Last modified: February 20, 2025
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Probability trees are a visual representation of all possible outcomes of a probabilistic experiment and the paths leading to these outcomes. They are especially helpful in understanding sequences of events, particularly when these events are conditional on previous outcomes.
A probability tree starts with a root node that branches out into possible outcomes of the first event. Each of these outcomes becomes a node that branches into further outcomes, continuing for as many events as are being considered. The branches are labeled with probabilities, and the path to each outcome shows the sequence of events leading to that outcome.
Consider a simplified example where we draw one card from a standard deck of 52 cards, and we're interested in the event of drawing a heart or not drawing a heart. The event space for this problem can be represented using a probability tree, which helps visualize the possible outcomes and their associated probabilities.
Tree:
__Heart (1/4)
|
Initial ──┤
|
|__Not Heart (3/4)Here is a step-by-step explanation of the tree:
$$ P(\text{Heart}) = \frac{13}{52} = \frac{1}{4} $$
$$ P(\text{Not Heart}) = \frac{39}{52} = \frac{3}{4} $$
This structure is consistent with the principle that the total probability of all possible outcomes must sum to 1, i.e.:
$$ P(\text{Heart}) + P(\text{Not Heart}) = 1 $$
$$ \frac{1}{4} + \frac{3}{4} = 1 $$
Thus, the probability tree correctly reflects the outcomes and their associated probabilities.
In this example, we examine two independent events: first, tossing a coin, and then rolling a die. The probability tree for this scenario helps to visualize the sequential outcomes and probabilities at each stage of the experiment.
Tree:
__Head (1/2)───1 (1/6)
| .
| .
Initial ──┤ 6 (1/6)
|
|__Tail (1/2)───1 (1/6)
.
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6 (1/6)I. Initial Node (Coin Toss): The experiment begins with a coin toss. The initial node represents the state before any action is taken.
II. First Level Branches (Coin Toss):
$$ P(\text{Head}) = \frac{1}{2} $$
Similarly, the probability of getting a "Tail" is:
$$ P(\text{Tail}) = \frac{1}{2} $$
These probabilities reflect the equal likelihood of both outcomes for a fair coin.
III. Second Level Branches (Die Roll):
$$ P(\text{Die Roll}) = \frac{1}{6} $$
since each face of the die is equally likely to appear.
The probability of any combined outcome (e.g., tossing a "Head" followed by rolling a "4") is calculated by multiplying the probabilities of the individual events along the relevant path in the tree. For example:
$$ P(\text{Head and 4}) = P(\text{Head}) \times P(4) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} $$
Similarly, the probability of getting a "Tail" followed by rolling a "2" would be:
$$ P(\text{Tail and 2}) = P(\text{Tail}) \times P(2) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} $$
The total probability for any combined event is always the product of the individual probabilities along its branch. Since the events are independent, the combined probability of any outcome from the coin toss followed by a die roll is given by:
$$ P(\text{Combined Event}) = P(\text{Coin Outcome}) \times P(\text{Die Outcome}) $$
The probability tree can be used to calculate all possible combinations of outcomes, and since there are 12 possible combinations (2 coin outcomes × 6 die outcomes), the total probabilities sum to 1. This ensures that the model accounts for all potential outcomes:
$$ \sum P(\text{Outcome}) = 1 $$
In this example, we explore the scenario of rolling two six-sided dice simultaneously. A probability tree provides a clear visualization of all possible outcomes and their associated probabilities.
Tree:
_____1 (1/6)──1 (1/6)
| .
| .
| 6 (1/6)
|
|_____2 (1/6)──1 (1/6)
Initial ──┤ .
| .
| 6 (1/6)
|
|__ ... (1/6)──1 (1/6)
| .
| .
| 6 (1/6)
|
|_____6 (1/6)──1 (1/6)
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6 (1/6)I. Initial Node (Before Rolling Dice): The initial node represents the state before any dice are rolled.
II. First Level Branches (First Die Roll):
$$ P(\text{First Die Outcome}) = \frac{1}{6} $$
III. Second Level Branches (Second Die Roll):
$$ P(\text{Second Die Outcome}) = \frac{1}{6} $$
To compute the probability of rolling a specific combination (e.g., rolling a "3" on the first die and a "5" on the second), we multiply the probabilities along the path that represents the combined event. For example:
$$ P(3 \text{ on first die and } 5 \text{ on second die}) = P(3) \times P(5) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$
Since the outcomes of the two dice are independent, the probability of any specific combination is always the product of the probabilities of the individual outcomes.
There are 36 possible combinations of outcomes (6 outcomes from the first die × 6 outcomes from the second die). Each combination has an equal probability of occurring:
$$ P(\text{Any Specific Combination}) = \frac{1}{36} $$
The probability of rolling any specific pair of numbers (e.g., a "2" on the first die and a "6" on the second) is calculated using the formula:
$$ P(\text{First Die Outcome and Second Die Outcome}) = P(\text{First Die Outcome}) \times P(\text{Second Die Outcome}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} $$
Thus, the total probability of all possible outcomes from rolling two dice must sum to 1:
$$ \sum P(\text{Outcome}) = 1 $$
This ensures that the probability model is complete and accounts for every possible pair of outcomes from rolling two dice.
Consider a scenario where a bag contains 6 red marbles (R) and 4 green marbles (G). Two marbles are drawn one after the other without replacement. We aim to calculate the probabilities of the following outcomes:
The following probability tree illustrates the outcomes and the conditional probabilities:
Tree:
┌── R (5/9)
┌── R (6/10)
│ └── G (4/9)
Initial Node
│ ┌── R (6/9)
└── G (4/10)
└── G (3/9)I. P(R then G) (Red marble followed by a Green marble):
The probability of drawing a red marble first is:
$$ P(\text{R first}) = \frac{6}{10} $$
Given that a red marble is drawn first, the probability of drawing a green marble second is:
$$ P(\text{G second | R first}) = \frac{4}{9} $$
Therefore, the probability of drawing a red marble followed by a green marble is:
$$ P(\text{R then G}) = P(\text{R first}) \times P(\text{G second | R first}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} \approx 0.267 $$
II. P(G then R) (Green marble followed by a Red marble):
The probability of drawing a green marble first is:
$$ P(\text{G first}) = \frac{4}{10} $$
Given that a green marble is drawn first, the probability of drawing a red marble second is:
$$ P(\text{R second | G first}) = \frac{6}{9} $$
Therefore, the probability of drawing a green marble followed by a red marble is:
$$ P(\text{G then R}) = P(\text{G first}) \times P(\text{R second | G first}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} \approx 0.267 $$
III. P(both R) (Both marbles are red):
The probability of drawing a red marble first is:
$$ P(\text{R first}) = \frac{6}{10} $$
After drawing one red marble, there are 5 red marbles left out of 9, so the probability of drawing a second red marble is:
$$ P(\text{R second | R first}) = \frac{5}{9} $$
Therefore, the probability of drawing two red marbles is:
$$ P(\text{both R}) = P(\text{R first}) \times P(\text{R second | R first}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = 0.333 $$
IV. P(both G) (Both marbles are green):
The probability of drawing a green marble first is:
$$ P(\text{G first}) = \frac{4}{10} $$
After drawing one green marble, there are 3 green marbles left out of 9, so the probability of drawing a second green marble is:
$$ P(\text{G second | G first}) = \frac{3}{9} $$
Therefore, the probability of drawing two green marbles is:
$$ P(\text{both G}) = P(\text{G first}) \times P(\text{G second | G first}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} \approx 0.133 $$