order of these events is such that the question of whether the launch occurs always precedes the impact of this decision on the stock price.
A: Yes you are right. $L$ and $S$ are not independent events here, hence what they have done is incorrect. Bayes' Theorem is the correct procedure, as you've done.
A: Another way to see that your reasoning is absolutely correct is to express the data in a tabular way

As you can see, $P(L|S)=\frac{45}{57}\approx78.95\%$
Actually what they state as $P(L|S)=0.45$ is instead $P(S\cap L)$
