Our next rule for probability concerns the probabilities of disjunctions.
That is, the probability that either one thing or another will happen.
And as with the case of the probability of conjunctions, we have two cases.
You need a different rule when these two events are mutually exclusive than when
they're not mutually exclusive. We're going to look first at cases that
are mutually exclusive. And what that means is simply that they
cannot occur together. The probability of them both occurring is
zero. It can't happen.
Now, the formula for that simpler case says that if two events are mutually
exclusive, then the probability of either one event or the other event occurring is
simply the sum of the probability of the first event occurring and the probability
of the second event occurring. And we can symbolize it using the same
symbols we've used before as the probability of hypothesis one or
hypothesis two is equal to the probability of hypothesis one plus the
probability of hypothesis two. Simple enough.
Let's look at an example. What's the probability that I will pick
either an ace of diamonds or an ace of hearts out of this deck?
Well, the probability that I pick an ace of diamonds is one in four.
The probability that I'll pick an ace of hearts is one in four.
So, the probability that I'll pick either one or the other is one in four plus one
in four, which is two in four, which is 0.5.
Two times out of four, 0.5. So, this formula makes perfect sense when
it's exclusive, when you cannot pick both an ace of
diamonds and an ace of hearts on a single pick.
Here's another example. It's a special case.
We've been talking so far about permutations which are ordered sets so
the two events in a conjunction would flipping heads first and then, tails
second. We can also look at combinations where
order doesn't matter. Flipping one heads and one tails out of
two flips but it doesn't matter what the order is.
Then, we have heads and tails in two flips, which means either a heads and
then a tails or a tails and then a heads. So, we've got a disjunction and we can
figure out that the probability of this combination, which is unordered, by
looking at the disjunction of those two possibilities.
Here's a chart showing the possibilities. If you take heads on flip one, you are in
the left-hand column. Tails on the flip one, you are in the
right-hand column. Head on flip two, you are in the top row.
Tails on flip two, you are in the bottom row.
Which of these, for the two flips, includes a heads and a tails where the
order doesn't matter? And the answer is, the upper right has
tails on flip one and heads on flip two. And the lower left has tails on flip two
and heads on flip one. So, it's those two possibilities, this
one and this one, out of the four that include the combination, one heads and
one tails in any other order. And the disjunction is calculated by
saying, what's the probability of this, 0.25, one in four.
And the probability of this, 0.25, one in four and you add them together.
Because the probability of doing either one or the other is going to be the sum,
just like the formula says. Next, we have to look at the general
formula that applies whether or not the two events are mutually exclusive.
And that's pretty simple. We're going to call it 3G for the
generalized version of rule three. What it says is that the probability of
either one event occurring or the other event occurring is the sum of the
probability of the first event occurring plus the probability of the second event
occurring minus the probability that they both occur.
Or in symbols, the probability of h1 or h2 is equal to the probability of h1 plus
the probability of h2 minus the probability of the conjunction of h1 and
h2. So, we're going to have to use that previous rule for conjunction to figure
out what gets subtracted. The more basic question is why are we
subtracting at the first place? And you can think about that in this
example. What's the probability of getting heads
on either the first flip or the second flip in two flips of a fare coin? Well,
one way to figure this out might be to say, well, what's the probability of
getting on the first flip? Well, that's 0.5.
What's the probability of getting on the second flip?
Well, that's 0.5. So, let's add them together and the
probability of getting it on either the first flip or the second flip is.5=1.
0.5 + 0.5 = 1. So, it's absolutely certain that were
going to get a heads on either the first flip or the second flip of two flips of a
fair coin. But you know that's not right because
sometimes you get two tails in a row. So, that can't be the right way to
calculate the probability. That's why you need to subtract the
conjunction because otherwise, you're going to get the wrong answer.
The right way then is 0.5 heads on the first flip, 0.5 heads on the second flip
minus the probability of both. But the question is, why do you have to
subtract? And this little chart should show you why
you have to subtract. Because the probability of getting heads
on the first flip is the left-hand column,
that's the upper one. You get heads on the first flip.
The lower one, you get heads on the first flip.
The probability of getting heads on the second flip is given by the top row.
The first one, you get heads on the second flip.
The second one, we get the heads on the second flip.
So, if you want to add them together, you are adding together this one and you're
adding it to this one, that gives you 0.5 of getting heads on the first flip. Then,
you add it to the probability of this one plus this one and that gives you 0.5 of
getting heads on the second flip. But then, if you add them together to get
the probability of getting heads on either the first flip or the second flip,
you've double counted the one in the upper left in red.
You counted it in the probability of the first flip and also, in the probability
of the second flip. And you don't want to double count it,
because then you'll get the wrong answer. So, I hope this chart illustrates why in
order to generalize the formula for disjunction, we need to subtract the
probability that both the events will occur.
Here's another example. Out of a standard deck of cards, what's the probability of
picking either a six or a club? Well,
we know that there're 4 sixes and 13 different types of cards, so the
probability of picking a six is one in 13.
What's the probability of picking a club? Well, there are 13 clubs and 4 suits, so
the probability of picking a club is one in four.
And if we add those probabilities together, we'll get 4 + 13 makes 17 out
of 52 cards, but that's not right because you've just double-counted the six of
clubs. You counted it as a club and you also
counted it as a six. So, in order to get the probability of
picking either a six or a club, you need to add the probability of getting a six,
one in thirteen, to the probability of getting a club, one in four, and subtract
the probability of them both happening namely, getting the six of clubs, one in
52, because otherwise, you'll be double-counting that six of clubs.
So now, I hope you understand the rationale behind this generalized version
of the rule for probabilities and disjunctions.
And I hope you are able to do a few exercises using that rule.