Scenario-3:
Suppose that you have with you a standard deck of
cards, and you shuffle it really well. You now draw a card at random from this deck. It is obvious that for a well-shuffled deck, you are equally likely to draw any of the
cards. For example, if
is the event that the card drawn is the king of Hearts while
is the event that the card drawn is the seven of Diamonds, we have
Since there are
cards and each is equally likely to turn up, and also, one of the
cards must turn up, the numerical value of the probability of any card being drawn is therefore
. For example,
Now, let
be the event that a spades is drawn. Since
can occur in
ways (there are
cards in any suit), or in other words, the number of cases favorable to
is
out of the total
possibilities, we must have
This value of
should have been obvious directly, since there are only
suits in the deck, and any suit is equally likely to turn up.
Going further, let
be the event that a black card is drawn. We have (
represents ‘Number of’),
We see that
, since there are
cases favorable to
. Again, the value of
should have been obvious directly, since there are only
colors in the deck (red, black), and each color is equally likely to turn up.
From this discussion, you should by now have understood the gist of the concept of probability. We can, taking cue from the preceeding examples, define the probability of an event as the number of outcomes favorable to that event, divided by the total number of possible outcomes (assuming each outcome to be equally likely).
Extending this further, the general idea is that if we have two events
and
with no outcomes in common, the first consisting of
outcomes and the second of
outcomes, then the event
which consists of all the outcomes in
and
consists of
outcomes, i.e,
An example of this is what we just saw:
Thus, if the total number of outcomes is
,
we have,
That the probabilities add when we consider the event
consisting of all outcomes in
and
is justified only if
and
have no outcomes in common, as we’ve been seeing till now.
What about the probability of an event consisting of all outcomes in two events
and
which may have some outcomes in common? For example, let
and
be events defined as follows:
The number of outcomes in
is
(why ? ). The number of outcomes in
is
. However, if we now consider the event
which consists of all the outcomes in
and
, we find that the relation
is not satisfied. Why? Because the events
and
have some outcomes in common. Let us understand this visually.
We let the large rectangle below represent all the
possible outcomes. The events
and
are then subsets of this large rectangle as shown.
Note that
and
have some elements in common, which is technically stated by saying that
and
are not mutually exclusive. (Thus, mutually exclusive events have no outcomes in common).
As stated earlier, the event
is defined to consist of all the outcomes in
and
, which (in easier language!) means that
is the event that the card drawn is either even or black (or both). How many elements are there in
?
Note that when we write
the outcomes common to
and
(represented by the dark regions in Figure –
) are counted twice on the right side. The actual number of outcomes in
will thus be obtained by subtracting from the
the number of these common outcomes, and thus we can write
where
represents the number of common outcomes. In this case,
would represent the event that the card drawn is both even and black. There are obviously
such cards, as is evident from Figure –
, and thus,
You are urged to verify this result by explicitly counting the number of cards lying within the (total) shaded regions in Figure –
.
In passing, it should be mentioned that the event
which has been defined as the event consisting of all the outcomes in
and
is technically termed as the union of the events
and
, and this fact is symbolically written as
Union of events |
You can think of the union of two events to be the event which can be said to occur when either of the two given events (or both) occurs.
Analogously, the event
defined to consist of all outcomes common to
and
is technically termed as the intersection of the events
and
, and this fact is symbolically written as
Intersection of events |
You can think of the intersection of two events to be the event which can be said to occur when both the two given events occur simultaneously. Finally, as we’ve seen above, we always have
If
and
have no outcomes in common, that is, if
and
are mutually exclusive:
this relation reduces to
Here’s a visual depiction to help you get things straight:
The relations
and
, when written in terms of probability (by dividing both sides with the total number of outcomes), become,
General relation | ||
The particular case of being mutually exclusive |
If the concept of mutually exclusive events is now clear to you, the significance of the relation
on Page –
immediately becomes apparent, where we simply summed the six probabilities and wrote the sum as
. All the six outcomes are mutually exclusive, and they exhaust all possibilities, so
is justified.
Now is a good point to learn some more terminology. In Figure –
above, each of the six possible outcomes is the most elementary outcome possible. For example, if
is the event of a
showing up, the event
consists of only one outcome, the number
.
would therefore be called an elementary event. Now consider the event
of an odd number showing up.
consists of three outcomes, namely
,
and
.
would therefore be termed a compound event.
As a further example, in the drawing of a card at random from a standard deck of
cards, let
and
be events defined as follows:
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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