Question

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

Answer 1

Answer:

P(B)=0.41, and P( that is the answer

Answer 2

Answer:

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

Step-by-step 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:

$\begin{aligned}& P(A \cup B) \\ =\; & P(A \backslash B) + P(A \cap B) + P(B \backslash A) \\ = \; & P(A) - P(A \cap B) + P(A \cap B) \\ &+ P(B) - P(A \cap B) \\ =\; & P(A) + P(B) - P(A \cap B) \end{aligned}$.