4.8 Tree Diagrams

1. Visualizing Conditional Probabilities

Sometimes, an experiment entails sequential choices or actions spanning two or more stages. In such scenarios, tree diagrams present an exceptionally effective method for organizing the sample space and clearly illustrating conditional probabilities.

EXAMPLE 1

Suppose a box contains 3 red balls and 5 green balls. If we select two balls successively without replacement, what is the probability that both balls are red?

Let $R_1$ denote the event "the 1st ball is Red". Clearly, $P(R_1) = \dfrac{3}{8}$.
Let $R_2$ denote the event "the 2nd ball is Red". Because the first ball is not replaced, the probability $P(R_2)$ strictly depends on the outcome of the first draw.
If the 1st ball is Red, we are left with exactly 2 red balls out of 7 total. Thus, the conditional probability is:

$P(R_2 | R_1) = \dfrac{2}{7}$

The requested combined event is $R_1 \cap R_2$. Using the fundamental rule for independent/dependent events:

$P(A \cap B) = P(A) \cdot P(B|A)$

We calculate the probability of the path RR explicitly:

$P(RR) = P(R_1) \cdot P(R_2 | R_1) = \dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{6}{56}$

Two Golden Rules of Tree Diagrams

Rule 1 (Multiply along branches): We trace a single continuous path from left to right to locate a specific sequence of events (an outcome). To evaluate the probability of this final outcome, we multiply all branch probabilities along that chosen path.
Rule 2 (Add parallel paths): Often, multiple distinct paths lead to a satisfying outcome (e.g., getting exactly one red ball). Since each complete path represents a mutually exclusive outcome within the sample space, we add their individual probabilities together.

EXAMPLE 2

Using the exact tree diagram constructed in Example 1, calculate the following probabilities:

a) Exactly one red ball is selected
Two distinct paths satisfy this condition: RG (Red then Green) and GR (Green then Red). $$ \begin{aligned} P(\text{exactly one red}) &= P(RG) + P(GR) \\ &= \left(\dfrac{3}{8} \cdot \dfrac{5}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{15}{56} + \dfrac{15}{56} \\ &= \mathbf{\dfrac{30}{56}} \end{aligned} $$

b) At least one red ball is selected
Three paths satisfy this condition: RR, RG, and GR. $$ \begin{aligned} P(\text{at least one red}) &= P(RR) + P(RG) + P(GR) \\ &= \left(\dfrac{3}{8} \cdot \dfrac{2}{7}\right) + \left(\dfrac{3}{8} \cdot \dfrac{5}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{6}{56} + \dfrac{15}{56} + \dfrac{15}{56} \\ &= \mathbf{\dfrac{36}{56}} \end{aligned} $$

Alternative "Complementary" Method: We can subtract the probability of the only failing path (GG) from the total sample space (1).
$P(\text{at least one red}) = 1 - P(GG) = 1 - \left(\dfrac{5}{8} \cdot \dfrac{4}{7}\right) = 1 - \dfrac{20}{56} = \mathbf{\dfrac{36}{56}}$

c) The second ball is red
This outcome relies on evaluating all possible paths that terminate with a red ball in the second stage (RR and GR). Notice that the unconditional probability $P(R_2)$ is generally equal to $P(R_1)$. $$ \begin{aligned} P(R_2) &= P(RR) + P(GR) \\ &= \left(\dfrac{3}{8} \cdot \dfrac{2}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{6}{56} + \dfrac{15}{56} \\ &= \dfrac{21}{56} \\ &= \mathbf{\dfrac{3}{8}} \end{aligned} $$

EXAMPLE 3

Suppose the two balls are selected with replacement (the first ball is returned to the box before the second is drawn).

In this scenario, the outcome of the first draw has no physical effect on the second. Therefore, the events are completely independent, and the probabilities remain static across all stages.

Using the rules of multiplication and addition over the modified tree:

$P(RR) = \dfrac{3}{8} \cdot \dfrac{3}{8} = \mathbf{\dfrac{9}{64}}$
$P(\text{exactly one red}) = P(RG) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{5}{8}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{8}\right) = \mathbf{\dfrac{30}{64}}$

2. Evaluating Complex Conditional Probabilities (Bayes' Rules)

Evaluating conditional probabilities spanning across multiple stages generally requires establishing the intersection relative to the conditional subspace. The fundamental formula remains our primary analytical tool:

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

A "Bayes' style" problem occurs when the conditioning event $B$ occurs temporally after the event $A$ in the experiment sequence. The tree diagram must be evaluated "backwards."

EXAMPLE 4

Based on the initial non-replacement experiment (Example 1), we know that exactly one red ball has been drawn. What is the probability that it was selected in the first draw?

We define the requisite events explicitly:
Condition (Given): $B = \text{"exactly one red"}$ (The paths RG and GR)
Target Event: $A = \text{"the first draw was red"}$ (The path RG)

We seek $P(A|B)$. By applying the formal definition of conditional probability:

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

Now, evaluate the intersection carefully. What represents the combined event $A \cap B$? We require a path that satisfies both condition $B$ (exactly one red) AND target $A$ (first draw is red). The only path strictly satisfying both is RG. Therefore, $P(A \cap B) = P(RG)$. $$ \begin{aligned} P(A|B) &= \dfrac{P(RG)}{P(\text{exactly one red})} \\ &= \dfrac{\dfrac{15}{56}}{\dfrac{30}{56}} \\ &= \mathbf{\dfrac{1}{2}} \end{aligned} $$

EXAMPLE 5

Assume again the non-replacement experiment from Example 1. Given that the second ball drawn was red, find the probability that the first ball was also red.

Define the events:
Condition (Given): $B = \text{"second ball is red"}$ (Paths RR and GR)
Target Event: $A = \text{"first ball is red"}$

We evaluate the intersection $A \cap B$. The path satisfying both "second ball red" AND "first ball red" is exclusively RR. $$ \begin{aligned} P(A|B) &= \dfrac{P(A \cap B)}{P(B)} \\ &= \dfrac{P(RR)}{P(R_2)} \\ &= \dfrac{\dfrac{6}{56}}{\dfrac{21}{56}} \\ &= \mathbf{\dfrac{6}{21}} = \mathbf{\dfrac{2}{7}} \end{aligned} $$

Alternative Insight: We could conceptually reverse the temporal logic. If the final draw yields a red ball from the box initially containing 3 red and 5 green balls, then practically, it is as if we removed one red ball prior to the first draw. The probability the first draw was red is exactly the conditional state resulting from this mental removal: $\dfrac{2}{7}$.