Question

a tile is selected at random from 8 tiles. each tile is labeled with a different letter from the first 8 letters of the alphabet.
the 8 possible outcomes are listed below.
note that each outcome has the same probability.
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.

	outcomes	probability
event x: selecting a vowel	☐	☐	☐	☐	☐	☐	☐	☐	☐
event y: selecting a letter from a to c	☐	☐	☐	☐	☐	☐	☐	☐	☐
event x and y: selecting a vowel and a letter from a to c	☐	☐	☐	☐	☐	☐	☐	☐	☐
event x or y: selecting a vowel or a letter from a to c	☐	☐	☐	☐	☐	☐	☐	☐	☐

(b) compute the following.
\\( p(x) + p(y) - p(x \text{ and } y) = \square \\)

(c) select the answer that makes the equation true.
\\( p(x) + p(y) - p(x \text{ and } y) = \\) select dropdown

Explanation:

Part (a)

Event X: Selecting a vowel

The first 8 letters of the alphabet are A, B, C, D, E, F, G, H. The vowels among these are A and E. So there are 2 favorable outcomes.
The total number of outcomes is 8.
Probability of Event X, \( P(X) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4} \)

Event Y: Selecting a letter from A to C

The letters from A to C are A, B, C. So there are 3 favorable outcomes.
Probability of Event Y, \( P(Y) = \frac{3}{8} \)

Event X and Y: Selecting a vowel and a letter from A to C

The vowels from A to C is only A. So there is 1 favorable outcome.
Probability of Event X and Y, \( P(X \text{ and } Y) = \frac{1}{8} \)

Event X or Y: Selecting a vowel or a letter from A to C

The favorable outcomes are A, B, C, E (since A and E are vowels, and B, C are from A to C; but A is counted once). So there are 4 favorable outcomes.
Probability of Event X or Y, \( P(X \text{ or } Y) = \frac{4}{8} = \frac{1}{2} \)

Part (b)

We know \( P(X) = \frac{2}{8} \), \( P(Y) = \frac{3}{8} \), and \( P(X \text{ and } Y) = \frac{1}{8} \)
\[

$$\begin{align*} P(X) + P(Y) - P(X \text{ and } Y) &= \frac{2}{8} + \frac{3}{8} - \frac{1}{8}\\ &= \frac{2 + 3 - 1}{8}\\ &= \frac{4}{8}\\ &= \frac{1}{2} \end{align*}$$

\]

Part (c)

From part (b), we found that \( P(X) + P(Y) - P(X \text{ and } Y) = \frac{1}{2} \), which is equal to \( P(X \text{ or } Y) \) (from part (a)). So the equation \( P(X) + P(Y) - P(X \text{ and } Y) = P(X \text{ or } Y) \) holds.

Final Answers

(a)

• Event X: Probability = \(\frac{1}{4}\)
• Event Y: Probability = \(\frac{3}{8}\)
• Event X and Y: Probability = \(\frac{1}{8}\)
• Event X or Y: Probability = \(\frac{1}{2}\)

(b)

\( P(X) + P(Y) - P(X \text{ and } Y) = \boxed{\frac{1}{2}} \)

(c)

\( P(X) + P(Y) - P(X \text{ and } Y) = \boxed{P(X \text{ or } Y)} \) (or \(\boxed{\frac{1}{2}}\) depending on the options, but based on the calculation, it equals \( P(X \text{ or } Y) \))

Answer:

Part (a)

Event X: Selecting a vowel

The first 8 letters of the alphabet are A, B, C, D, E, F, G, H. The vowels among these are A and E. So there are 2 favorable outcomes.
The total number of outcomes is 8.
Probability of Event X, \( P(X) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{8} = \frac{1}{4} \)

Event Y: Selecting a letter from A to C

The letters from A to C are A, B, C. So there are 3 favorable outcomes.
Probability of Event Y, \( P(Y) = \frac{3}{8} \)

Event X and Y: Selecting a vowel and a letter from A to C

The vowels from A to C is only A. So there is 1 favorable outcome.
Probability of Event X and Y, \( P(X \text{ and } Y) = \frac{1}{8} \)

Event X or Y: Selecting a vowel or a letter from A to C

The favorable outcomes are A, B, C, E (since A and E are vowels, and B, C are from A to C; but A is counted once). So there are 4 favorable outcomes.
Probability of Event X or Y, \( P(X \text{ or } Y) = \frac{4}{8} = \frac{1}{2} \)

Part (b)

We know \( P(X) = \frac{2}{8} \), \( P(Y) = \frac{3}{8} \), and \( P(X \text{ and } Y) = \frac{1}{8} \)
\[

$$\begin{align*} P(X) + P(Y) - P(X \text{ and } Y) &= \frac{2}{8} + \frac{3}{8} - \frac{1}{8}\\ &= \frac{2 + 3 - 1}{8}\\ &= \frac{4}{8}\\ &= \frac{1}{2} \end{align*}$$

\]

Part (c)

From part (b), we found that \( P(X) + P(Y) - P(X \text{ and } Y) = \frac{1}{2} \), which is equal to \( P(X \text{ or } Y) \) (from part (a)). So the equation \( P(X) + P(Y) - P(X \text{ and } Y) = P(X \text{ or } Y) \) holds.

Final Answers

(a)

• Event X: Probability = \(\frac{1}{4}\)
• Event Y: Probability = \(\frac{3}{8}\)
• Event X and Y: Probability = \(\frac{1}{8}\)
• Event X or Y: Probability = \(\frac{1}{2}\)

(b)

\( P(X) + P(Y) - P(X \text{ and } Y) = \boxed{\frac{1}{2}} \)

(c)

\( P(X) + P(Y) - P(X \text{ and } Y) = \boxed{P(X \text{ or } Y)} \) (or \(\boxed{\frac{1}{2}}\) depending on the options, but based on the calculation, it equals \( P(X \text{ or } Y) \))