Event A occurs with probability 0.2. Event B occurs with probability 0.3. Event C occurs with probability 0.4. If A, B, and C are disjoint, then:
P(A or B or C) = 0.9.
P(A or B) = 0.5.
P(A or C) = 0.6.
All of the answer choices are correct.
The correct answer and explanation is :
To solve this problem, let’s carefully go through the information provided and the question:
Given Information:
- ( P(A) = 0.2 )
- ( P(B) = 0.3 )
- ( P(C) = 0.4 )
- Events A, B, and C are disjoint (meaning that no two events can happen simultaneously, i.e., the probability of the intersection of any two events is zero: ( P(A \cap B) = P(A \cap C) = P(B \cap C) = 0 )).
What we need to find:
- ( P(A \text{ or } B \text{ or } C) ), which is ( P(A \cup B \cup C) ).
- ( P(A \text{ or } B) ), which is ( P(A \cup B) ).
- ( P(A \text{ or } C) ), which is ( P(A \cup C) ).
Using the formula for the union of disjoint events:
Since events A, B, and C are disjoint, we can apply the formula for the probability of the union of disjoint events, which is simply the sum of the individual probabilities:
[
P(A \cup B \cup C) = P(A) + P(B) + P(C)
]
[
P(A \cup B \cup C) = 0.2 + 0.3 + 0.4 = 0.9
]
So, the first statement ( P(A \cup B \cup C) = 0.9 ) is correct.
Finding ( P(A \cup B) ):
For disjoint events A and B:
[
P(A \cup B) = P(A) + P(B) = 0.2 + 0.3 = 0.5
]
Thus, the second statement ( P(A \cup B) = 0.5 ) is also correct.
Finding ( P(A \cup C) ):
For disjoint events A and C:
[
P(A \cup C) = P(A) + P(C) = 0.2 + 0.4 = 0.6
]
So, the third statement ( P(A \cup C) = 0.6 ) is correct as well.
Conclusion:
Since all the statements given in the problem are correct, the correct answer is “All of the answer choices are correct.”
Explanation:
- Disjoint events (mutually exclusive events) are events that cannot happen simultaneously. Therefore, the probability of the intersection of any two disjoint events is zero.
- When calculating the union of disjoint events, we simply add the probabilities of the individual events, since there is no overlap between them. This property makes solving for the union of disjoint events straightforward, as we don’t need to subtract any overlap (intersection) probabilities.
Hence, the solution involves straightforward addition, leading to the final answer that all of the provided statements are true.