Question

a tile is selected from seven tiles, each labeled with a different letter from the first seven letters of the alphabet. the letter selected will be recorded as the outcome. consider the following events. event x: the letter selected comes before \e\. event y: the letter selected is found in the word \face\. give the outcomes for each of the following events. if there is more than one element in the set, separate them with commas. (a) event \x and y\: {} (b) event \x or y\: {} (c) the complement of the event x: {}

Explanation:

Step1: Identify the sample space

The first seven letters of the alphabet are \( A, B, C, D, E, F, G \). So the sample space \( S=\{A, B, C, D, E, F, G\} \).

Step2: Determine outcomes for Event X

Event X: The letter comes before "E". The letters before "E" (in the first seven letters) are \( A, B, C, D \). So \( X = \{A, B, C, D\} \).

Step3: Determine outcomes for Event Y

Event Y: The letter is in the word "FACE". The letters in "FACE" are \( F, A, C, E \). So \( Y=\{A, C, E, F\} \).

Step4: Solve part (a): Event "X and Y"

"X and Y" means the intersection of \( X \) and \( Y \), i.e., the letters that are in both \( X \) and \( Y \). From \( X = \{A, B, C, D\} \) and \( Y=\{A, C, E, F\} \), the common elements are \( A, C \). So \( X \cap Y=\{A, C\} \).

Step5: Solve part (b): Event "X or Y"

"X or Y" means the union of \( X \) and \( Y \), i.e., the letters that are in \( X \) or \( Y \) (or both). Combining \( X = \{A, B, C, D\} \) and \( Y=\{A, C, E, F\} \), we get \( \{A, B, C, D, E, F\} \) (we combine all elements, removing duplicates).

Step6: Solve part (c): Complement of Event X

The complement of \( X \) (denoted as \( X^c \)) is the set of elements in the sample space \( S \) that are not in \( X \). Since \( X = \{A, B, C, D\} \) and \( S=\{A, B, C, D, E, F, G\} \), the elements not in \( X \) are \( E, F, G \). So \( X^c = \{E, F, G\} \).

Answer:

s:
(a) \(\{A, C\}\)
(b) \(\{A, B, C, D, E, F\}\)
(c) \(\{E, F, G\}\)