TREE DIAGRAMS
Tree Diagrams allows one to graphically represent complex probability situations involving conditional
probabilities. I will assume that you have read your text so this will allow me to quickly move onto the main
portion of this writing which is understanding how tree diagrams should be read.
Case I - The classic exampleNotice that this represents a series of events that start at the left and end at the
right. Thus, the first step is either event A occurs or it does not, Ac. Once that
has occurred event B follows. Event B occurs or does not occur. Notice that a
tree is laid out like a path, and you should think of it that way. Each fork on
the road represents another possible ending. The way I just presented this it
would seem like tree diagrams fit the situation of two events occurring, one
after another. But this not need be the case. It could represent one event and
what we are merely suggesting is that this one event incorporates events A, B,
A and B, neither or some combination of the two.
P(B|A)
B
P(BC|A)
Bc
A
P(A)
P(B|AC )
C
P(A )
B
c
A
c
P(BC|AC ) B
Suppose the question becomes, what is the probability that a events A and B occur, P(A and B)? This is
found by starting at the start of the path and ending at B, as shown. Notice that we use the general form of
calculating the AND statement, P(A AND B) = P(A)P(B|A).
P(B|A)
B
P(BC|A)
Bc
finish
P(A and B) = P(A)P(B| A)
A
P(A)
start
P(B|AC )
C
P(A )
B
Ac
c
P(BC|AC ) B
Suppose I wanted to calculate P(Ac AND B) , I would multiply P(Ac) P(B | Ac).
Facts
•
•
•
•
P(A) + P(Ac) = 1
P(B|A) + P(Bc|A) = 1
P(B|Ac) + P(Bc|Ac) = 1
P(A and B) + P(A and Bc) + P(Ac and B) +P(Ac and Bc) = 1
See the diagram above.
Suppose I wanted to know P(B). This says what is the probability of B occurring. Notice is does not bar any
other event from also happening. Notice the diagram does not have P(B). It does have P(B |Ac) and P(B | A).
These two probabilities state the probabilities of B occurring. That is if we follow the paths on the tree, there
are two routes that end in B occurring. So B occurs if we follow the two paths, shown in red and blue. In
other words B occurs if we follow the path
P(Ac)P(B | Ac) or the path P(A) P(B |A). So
P(B) = P(Ac)P(B | Ac) + P(A) P(B |A). Notice that the two paths are disjoint ( in
terms of the picture, we notice that we can not be at both paths simultaneously), so
we need only add the two probabilities.
P(B|A)
B
P(BC|A)
Bc
A
P(A)
P(B|AC )
P(AC )
B
Ac
c
P(BC|AC ) B
Case 2 - A general viewWe can show a tree diagram having multiple paths, with some paths ending while others continue.
P(E |D and A)
D
P(D|A)
c
P(Ec |D and A) E
A
P(Dc|A)
P(A)
P(B)
E
Dc
B
P(F| D and C)
P(C)
P(D|C)
F
D
C
P(Dc|C)
Dc
Suppose I wanted to calculate P(A and D and E) this would entail passing through the three branches A, D
and E, so I need to multiply the three probabilities P(A)P(D|A)P(E|D and A).
Fact list
•
•
•
•
•
•
P(A) + P(B) +P(C) = 1
P(D|A) + P(Dc|A) = 1
P(E |A and D) +P(Ec| A and D) = 1
P(D | C) + P(Dc |C) = 1
P(F| D and C) = 1
P(B) + P(A and D and E) + P(A and Dc) + P(A and D and Ec) + P(C and Dc) + P(C and D and F) = 1
BAYES RULE
When anyone first encounter's the formula for Baye's Rule in any textbook it looks intimidating and confusing
to the untrained reader. However, once we understand what the formula is trying to convey then, it becomes
as easy to read, and furthermore, we are free from needing to remember the notation of the rule, but rather its
function.
A book I read not long ago titled, "Calculated Risks" implores the reader to understand the importance of
Baye's Rule and also presents an easier view of this rule. I will write here the premise proposed by the author.
Case 1 - book formula.
P(B|A)
B
P(BC|A)
Bc
A
P(A)
P(B|AC )
C
P(A )
P(A|B) =
P(B|A)P(A)
P( A and B)
=
c
c
P(B|A)P(A) + P(B|A )P(A )
P( B)
B
c
A
c
P(BC|AC ) B
If you look closely this is what the formula is indicating. Event B has occurred, thus you could only have
traveled down two paths- red path or blue path. Only one of the two paths contains event A, the blue path. So
our formula says, "Find the probability of traveling those two paths- total probability see
(http://spot.pcc.edu/~hmesa/math243/lecture/independent_events/conditional_probabilities.htm) B
P(B|A)
which is encrypted in the denominator of the formula as
A
P(BC|A)
P(A)
P(B|AC )
P(AC )
c
B
B
Ac
c
P(BC|AC ) B
P(B | A)P(A) + P(B|Ac)P(Ac).
The only path you want is the blue path and the probability of landing there is
P(B|A)P(A). So this is the same finding the probability of tossing a die and having it
land on a four. There are six possibilities (six paths), and only one contains the
1
Probability of landing a four
1
number four. Using probabilities it would be stated as
=6=
Probability of a number apearring 6 6
6
Look at the formula again, P(A|B) =
denominator indicate.
P(B|A)P(A)
, and look at the paths the numerator and
P(B|A)P(A) + P(B|A c )P(A c )