Why the probability of two independent events happening is the multiplication of the probability of each?

Question

For example, p(getting two heads from tossing a coin twice) = 0.5 * 0.5...

I passed my probability course in college, but I am still having trouble getting the intuition for this.

Answer 1

Suppose we have a building where an exam is to be given. The building has two floors, and on each floor there is an exam room on the north side and another exam room on the south side.

Now among several thousand students who have to take the exam, we assign half of them to go to the first floor of the building, the other half to the second floor. Then of all the students who were assigned to the second floor, we assign half of them to go to the room on the north side.

We will end up sending one quarter of all the students to the north room on the second floor. (These are very large exam rooms!)

Now suppose as each student enters we flip a coin to see whether they should go to the first or second floor (first if tails, second if heads), and once a student arrives on a floor someone flips another coin to decide whether the student should go to the north room or the south room (south if tails, north if heads).

It is unlikely that we will end up with exactly the same number of students in each room, or even that the number of students in one room will be just one more or less than the number in another room, but the expected number of students in the north room on the second floor (the ones for whom we flipped heads twice) is still one quarter of all the students, because it is half of one half of the students: $$ \frac12 \times \frac12 = \frac14. $$

The notion here is that the assignment to first or second floor and the assignment to north or south room are independent events, because regardless of which floor you go to, we choose north or south the same way. If the coin on the first floor were a fair coin but the coin on the second floor was a trick coin with two heads, then the assignments would not be independent and the probability to go to the north room on the second floor would not just be the product of the probability to go to the second floor (which is still $\frac12$) and the probability to go north (which is $\frac34$, since $\frac14$ of all students end up in the north on the first floor and $\frac12$ end up in the north on the second floor).

Answer 2

Assuming you have a fair coin. You toss it once. Let us denote this event of tossing head as $A$. Then $P(A)=0.5$ After you have tossed head once you toss head again. This is a conditional event. The probability for is $P(B|A)$. In written words:

"The probability that tossing a head in the second trial given that the first trial is head".

Indenpendence of two events

The probabiltity of tossing in the second trial is $P(B)=0.5$. The probability of tossing a head in the second trial does not depend (independence) on the outcome of the first trial. Thus $P(B|A)=P(B)=0.5$

Now you can calculate the probability that you to toss head in the first trial and in the second trial. It is denoted as $P(A\cap B)$. $A\cap B$ is the union of event $A$ and event $B$.

Using Bayes theorem we get

$P(A\cap B)=P(A)\cdot P(B|A)=P(A)\cdot P(B)=0.5\cdot 0.5=0.25=\frac14$