Answer:
[tex]P(A \cup B) = 0.52[/tex] (assuming that [tex]P(A \cap B) = 0.12[/tex].)
Step-by-step explanation:
The question is asking for the probability of the event [tex]A \cup B[/tex] ([tex]A[/tex] or [tex]B[/tex], or both.)
Refer to the diagram attached. There are three mutually-exclusive ways to satisfy [tex]A \cup B[/tex]:
- [tex]A[/tex] is satisfied but [tex]B[/tex] isn't.
- [tex]A[/tex] and [tex]B[/tex] are both satisfied.
- [tex]B[/tex] is satisfied but [tex]A[/tex] isn't.
Probability that [tex]A[/tex] is satisfied but [tex]B[/tex] isn't:
[tex]\begin{aligned} & P(A \backslash B) \\ =\; & P(A) - P(A \cap B) \\ =\; & 0.23 - 0.12 \\ =\; & 0.11 \end{aligned}[/tex].
Probability that [tex]A[/tex] and [tex]B[/tex] are both satisfied:
[tex]P(A \cap B) = 0.12[/tex].
Probability that [tex]B[/tex] is satisfied but [tex]A[/tex] isn't:
[tex]\begin{aligned} & P(B \backslash A) \\ =\; & P(B) - P(A \cap B) \\ =\; & 0.41 - 0.12 \\ =\; & 0.29 \end{aligned}[/tex].
There's no intersection between these three ways for satisfying [tex]A \cup B[/tex]. Hence, the probability [tex]P(A \cup B)[/tex] would be the sum of the probability of each of the three ways:
[tex]\begin{aligned}& P(A \cup B) \\ =\; & P(A \backslash B) + P(A \cap B) + P(B \backslash A) \\ =\; & 0.11 + 0.12 + 0.29 \\ =\; & 0.52\end{aligned}[/tex].
Instead of calculating [tex]P(A \backslash B)[/tex] and [tex]P(B \backslash A)[/tex] separately, the work above may be condensed into one equation:
[tex]\begin{aligned}& P(A \cup B) \\ =\; & P(A \backslash B) + P(A \cap B) + P(B \backslash A) \\ = \; & P(A) - P(A \cap B) + P(A \cap B) \\ &+ P(B) - P(A \cap B) \\ =\; & P(A) + P(B) - P(A \cap B) \end{aligned}[/tex].
