The rule for the probability of 
conjunctions is a little more 
complicated. 
One reason is that, there're really two 
cases we have to keep separate. 
We're going to look first at the case of 
independent events, and we have to begin 
by saying what it means to call them 
independent. 
To call them independent is simply to say 
that the probability of one does not 
affect the probability of the other. 
So, if you flip a coin or roll a die, and 
get a certain number, and then you pick 
up the coin or the die, and you flip it 
or roll it again, then the second result 
is not affected by the first result. 
So, those are independent events, 
and with cards, if you take a deck of 
cards, 
and you pick out a card, 
two of spades. Then you take the two of 
spades, and you put the pack in the deck, 
you shuffle, it's going to be independent 
of whether you get it again, because what 
you picked up the first tine doesn't 
affect what you pick up the second time. 
But, 
take a deck of cards and you pull out the 
king of spades, and you take that card, 
and you throw it away, and don't put it 
back in the deck, 
then it affects all the probabilities of 
the remaining cards, because now you have 
fewer cards than you had before. 
So that's the difference between 
independent events and dependent events. 
So we're going to look at independent 
events first, and look at the rule for 
the conjunction of independent events. 
And the rule for, for this case is pretty 
simple. 
If two events are independent, then the 
probability that both events will occur 
in that order, because we're talking 
about this one and then that one, we'll 
talk a little bit more about that later, 
but we're assuming that. The product that 
they both will occur is the product of 
the probability of the first event times 
the probability of the second event. 
And notice that the both occur is tied to 
the product, or multiplying the two 
probabilities together. 
And the reason for that is that it's less 
likely that they will both to occur, than 
that one of them will occur separately. 
And when you multiply the two 
probabilities between zero and one, 
you're going to get a lower probability 
that they both occur, than for either one 
of them occurring, you know, all by 
itself. 
We can restate this rule symbolically 
using the symbols that we used before for 
the probability of negation. 
Namely, the probability of hypothesis 1 
and hypothesis 2 both being true is the 
probability of hypothesis 1 times the 
probability of hypothesis 2. 
Now, let's apply it to some cases. 
Suppose you flip a coin, and you get 
heads? 
And you flip another coin, and you get 
heads again? 
What's the probability of getting heads 
twice at two flips of a coin? 
Well the probability is 5. for the first 
flip of getting heads. 
It's 5. for the second flip of getting 
heads and then the probability of getting 
both is going to be 5. times 5,. or 25.. 
So the probability of getting two heads 
in a row is 25. on two flips of a fair 
coin. 
A little more complicated example uses 
cards. 
Here we have a standard deck of cards. 
Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, 
Queen, King. 
So there are 13 different types of cards, 
and there are 4 different suits, spades, 
hearts, diamonds and clubs. 
So we can remember that. 
The probability of having one of these 
types of cards chosen, assuming that 
they're shuffled properly, is one out of 
thirteen, and the probability of getting 
a particular suit, say a club, is one out 
of four, because there are 13 clubs out 
of 52 cards. 
Clubs are one suit out of four, so the 
probability of picking a club is one in 
four. 
and the probability of picking a 
particular card, say a 3, is one in 
thirteen. 
Again, a deck of cards. 
Suppose I pull out a 4, and then I pull 
out a club. 
Now in a standard deck of cards, there's 
thirteen cards, we're assuming no jokers. 
So the odds of getting a four are one in 
thirteen. 
There're four suits, spades, hearts, 
diamonds and clubs. 
So the odds of picking a card that's a 
club is one in four. 
What's the probability of picking a four, 
and then picking a club? 
Well it's going to be 1 in 13 times 1 in 
4 And that means, this can be 1 in 52. 
So the odds of that combination 
occurring, right, in that order, is going 
to be, 1 over 13 times 1 over 4 equals 1 
in 52. 
That example shows you how to calculate 
the probability of a conjunction with 
independent events, but what if the 
events are not independent? That is, what 
if the probability of one effects the 
probability of the other? Here is a 
question. 
Joe can run a mile in less than five 
minutes, about half the time. 
When he runs and he's timed, half of the 
time he's under five minutes, half of the 
time he's over five minutes. 
So what's the probability he can run a 
mile in less than five minutes on a given 
occasion? .5. 
But, what's the probability he can run 
one mile in under five minutes, and then 
run another mile in under five minutes 
right after that? 
Well, if you did this calculation you 
will have 5. times 5.. 
And that means 25,. but you know that's 
not right, because he's going to be dead 
tired after the first mile. 
There's no way that the probability of 
running the second mile in under five 
minutes is just as high, 
whether or not he would have just 
finished running a mile. 
