4.8 Tree Diagrams
1. Visualizing Conditional Probabilities
When an experiment involves choices across multiple stages, tree diagrams organize the sample space and show conditional probabilities.
EXAMPLE 1
A box contains 3 red balls and 5 green balls. If we select two balls without replacement, what is the probability that both are red?
Let $R_1$ be the event "the 1st ball is Red". Here, $P(R_1) = \dfrac{3}{8}$.
Let $R_2$ be the event "the 2nd ball is Red". Because the first ball is not replaced, $P(R_2)$ depends on the outcome of the first draw.
If the 1st ball is Red, 2 red balls remain out of 7 total. The conditional probability is:
Let $R_2$ be the event "the 2nd ball is Red". Because the first ball is not replaced, $P(R_2)$ depends on the outcome of the first draw.
If the 1st ball is Red, 2 red balls remain out of 7 total. The conditional probability is:
$P(R_2 | R_1) = \dfrac{2}{7}$
The combined event is $R_1 \cap R_2$. Using the multiplication rule:
$P(A \cap B) = P(A) \cdot P(B|A)$
Calculate the probability of path RR:
$P(RR) = P(R_1) \cdot P(R_2 | R_1) = \dfrac{6}{56}$
Rules of Tree Diagrams
- Rule 1 (Multiply along branches): Trace a path from left to right to find the probability of a specific outcome. Multiply the probabilities along that path.
- Rule 2 (Add parallel paths): If multiple paths satisfy a condition, add their individual probabilities.
EXAMPLE 2
Using the tree diagram from Example 1, calculate these probabilities:
a) Exactly one red ball is selected
Two paths satisfy this condition: RG and GR. $$ \begin{aligned} P(\text{exactly one red}) &= P(RG) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{5}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{15}{56} + \dfrac{15}{56} = \mathbf{\dfrac{30}{56}} \end{aligned} $$
Two paths satisfy this condition: RG and GR. $$ \begin{aligned} P(\text{exactly one red}) &= P(RG) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{5}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{15}{56} + \dfrac{15}{56} = \mathbf{\dfrac{30}{56}} \end{aligned} $$
b) At least one red ball is selected
Three paths satisfy this condition: RR, RG, and GR. $$ \begin{aligned} P(\text{at least one red}) &= P(RR) + P(RG) + P(GR) \\ &= \left(\dfrac{3}{8} \cdot \dfrac{2}{7}\right) + \left(\dfrac{3}{8} \cdot \dfrac{5}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) \\ &= \dfrac{6}{56} + \dfrac{15}{56} + \dfrac{15}{56} \\ &= \mathbf{\dfrac{36}{56}} \end{aligned} $$
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 Method: Subtract the probability of the failing path (GG) from 1.
$$
\begin{aligned}
P(\text{at least one red}) &= 1 - P(GG) = 1 - \left(\dfrac{5}{8} \cdot \dfrac{4}{7}\right) \\
&= 1 - \dfrac{20}{56} = \mathbf{\dfrac{36}{56}}
\end{aligned}
$$
c) The second ball is red
Add the paths that end with a red ball (RR and GR). Note that $P(R_2) = P(R_1)$. $$ \begin{aligned} P(R_2) &= P(RR) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{2}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) = \dfrac{6}{56} + \dfrac{15}{56} \\ &= \dfrac{21}{56} = \mathbf{\dfrac{3}{8}} \end{aligned} $$
Add the paths that end with a red ball (RR and GR). Note that $P(R_2) = P(R_1)$. $$ \begin{aligned} P(R_2) &= P(RR) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{2}{7}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{7}\right) = \dfrac{6}{56} + \dfrac{15}{56} \\ &= \dfrac{21}{56} = \mathbf{\dfrac{3}{8}} \end{aligned} $$
EXAMPLE 3
Suppose the two balls are selected with replacement (the first ball is returned before the second draw).
The first draw does not affect the second. The events are independent, so probabilities remain the same across stages.
Using the multiplication and addition rules:
- $P(RR) = \dfrac{3}{8} \cdot \dfrac{3}{8} = \mathbf{\dfrac{9}{64}}$
- $P(\text{exactly one red}) = P(RG) + P(GR) = \left(\dfrac{3}{8} \cdot \dfrac{5}{8}\right) + \left(\dfrac{5}{8} \cdot \dfrac{3}{8}\right) = \mathbf{\dfrac{30}{64}}$
2. Evaluating Conditional Probabilities (Bayes' Theorem)
Evaluating multi-stage conditional probabilities requires finding the intersection within the conditional subspace. Use the formula:
$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$
A Bayes' problem occurs when the given event $B$ happens after event $A$. The tree diagram is evaluated backwards.
EXAMPLE 4
In the non-replacement experiment, given that exactly one red ball was drawn, find the probability it was selected first.
Define the events:
Given: $B = \text{"exactly one red"}$ (Paths RG and GR)
Target: $A = \text{"first draw was red"}$ (Path RG)
Given: $B = \text{"exactly one red"}$ (Paths RG and GR)
Target: $A = \text{"first draw was red"}$ (Path RG)
Find $P(A|B)$ using the definition of conditional probability:
$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$
The intersection $A \cap B$ requires a path that satisfies both conditions. Only path RG qualifies, so $P(A \cap B) = P(RG)$.
$$
\begin{aligned}
P(A|B) &= \dfrac{P(RG)}{P(\text{exactly one red})} = \dfrac{\dfrac{15}{56}}{\dfrac{30}{56}} = \mathbf{\dfrac{1}{2}}
\end{aligned}
$$
EXAMPLE 5
In the non-replacement experiment, given that the second ball was red, find the probability that the first ball was also red.
Define the events:
Given: $B = \text{"second ball is red"}$ (Paths RR and GR)
Target: $A = \text{"first ball is red"}$
Given: $B = \text{"second ball is red"}$ (Paths RR and GR)
Target: $A = \text{"first ball is red"}$
The path satisfying both conditions is RR.
$$
\begin{aligned}
P(A|B) &= \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(RR)}{P(R_2)} = \dfrac{\dfrac{6}{56}}{\dfrac{21}{56}} = \mathbf{\dfrac{2}{7}}
\end{aligned}
$$
Alternative Insight: If the second draw is red, it is as if one red ball was removed before the first draw. The probability the first draw was red is $\dfrac{2}{7}$.