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 a: rolling a number from 5 to 6
event b: rolling an even number
event a or b: rolling a number from 5 to 6 or rolling an even number
event a and b: rolling a number from 5 to 6 and rolling an even number
(b) compute the following.
$p(a) + p(b) - p(a \text{ and } b)$
(c) select the answer that makes the
$p(a) + p(b) - p(a \text{ and } b) = $ dropdown with options: $p(a \text{ and } b)$, $p(a)$, $p(a \text{ or } b)$, $p(b)$

Explanation:

Part (a) - Identifying Outcomes

We assume we are rolling a standard six - sided die (numbers 1 - 6).

• Event A (Rolling 5 - 6): The outcomes are 5, 6. So we check the boxes for 5 and 6.
• Event B (Rolling even): Even numbers on a die are 2, 4, 6. So we check the boxes for 2, 4, 6.
• Event A or B: Outcomes that are in A or B. From A (5, 6) and B (2, 4, 6), the combined set is 2, 4, 5, 6. So we check the boxes for 2, 4, 5, 6.
• Event A and B: Outcomes that are in both A and B. A has 5, 6 and B has 2, 4, 6. The intersection is 6. So we check the box for 6.

Part (b) - Calculating Probabilities

A standard die has 6 possible outcomes.

• Step 1: Find \(P(A)\)

Event A has 2 outcomes (5, 6). So \(P(A)=\frac{\text{Number of outcomes in A}}{\text{Total outcomes}}=\frac{2}{6}\)

• Step 2: Find \(P(B)\)

Event B has 3 outcomes (2, 4, 6). So \(P(B)=\frac{\text{Number of outcomes in B}}{\text{Total outcomes}}=\frac{3}{6}\)

• Step 3: Find \(P(A\text{ and }B)\)

Event \(A\text{ and }B\) has 1 outcome (6). So \(P(A\text{ and }B)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total outcomes}}=\frac{1}{6}\)

• Step 4: Calculate \(P(A)+P(B)-P(A\text{ and }B)\)

\[

$$\begin{align*} P(A)+P(B)-P(A\text{ and }B)&=\frac{2}{6}+\frac{3}{6}-\frac{1}{6}\\ &=\frac{2 + 3-1}{6}\\ &=\frac{4}{6}\\ &=\frac{2}{3} \end{align*}$$

\]

Part (c) - Identifying the Formula

By the principle of inclusion - exclusion for probability, \(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\). So the correct option to fill in the blank is \(P(A\text{ or }B)\)

Final Answers

• (a) Outcomes for A: 5, 6; Outcomes for B: 2, 4, 6; Outcomes for A or B: 2, 4, 5, 6; Outcomes for A and B: 6 (check the respective boxes).
• (b) \(P(A)+P(B)-P(A\text{ and }B)=\frac{2}{3}\)
• (c) The answer is \(P(A\text{ or }B)\)

Answer:

Part (a) - Identifying Outcomes

We assume we are rolling a standard six - sided die (numbers 1 - 6).

• Event A (Rolling 5 - 6): The outcomes are 5, 6. So we check the boxes for 5 and 6.
• Event B (Rolling even): Even numbers on a die are 2, 4, 6. So we check the boxes for 2, 4, 6.
• Event A or B: Outcomes that are in A or B. From A (5, 6) and B (2, 4, 6), the combined set is 2, 4, 5, 6. So we check the boxes for 2, 4, 5, 6.
• Event A and B: Outcomes that are in both A and B. A has 5, 6 and B has 2, 4, 6. The intersection is 6. So we check the box for 6.

Part (b) - Calculating Probabilities

A standard die has 6 possible outcomes.

• Step 1: Find \(P(A)\)

Event A has 2 outcomes (5, 6). So \(P(A)=\frac{\text{Number of outcomes in A}}{\text{Total outcomes}}=\frac{2}{6}\)

• Step 2: Find \(P(B)\)

Event B has 3 outcomes (2, 4, 6). So \(P(B)=\frac{\text{Number of outcomes in B}}{\text{Total outcomes}}=\frac{3}{6}\)

• Step 3: Find \(P(A\text{ and }B)\)

Event \(A\text{ and }B\) has 1 outcome (6). So \(P(A\text{ and }B)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total outcomes}}=\frac{1}{6}\)

• Step 4: Calculate \(P(A)+P(B)-P(A\text{ and }B)\)

\[

$$\begin{align*} P(A)+P(B)-P(A\text{ and }B)&=\frac{2}{6}+\frac{3}{6}-\frac{1}{6}\\ &=\frac{2 + 3-1}{6}\\ &=\frac{4}{6}\\ &=\frac{2}{3} \end{align*}$$

\]

Part (c) - Identifying the Formula

By the principle of inclusion - exclusion for probability, \(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\). So the correct option to fill in the blank is \(P(A\text{ or }B)\)

Final Answers

• (a) Outcomes for A: 5, 6; Outcomes for B: 2, 4, 6; Outcomes for A or B: 2, 4, 5, 6; Outcomes for A and B: 6 (check the respective boxes).
• (b) \(P(A)+P(B)-P(A\text{ and }B)=\frac{2}{3}\)
• (c) The answer is \(P(A\text{ or }B)\)