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Defining Independent Events

Date: 07/14/2009 at 18:03:55
From: Amy
Subject: Independent events

My book defines independence in the following way...

Two events E and F are independent if P(E|F) = P(E).  My question is, 
do we also have to have P(F|E) = P(F)?  Or does this automatically 
follow?  Is there any situation where we could have P(E|F) = P(E) but 
P(F|E) =/= P(F)?

It just seems like when checking for independence, they usually just 
check the one.  For example, T: a person is tall and R: a person has 
red hair.  They say since P(T|R) = P(T) the two events are 
independent.  How come they don't consider P(R|T)?  

I am thinking that if you know P(T|R) = P(T), then you also know that 
P(R|T) = P(R) but I'm not sure why this is.  Also, it seems there are 
some instances where you cannot even consider one way over the other 
because it doesn't make sense.  For example, if E = drawing a club on 
the first draw and F = drawing a club on the second draw (without 
replacement) then P(E|F) doesn't even make sense (we cannot calculate 

Date: 07/14/2009 at 23:50:35
From: Doctor Peterson
Subject: Re: Independent events

Hi, Amy.

A more symmetrical definition is

  Two events A and B are independent if P(A and B) = P(A) P(B).

Since we define P(A|B) as P(A and B) / P(B), this is equivalent to
your text's definition, AS LONG AS P(B) IS NOT ZERO!  That is, we can
divide my equation by P(B) to get

  P(A and B) / P(B) = P(A)

which says that

  P(A | B) = P(A)

Or, we could divide by P(A), if it is not zero, to get

  P(B | A) = P(B)

So, again, as long as P(A) and P(B) are non-zero, all these statements
are equivalent, and if one is true the others must be.

What if one of the events can't happen?  That's where you might have 
to watch out.

What is the probability that I toss heads on this coin, given that I
roll a 7 on that die?  Since we'd be dividing by zero, it's undefined
(which sounds like the right answer!).  Are the events independent?
Since tossing heads has a defined probability, it sounds by your
definition as if they must not be independent.

What is the probability that I roll a 7 on the die, given that I toss
heads?  It's 0; and this is equal to the probability of rolling a 7 in
the first place, so they ARE independent.  So this is a case where the
two directions give different results.  (But it's pretty special, 
isn't it?)

If we take my definition, P(heads) = 1/2, P(7) = 0, and P(heads and 7)
= 0, and it is true that

  1/2 * 0 = 0

So the events really are independent.  Clearly this is a better
definition to use.  Your definition is fine as long as you don't use 
it when F has zero probability.

As for your comment that it doesn't make sense to consider P(club on
first draw | club on second draw); that's not really true.  You could
still find P(club on first draw and club on second draw)/P(club on
second draw); or you could enumerate all the outcomes and divide,
following the more basic definition of conditional probability.  It's
just not in agreement with the common sense notion that "given"
implies a time sequence.  It doesn't in probability!

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 07/15/2009 at 11:50:30
From: Amy
Subject: Independent events

Thank you for your quick response--it made a lot of sense to me.  I 
do have one other question (I struggle with independence and 
conditional probabilities).  Consider that we are drawing 2 cards 
(without replacement) and that event A = drawing a jack on the first 
card and event B = drawing a jack on the second card.  I can see 
that A and B are dependent because one outcome affects the 
probability of the other, but how would you show this 
mathematically?  I know you could show P(A and B) =/= P(A)P(B), but 
I am having trouble finding P(B) in this case...or could you show P
(A|B) =/= P(A) or P(B|A) =/= P(B)?  Either way I am having trouble 
computing P(B).  Thank you in advance for your help, it is really 
clearing up these ideas for me.

Date: 07/15/2009 at 20:26:08
From: Doctor Peterson
Subject: Re: Independent events

Hi, Amy.

One approach is just to recognize that it really doesn't make any 
difference in which order the cards are chosen, because we'd have the 
same set of outcomes either way, so P(B) is just the probability of 
choosing a jack, period. 

This doesn't entirely convince ME, so let's try revisualizing it.  We 
draw two cards and put the first on a spot on the table labeled A, 
and the second on a spot labeled B.  Any result I get could just as 
well have been obtained by first putting a card on B and then putting 
a second card on A!  So the probability that card B is a jack is 
1/13, regardless of whether card B was drawn first or second.

If that's still not convincing, or you want a very direct way to 
calculate P(B), you can just make a table listing all ways you can 
choose two cards without replacement.  The table might look like 
this, if we start with the ace of spades, two of spades, etc.:

     AS 2S 3S 4S ...
  AS  -  x  x  x
  2S  x  -  x  x
  3S  x  x  -  x
  4S  x  x  x  -

That is, we'd have 52 rows (the first card) and 52 columns (the 
second card), and any of the 52*52 spaces EXCEPT the diagonal (on 
which the two cards would be the same) would be equally likely.  That 
means each "x" in my table represents a probability of 1/(52*51).

Now, P(A) is the probability that the row is a jack.  There are 4 
such rows, each of which has 51 "x"s in it, so the probability is 
(4*51)/(52*51) = 1/13. But P(B) is the probability that the COLUMN is 
a jack--and all the reasoning is exactly the same!  This is what I 
mean by saying that it makes no difference which is A and which is B: 
The whole table is symmetrical.

By the way, a table like this is a good way to visualize conditional 
probability, too.  To make this clearer, let's make a smaller 
problem.  We'll have a deck that consists only of the four aces; and 
our A and B will be "the first card is the ace of spades" and "the 
second card is the ace of spades".  Here's our table:

     AS AC AH AD
  AS  -  x  x  x
  AC  x  -  x  x
  AH  x  x  -  x
  AD  x  x  x  -

I'll mark which outcomes belong to A, and to B (none can be both!):

     AS AC AH AD
  AS  -  A  A  A
  AC  B  -  x  x
  AH  B  x  -  x
  AD  B  x  x  -

Now, P(A) is the number of A's over the number of letters: 3/12 = 
1/4.  Likewise, P(B) is the number of B's over the number of letters, 
also 1/4.  And P(A and B) is 0, since no outcomes are in both A and 
B.  Since this is not P(A)*P(B), they are not independent.

P(B|A) means we restrict ourselves to only the row where the first 
card is AS (that is, this is given, or assumed):

     AS AC AH AD
  AS  -  A  A  A

Given this, there are NO ways the second card can be AS, so P(B|A) = 
0/3 = 0.

Similarly, P(A|B) means we consider only the column where the second 
card is AS:

  AS  -
  AC  B
  AH  B
  AD  B

Here again, there are 3 outcomes in all, but none of them are in A 
(that is, have AS as the first card), so the probability is 0.

Thus, P(B|A) = 0 while P(B) = 1/4, and P(A|B) = 0 while P(A) = 1/4, 
so on both counts A and B are not independent.

- Doctor Peterson, The Math Forum 
Associated Topics:
College Probability
High School Probability

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