Final answer:
The answer is True because we can calculate P(B) using the given information on P(A and Bc) and P(Ac). The probability of event B is calculated using the complement rule and mutual exclusivity of probabilities within a sample space. P(A) is found to be 0.65, and since P(A and Bc) is 0.25, the remainder, P(B), is 0.40.
Step-by-step explanation:
The question involves calculating the probability of the event B given that A union B is equal to the sample space S, the probability of A and the complement of B (Bc) is given as 0.25, and the probability of the complement of A (Ac) is 0.35. We can solve the problem by using the basic properties of probabilities and the knowledge that the sum of the probabilities of an event and its complement is always equal to one.
Firstly, we note that since A union B equals the sample space, the events complement each other, meaning P(A) + P(B) = 1. Given P(Ac) = 0.35, we can find P(A) by subtracting from 1:
P(A) = 1 - P(Ac) = 1 - 0.35 = 0.65
Now, we look at P(A and Bc) which is given as 0.25. Since A and Bc are mutually exclusive, P(A) includes P(A and Bc). Therefore, P(A and B) is the remainder of P(A) when P(A and Bc) is subtracted:
P(A and B) = P(A) - P(A and Bc) = 0.65 - 0.25 = 0.40
Since, P(A and B) is also P(B) in this case because A and B are complements and the occurrence of A ensures B. Thus, P(B) equals P(A and B), and so:
P(B) = 0.40
Therefore, the probability of event B is 0.40.