Use the Venn diagram of Figure 7 with the prob-abilities a, b, c, d, e, f, and g assigned to ANBNC, A∩B∩C’,…, and A∩B’C’ to show that if P(A) = P(B) = P(C) = 1, then P(A∩B∩C) = 1. [Hint: Start with the argument that since P(A) = 1, it follows that e = c = f = 0.]
The Correct Answer and Explanation is :
Given that ( P(A) = P(B) = P(C) = 1 ), we are dealing with three events A, B, and C, each having a probability of 1. This means that each event must occur with certainty. In a Venn diagram of three events, we assign probabilities to various intersections and exclusive parts of these events. Let’s break it down step by step.
Venn Diagram Intersections:
- ( a ) represents the probability of ( A \cap B \cap C’ ): the part where A and B happen, but not C.
- ( b ) represents ( A \cap B’ \cap C ): A and C happen, but not B.
- ( c ) represents ( A \cap B’ \cap C’ ): A happens, but neither B nor C.
- ( d ) represents ( A’ \cap B \cap C’ ): B happens, but not A or C.
- ( e ) represents ( A \cap B \cap C ): All three events occur.
- ( f ) represents ( A’ \cap B \cap C ): B and C happen, but not A.
- ( g ) represents ( A’ \cap B’ \cap C ): C happens, but not A or B.
Key Argument:
- Given ( P(A) = P(B) = P(C) = 1 ), it implies that events A, B, and C are guaranteed to happen.
- Therefore, there is no room for parts of the diagram where A, B, or C do not occur:
- ( e = c = f = 0 ), meaning that the areas representing only A, only C, and only B and C without A are not possible.
- Since ( P(A) = 1 ), the entire space for A must be covered by intersections with B or C, meaning ( a + b + e = 1 ).
- Similarly, for B and C to also be 1, we must have all of their intersections filled. This means that the total probability for the intersection ( A \cap B \cap C ) must be 1, i.e., ( P(A \cap B \cap C) = 1 ).
Thus, the conclusion is that, given the assumption that all three events are certain to occur, their intersection must also occur with probability 1.