Given that S = {1, 2, 3, 4, 5, 6) and two events A = (1, 2, 3, 4) and B (2, 3) are observed on S, the P (A|B) is equal to:
The notation P(A|B) represents the conditional probability of event A occurring given that event B has already occurred.
In this case, event A consists of the numbers {1, 2, 3, 4} and event B consists of the numbers {2, 3}.
To calculate P(A|B), we need to determine the probability of event A occurring, given that event B has already occurred.
Since event B consists of the numbers {2, 3}, we know that both of these numbers are included in event A as well.
Therefore, all the outcomes in event B are also outcomes in event A.
In other words, if event B occurs, event A is guaranteed to occur as well.
So, P(A|B) = 1 (since the probability of event A occurring given that event B has already occurred is equal to 1 or 100%).
To find the probability of event A given that event B has occurred, we use the formula:
P(A|B) = P(A ∩ B) / P(B)
1. First, we need to find the probability of A intersect B (A ∩ B). This is the probability that both A and B occur at the same time.
A ∩ B = {2, 3}
2. Next, we find the probability of event B occurring (P(B)). This is the probability that event B occurs.
P(B) = Number of outcomes in B / Number of outcomes in S = 2 / 6 = 1/3
3. Now, we can substitute the values we found into the formula for conditional probability:
P(A|B) = P(A ∩ B) / P(B)
= 2 / 6 ÷ 1/3
= 2/6 * 3/1
= 1/3 * 3/1
= 1
Therefore, P(A|B) is equal to 1.