In the experiment of drawing one card at random from a standard well-shuffled deck of
cards, consider the event
defined as:
Obviously, we have
However, if you are now asked to find the probability of event
occurring given that the card drawn has a number greater than
, what would you say?
To answer correctly, you must understand that now the situation is entirely different from earlier. Now we already know that the card drawn has a number greater than
, which means that the number of possibilities for the card has reduced; earlier, there were
possibilities, but now the number of possible cards are the
Jacks, the
Queens and the
Kings, which means that now there are only
possibilities for the card. Of these, the favorable possibilities are
; thus,
Let us denote by
the event that the card drawn has a number greater than
. The probability just calculated above is then written in standard notation as
which is read as
i.e. “the probability of event
occurring given that
has occurred”.
Let us consider another example. In the random experiment of throwing two dice, let events
and
be defined as
First, let us find the probability of
occurring. This is simply
Now, suppose we are given that
has already occurred, meaning we now know that one of the numbers on top is
. What is the probability of
now when we already possess this information?
The number of total possibilities now are
; we list them explicitly:
Out of these, the favorable possibilities are only
, namely:
and
. Thus
In both the preceding examples, what happens is that with the possession of some information, the number of possibilities reduce, i.e., the sample space reduces. In the first example, with the information that
has occurred, the number of possibilities reduces from
to
. We then looked for favorable cases only within this reduced sample space of
outcomes. Similarly, in the second example, when we possess the information that
has occurred, the number of possibilities reduces from
to
; it is in this reduced sample space of outcomes that we then looked for the favorable possibilities.
In passing we should mention that probabilities of the form
are called conditional probabilities.
is said to be the conditional probability of
given that
has occurred.
Let us understand all this somewhat more deeply. In particular, let us ponder over the following question: Let
and
be two events. Let the probability of
occurring be
. Now, if we come to know that
has occurred, will the probability of
occurring always increase, like it did in the last two examples, or can it decrease too? Can it remain the same?
It turns out that all the three are possible, which should even be obvious to an alert reader.
Let us consider the case where the information that
has occurred decreases the probability of
occurring. Let
and
be, in the card drawing experiment of the first example
The original probability of
occuring is
But when
has already occurred, the probability of
occurring is
which is lesser than the original probability of
occurring. This should be intuitively obvious: In the first case, there are
multiples of four per suit of
cards. In the second, when we are told that the number of the card is greater than eight, the multiple of four possible is only
per suit which is the queen. Thus the favorable possibilities decrease to one-third. The total possibilities also decrease i.e, from
to
per suit, but it can be easily appreciated that the percentage reduction in favorable possibilities is greater than the percentage reduction in total possibilities.
We now come to the third very important question: for two events
and
, can
be the same as
? You must appreciate that this is equivalent to saying that the probability of
occurring is not affected by the occurrence or non-occurrence of
.
As we said earlier, this is possible, and such events are called independent events:
If | |
Let us see an example. In the card-drawing experiment, let events
and
be defined as
The alert reader will immediately realize that
is the same as
. Why? Because, the knowledge that the card is black does not change the number of cards per suit that are greater than eight. Stated explicitly,
and | because there are only two black suits | |
There is one important point the reader must notice and appreciate:
If 
Let us verify this in the example above. We have,
and,
which confirms the assertion
Note that the favorable cases while calculating
are those cases in
that are common to
; the total cases are all cases in
. Thus
If the entire sample space of the experiment consists of
out comes, we can write
as
Similarly,
This means that if
and
are independent, then
From | ||
Thus, the probability of the intersection of two independent events is simply the product of the individual probabilities.
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