Question

complete parts (a) through (c). write the probabilities as fractions.

(a) check the outcomes for each event below. then, enter the probability of the event.

event	outcomes (1, 2, 3, 4, 5, 6)	probability
event b: the selected ball has an even number on it	1 (□), 2 (✔), 3 (□), 4 (✔), 5 (□), 6 (✔)	\\(\frac{1}{2}\\)
event a or b: the selected ball is white or has an even number on it	1 (✔), 2 (✔), 3 (✔), 4 (✔), 5 (✔), 6 (✔)	1
event a and b: the selected ball is white and has an even number on it	1 (✔), 2 (□), 3 (✔), 4 (□), 5 (✔), 6 (□)	0

(b) compute the following.
\\( p(a) + p(b) - p(a \text{ and } b) = 1 \\)

(c) select the answer that makes the equation true.
\\( p(a) + p(b) - p(a \text{ and } b) = \\) dropdown with ( p(a \text{ or } b) )

Explanation:

Part (b)

Step 1: Identify \( P(A) \), \( P(B) \), and \( P(A \text{ and } B) \)

From part (a), we have:

• \( P(A) = \frac{1}{2} \)
• \( P(B) = \frac{1}{2} \)
• \( P(A \text{ and } B) = 0 \)

Step 2: Substitute into the formula \( P(A) + P(B) - P(A \text{ and } B) \)

Substitute the values:
\[
P(A) + P(B) - P(A \text{ and } B) = \frac{1}{2} + \frac{1}{2} - 0
\]

Step 3: Simplify the expression

\[
\frac{1}{2} + \frac{1}{2} = 1 \quad \text{and} \quad 1 - 0 = 1
\]

The formula for the probability of the union of two events \( A \) and \( B \) (i.e., \( P(A \text{ or } B) \)) is given by the addition rule of probability: \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \). This is a fundamental rule in probability theory that accounts for the overlap (intersection) of the two events to avoid double - counting.

Answer:

\( 1 \)

Part (c)