So those events are not independent, 
and that example shows you why you need a 
new formula to calculate the probability 
of a conjunction, 
when events are not independent, 
but instead their probabilities depend on 
each other. 
The same distinction between independence 
and dependence applies to our old friend, 
cards. 
So just to simplify matters let's start 
with a very limited deck. 
It's just got four aces It's got the ace 
of spades, 
the ace of hearts, 
the ace of diamonds, and the ace of 
clubs. 
And notice that the ace of hearts and the 
ace of diamonds are red, and the ace of 
spades and the ace of clubs are black. 
So, knowing that, out of these cards, if 
we shuffle them, and don't look at them, 
what's the probability of picking a red 
ace? 
Well, 1 in 2, right, because 2 of them 
are red out of 4. Now. 
Sorry, Oh, 
a red one, I've picked the red one. 
Now, 
let's put it back in the deck and shuffle 
them together, 
okay. 
What's the probability now of picking 
another red Ace? 
Again 1 in 2. 
So, if you look at the possibilities all 
laid down, the first pick is in the 
columns, right? 
So the leftmost column is if you pick an 
ace of spades on your first pick, 
second column is ace of hearts on your 
first pick. 
Third column is ace of diamonds on your 
first pick, 
fourth column is ace of clubs on your 
first pick. 
And then your second pick is this rows, 
and again it's either spades, hearts, 
diamonds or clubs. 
So there are 16 possible combinations of 
picks here, in order, and how many of 
them are picks where you have two red 
aces? 
Well, this is one, this is one, this is 
one, and this is one. 
So, the 4 possibilities in the middle are 
the possibilities where you've picked an 
ace that's red, both on your first pick 
and also on your second pick. So the odds 
of doing that are 4 out of 16 or one out 
of four, or 25,. just like we calculated 
using the formula. 
But, what if there's no independence? 
Let's start with the same four aces. 
Okay, 
and what's the probability of picking a 
red ace? 
Still one in two. 
So let's suppose I pick the ace of 
hearts, 
throw it away, 
do not put it back in the deck. 
And now I've only got three aces left. 
What's the probability of picking a 
second red ace without looking? 
Well it's going to be one in three, 
because one of them is red and the other 
two are not. 
There are three cards left, 
one of them is a red ace. 
We can also look at it this way. 
There are sixteen possibilities. 
How many of these include a red ace on 
the first pick and also a red ace in the 
second pick? 
Well, those are the four in the middle, 
but if we got rid of the ace of hearts by 
throwing it out and not putting it back 
in the deck, then, we've left out this 
whole row. 
And we know that we are in this column 
because we picked an ace of hearts the 
first time. 
And only one out of the three is picking 
a red ace the second time, so we know 
that there's one out of three. 
So when it's dependent like this, because 
you don't put the card back, you need a 
different formula. 
The probability is one out of two times 
one out of three equals one out of six, 
instead of the one out of four that we 
had when they were independent and you 
put the card back. 
So we need a different formula to 
generalize for this case where there is 
dependence, okay. 
And this formula's going to use the 
conditional probability, which is just a 
fancy way of saying what I've just 
mentioned to you. 
Right? 
The conditional probability of picking a 
red ace on the second pick given that I 
pick a red ace on the first pick and 
threw it away, is one in three. 
It's how many times does x occur picking 
a red ace on the second pick out of the 
cases where I pick a red ace on the first 
pick and threw it away and that's going 
to be your one in three. 
So that's the conditional probability, 
and now the formula. 
The probability of both of two of events 
occurring is the product of the 
probability of the first event occurring 
times the conditional probability of the 
second event occurring given that the 
first event occurred. 
That's where you use the conditional 
probability. 
Now notice, when the events are 
independent by conditional probability of 
the second event occurring, given that 
the first event occurred, is just 
going to be the same as the probability 
of the second event occurring, 
because the first event doesn't affect 
the second event. 
So that first formula for an independence 
is just an instance of this formula, 
where the conditional probability of the 
second, given the first, is identical 
with the probability of the second. 
So that's why it's a generalized formula 
and actually includes the first formula, 
but when they're independent it's just 
simpler to think of the first formula 
alone. 
We could also symbolize this rule. 
We can say that the probability of 
hypothesis 1 and hypothesis 2 is equal to 
the probability of hypothesis 1 times the 
probability of hypothesis 2, given 
hypothesis 1, and that straight up and 
down line in this formula is what 
indicates conditional probability. 
This little formula here means the 
probability of hypothesis 2 given 
hypothesis 1. 
So you can use this rule to calculate the 
probability of any old conjunction, 
whether they're independent or not, 
but when they are independent, you might 
want to use that simple rule, which is 
the first version of probability of 
conjunction.