Probability Multiplication Theorem
The probability multiplication theorem states that the probability of both events A and B occurring, \( P(A \cap B) \), equals the product of the probability of event A, \( P(A) \), and the conditional probability of event B given A, \( P(B|A) \), assuming that event A has already occurred. $$ P(A \cap B) = P(A) \cdot P(B|A) $$
The probability multiplication theorem, also known as the product rule, is a fundamental principle in probability and statistics.
What is it used for?
It provides a way to calculate the probability of multiple events happening at the same time, based on the probabilities of individual events and their interdependencies.
This theorem applies to both dependent and independent events.
- Dependent events
For dependent events, where the occurrence of one event affects the likelihood of the other, the theorem is used in its full form: $$ P(A \cap B) = P(A) \cdot P(B|A) $$ - Independent events
When two events are independent, meaning that the occurrence of one does not affect the other, the theorem simplifies to the product of the individual probabilities. In this case, the conditional probability p(B|A)=p(B) is simply the probability of event B. $$ P(A \cap B) = P(A) \cdot P(B) $$
Note: Understanding this theorem requires a solid grasp of how events are dependent or independent, along with knowledge of each event’s probability. For more details, see the difference between dependent and independent events.
A Practical Example
Let's consider rolling two dice and finding the probability of rolling a 4 on one die and a 6 on the other.
We have two events:
- A = "rolling a 4 on the first die"
- B = "rolling a 6 on the second die"
The probability of event A is p(A)=1/6 because a die has six faces, and only one of them shows the number 4.
$$ p(A)= \frac{1}{6} $$
The conditional probability p(B|A) of event B given A is also p(B|A)=1/6, since the outcome of the first roll doesn't influence the probability of the second roll.
$$ p(B|A)= p(B) = \frac{1}{6} $$
Therefore, the events are independent, as the result of the first roll does not change the likelihood of rolling a 6 on the second die.
In this case, the probability of both events occurring is the product of their individual probabilities:
$$ P(A \cap B) = P(A) \cdot P(B|A) $$
Since p(B|A)=p(B), we have:
$$ P(A \cap B) = P(A) \cdot P(B) $$
$$ P(A \cap B) = \frac{1}{6} \cdot \frac{1}{6} $$
$$ P(A \cap B) = \frac{1}{36} = 0.027 $$
Example 2
Now, let's consider an urn containing 20 balls, of which 12 are white and 8 are black.
We want to calculate the probability of drawing two white balls without replacing them in the urn.
So, we define two events:
- A = "the first drawn ball is white"
- B = "the second drawn ball is white"
The probability of drawing a white ball on the first try (event A) is the number of white balls divided by the total number of balls in the urn:
$$ p(A) = \frac{12}{20} $$
After drawing a white ball, there are now 11 white balls left and 19 balls in total in the urn.
Thus, the probability of drawing a second white ball (event B given A) is p(B|A)=11/19.
$$ p(B|A) = \frac{11}{19} $$
We use the multiplication theorem to find the probability that both events occur, which is the probability of the intersection event (A∩B):
$$ P(A \cap B) = P(A) \cdot P(B|A) $$
$$ P(A \cap B) = \frac{12}{20} \cdot \frac{11}{19} $$
$$ P(A \cap B) = 0.347 $$
The probability of drawing two white balls from an urn with 12 white and 8 black balls, without replacing the drawn balls, is approximately 0.347 or 34.7%.
This means you can expect to draw two white balls in sequence roughly one-third of the time.
And so on.
