A ⊆ B. Prove P(A|B)>= P(A)
To prove that P(A|B) >= P(A), we need to understand the definitions of conditional probability and probability.
Conditional Probability:
The conditional probability of an event A given an event B is denoted by P(A|B) and can be defined as the probability of event A occurring, given that event B has already occurred. It can be calculated using the formula:
P(A|B) = P(A ∩ B) / P(B)
where P(A ∩ B) represents the probability of events A and B both occurring, and P(B) represents the probability of event B occurring.
Probability:
The probability of an event A can be defined as the likelihood of event A occurring. It is denoted by P(A) and can be calculated by:
P(A) = Number of favorable outcomes for event A / Total number of outcomes in the sample space
Proof:
Given that A ⊆ B, it means that event A is a subset of event B. In other words, every outcome that belongs to event A also belongs to event B.
To prove that P(A|B) >= P(A), we can start by manipulating the definition of conditional probability:
P(A|B) = P(A ∩ B) / P(B)
Since A ⊆ B, it implies that A ∩ B = A, as every outcome in A also belongs to B. Therefore:
P(A|B) = P(A) / P(B)
To prove the inequality P(A|B) >= P(A), we can rearrange the equation as:
P(A|B) / P(A) >= 1
To further simplify, we know that the denominator P(A) is always positive, so we can multiply both sides of the inequality by P(A):
P(A|B) >= P(A)
This concludes the proof that P(A|B) >= P(A) when A ⊆ B.