Question

a tile is selected at random from 6 tiles. each tile is labeled with a different letter from the first 6 letters of the alphabet. the 6 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
	a	b	c	d	e	f
event x: selecting a vowel	□	□	□	□	□	□	□
event y: selecting a letter that comes after d	□	□	□	□	□	□	□
event x and y: selecting a vowel and a letter that comes after d	□	□	□	□	□	□	□
event x or y: selecting a vowel or a letter that comes after d	□	□	□	□	□	□	□

(b) compute the following.
$p(x) + p(y) - p(x \text{ and } y) = \square$
(c) select the answer that makes the equation true.

Explanation:

Part (a)

Event X: Selecting a vowel

The first 6 letters of the alphabet are A, B, C, D, E, F. The vowels among these are A and E. So there are 2 favorable outcomes. The total number of outcomes is 6.
The probability \( P(X) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6}=\frac{1}{3} \)

Event Y: Selecting a letter that comes after D

Letters after D (from A - F) are E and F. So there are 2 favorable outcomes.
The probability \( P(Y)=\frac{2}{6}=\frac{1}{3} \)

Event X and Y: Selecting a vowel and a letter that comes after D

We need to find the letters that are vowels and come after D. From the vowels (A, E) and letters after D (E, F), the common letter is E. So there is 1 favorable outcome.
The probability \( P(X \text{ and } Y)=\frac{1}{6} \)

Event X or Y: Selecting a vowel or a letter that comes after D

Using the principle of inclusion - exclusion, \( P(X \text{ or } Y)=P(X)+P(Y)-P(X \text{ and } Y) \)
We know \( P(X)=\frac{2}{6}, P(Y)=\frac{2}{6}, P(X \text{ and } Y)=\frac{1}{6} \)
\( P(X \text{ or } Y)=\frac{2 + 2- 1}{6}=\frac{3}{6}=\frac{1}{2} \)
We can also count the favorable outcomes directly. The letters that are vowels (A, E) or come after D (E, F) are A, E, F. So there are 3 favorable outcomes. \( P(X \text{ or } Y)=\frac{3}{6}=\frac{1}{2} \)

Part (b)

We know \( P(X)=\frac{2}{6}, P(Y)=\frac{2}{6}, P(X \text{ and } Y)=\frac{1}{6} \)
\( P(X)+P(Y)-P(X \text{ and } Y)=\frac{2}{6}+\frac{2}{6}-\frac{1}{6}=\frac{2 + 2- 1}{6}=\frac{3}{6}=\frac{1}{2} \)

Part (c)

From the principle of inclusion - exclusion for two events \( X \) and \( Y \), we have \( P(X \text{ or } Y)=P(X)+P(Y)-P(X \text{ and } Y) \)
So the equation \( P(X)+P(Y)-P(X \text{ and } Y)=\boldsymbol{P(X \text{ or } Y)} \)

Final Answers

(a)

• Event X: \( \frac{1}{3} \)
• Event Y: \( \frac{1}{3} \)
• Event X and Y: \( \frac{1}{6} \)
• Event X or Y: \( \frac{1}{2} \)

(b)

\( \frac{1}{2} \)

(c)

\( P(X \text{ or } Y) \)

Answer:

Part (a)

Event X: Selecting a vowel

The first 6 letters of the alphabet are A, B, C, D, E, F. The vowels among these are A and E. So there are 2 favorable outcomes. The total number of outcomes is 6.
The probability \( P(X) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6}=\frac{1}{3} \)

Event Y: Selecting a letter that comes after D

Letters after D (from A - F) are E and F. So there are 2 favorable outcomes.
The probability \( P(Y)=\frac{2}{6}=\frac{1}{3} \)

Event X and Y: Selecting a vowel and a letter that comes after D

We need to find the letters that are vowels and come after D. From the vowels (A, E) and letters after D (E, F), the common letter is E. So there is 1 favorable outcome.
The probability \( P(X \text{ and } Y)=\frac{1}{6} \)

Event X or Y: Selecting a vowel or a letter that comes after D

Using the principle of inclusion - exclusion, \( P(X \text{ or } Y)=P(X)+P(Y)-P(X \text{ and } Y) \)
We know \( P(X)=\frac{2}{6}, P(Y)=\frac{2}{6}, P(X \text{ and } Y)=\frac{1}{6} \)
\( P(X \text{ or } Y)=\frac{2 + 2- 1}{6}=\frac{3}{6}=\frac{1}{2} \)
We can also count the favorable outcomes directly. The letters that are vowels (A, E) or come after D (E, F) are A, E, F. So there are 3 favorable outcomes. \( P(X \text{ or } Y)=\frac{3}{6}=\frac{1}{2} \)

Part (b)

We know \( P(X)=\frac{2}{6}, P(Y)=\frac{2}{6}, P(X \text{ and } Y)=\frac{1}{6} \)
\( P(X)+P(Y)-P(X \text{ and } Y)=\frac{2}{6}+\frac{2}{6}-\frac{1}{6}=\frac{2 + 2- 1}{6}=\frac{3}{6}=\frac{1}{2} \)

Part (c)

From the principle of inclusion - exclusion for two events \( X \) and \( Y \), we have \( P(X \text{ or } Y)=P(X)+P(Y)-P(X \text{ and } Y) \)
So the equation \( P(X)+P(Y)-P(X \text{ and } Y)=\boldsymbol{P(X \text{ or } Y)} \)

Final Answers

(a)

• Event X: \( \frac{1}{3} \)
• Event Y: \( \frac{1}{3} \)
• Event X and Y: \( \frac{1}{6} \)
• Event X or Y: \( \frac{1}{2} \)

(b)

\( \frac{1}{2} \)

(c)

\( P(X \text{ or } Y) \)