Question

What is the probability of P(AUB), if P(A) = 0.23, P(B) = 0.41, and P(
AB) = 0.12?

Answer 1

`P(A \cup B) = 0.52` (assuming that
`P(A \cap B) = 0.12`.)

Explanation:

The question is asking for the probability of the event
`A \cup B` (
`A` or
`B`, or both.)

Refer to the diagram attached. There are three mutually-exclusive ways to satisfy
`A \cup B`:

• 
  `A` is satisfied but
  `B` isn't.
• 
  `A` and
  `B` are both satisfied.
• 
  `B` is satisfied but
  `A` isn't.

Probability that
`A` is satisfied but
`B` isn't:

`\begin{aligned} & P(A \backslash B) \\ =\; & P(A) - P(A \cap B) \\ =\; & 0.23 - 0.12 \\ =\; & 0.11 \end{aligned}`.

Probability that
`A` and
`B` are both satisfied:

`P(A \cap B) = 0.12`.

Probability that
`B` is satisfied but
`A` isn't:

`\begin{aligned} & P(B \backslash A) \\ =\; & P(B) - P(A \cap B) \\ =\; & 0.41 - 0.12 \\ =\; & 0.29 \end{aligned}`.

There's no intersection between these three ways for satisfying
`A \cup B`. Hence, the probability
`P(A \cup B)` would be the sum of the probability of each of the three ways:

`\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}`.

Instead of calculating
`P(A \backslash B)` and
`P(B \backslash A)` separately, the work above may be condensed into one equation:

`\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}`.