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.