Suppose $A, B, C$ are events in a common probability space with $P(A) = 0.2, \ P(B) = 0.2, \ P(C) = 0.3$, $P(A \cap B) = 0.1, \ P(A \cap C) = 0.1, \ P(B \cap C) = 0.1$. Which of the following are possible values of $P(A \cup B \cup C)$?
To determine the possible values of \(P(A \cup B \cup C)\), we can use the principle of inclusion-exclusion in probability. This formula is given by:
\(P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\)
We know:
- \(P(A) = 0.2\)
- \(P(B) = 0.2\)
- \(P(C) = 0.3\)
- \(P(A \cap B) = 0.1\)
- \(P(A \cap C) = 0.1\)
- \(P(B \cap C) = 0.1\)
Let's compute with the assumption that \(P(A \cap B \cap C) = 0\), as we are not given its value:
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0 \\ &= 0.6 - 0.3 \\ &= 0.3 \end{align*}\)
This result shows that if \(P(A \cap B \cap C) = 0\), then \(P(A \cup B \cup C) = 0.3\).
However, the problem doesn't provide \(P(A \cap B \cap C)\), so we need to consider possible values to find other potential values of \(P(A \cup B \cup C)\). Let's check potential maximum overlap when \(P(A \cap B \cap C) = 0.1\).
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0.1 \\ &= 0.6 - 0.2 \\ &= 0.4 \end{align*}\)
If we further assume \(P(A \cap B \cap C) = 0.2\), let's calculate:
\(\begin{align*} P(A \cup B \cup C) &= 0.2 + 0.2 + 0.3 - 0.1 - 0.1 - 0.1 + 0.2 \\ &= 0.6 - 0.1 \\ &= 0.5 \end{align*}\)
Therefore, the possible values of \(P(A \cup B \cup C)\) consistent with the provided information are \(0.4\) and \(0.5\).
Thus, the correct options are:
0.5
0.4
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