Conditional probability: Formula explanation I am reading on conditional probability and am trying to wrap my head around the formula:
P(A and B) = P(A) x P(B|A). Example, there are 5 marbles in a bag; 2 blue and 3 red. The probability of drawing a blue marble from the bag is 2/5. Assuming, I got a blue marble in the first draw, my probability of drawing another blue marble is 1/4. Now, my probability of drawing 2 successive blue marbles is denoted by the formula above, which is 2/5*1/4 = 1/10. I don't quite understand why are the 2 chances multiplied to derive the probability and what does 1/10 mean in this context. Appreciate your time.
Regards,
Aj
Source: https://www.mathsisfun.com/data/probability-events-conditional.html
 A: If you really want to make sense of the number  $\frac{1}{10}$, you should think about $\frac{2}{20}$ and then think of $20$ as the cardinality of the sample space and $2$ as the cardinality of favourable events.
This will also give you a clue about why the two probabilities are multiplied.
A: Imagine doing this experiment $1000$ times.  You would expect that on about $\frac25$ of the trials you would get a blue on draw 1.  That is about 400 times you would get a blue on draw 1.
Of these $400$ times, in about $\frac14$ of them, you would expect to also get a blue on draw 2.  That is, about $100$ times you would also get a blue for the second draw.
Thus in about $100$ of the $1000$ trials you would draw blue-blue, giving you your probability of $\frac1{10}$.
Shorter approach $\frac14$ of $\frac25$ of the time you should get blue-blue.  And $\frac14$ of $\frac25$ is $\frac14 \cdot \frac 25=\frac1{10}$.
A: To help you understand, I will discuss why we multiply two independent events, $A$ and $B$.
Independent events mean that the probability that one event occurs does not affect the probability that the other event occurs. So, if $A$ and $B$ are independent then $P(A)=P(A|B)$ and $P(B)=P(B|A)$. 
Now, the probably that two independent events occurs is given by $P(A \cap B) = P(A) P(B)$. Why is that you ask? To understand, you need to consider the sample space. For the sake of simplicity, lets consider your problem in the case that each marble is replaced after drawing it from the bag. This makes the events independent. 
Now, imagine that we write a number on each of the marbles. We are only doing this, so we can keep track of each marble. So, marble 1 is blue, marble 2 is blue, marble 3 is red, marble 4 is red, and marble 5 is red. Just as before, we have 2 blue marbles and 3 red marbles but they each have a unique label. 
How many possible ways can we draw two marbles from the bag with replacement? 
Well, we can draw marble 1 and then marble 2, we could also draw marble 2 then marble 1, etc... considering all possibilities there is 25 different ways (11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55). This is the sample space.
Now, what is the probability of drawing two blue marbles with replacement? 
The scenarios that this occurs is any of the scenarios that contain both marble 1 and marble 2, so there are 4 possible cases in our sample space where this can happen (11, 12, 21, 22), so we know the probability is $\frac{4}{25}$, which is $\frac{events \; of \; interest}{all \; possible \; events}$. You could also use our equation from before $$P(A \cap B) = P(A)P(B) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$$
Writing out the sample space for each problem is extremely time-consuming. Instead of doing this for every problem, we can use this logic to derive formulas. This logic is identical for conditional probabilities except the sample space is not as easy to write out.
