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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 is found in the word " \( B E A D^{\text {" }} \). Event \( Y: \) The letter selected comes after " \( D \) ". 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 \( Y \) ": \{[i]) (b) Event \( { }^{n} X \) or \( Y *:\{D\} \) (c) The complement of the event \( Y:\{\square\} \)

Ask by Stuart Marsh. in the United States
Mar 16,2025

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Answer

- (a) Event " \( X \) and \( Y \) ": \( E \) - (b) Event " \( X \) or \( Y \) ": \( B, E, A, D, F, G \) - (c) The complement of the event \( Y \): \( A, B, C, D \)

Solution

To solve the problem, we first identify the letters on the tiles and the events defined. The letters on the tiles are: \( A, B, C, D, E, F, G \). ### Event Definitions: - **Event \( X \)**: The letter selected is found in the word "BEAD". The letters in "BEAD" are \( B, E, A, D \). - **Event \( Y \)**: The letter selected comes after "D". The letters that come after "D" are \( E, F, G \). Now, let's find the outcomes for each of the specified events. ### (a) Event " \( X \) and \( Y \) " This event represents the intersection of events \( X \) and \( Y \). We need to find the letters that are both in \( X \) and \( Y \). - Outcomes of \( X \): \( \{B, E, A, D\} \) - Outcomes of \( Y \): \( \{E, F, G\} \) The intersection \( X \cap Y \) is: - Common letters: \( E \) Thus, the outcome for event " \( X \) and \( Y \) " is: \[ \{E\} \] ### (b) Event " \( X \) or \( Y \) " This event represents the union of events \( X \) and \( Y \). We need to find all unique letters that are in either \( X \) or \( Y \). - Outcomes of \( X \): \( \{B, E, A, D\} \) - Outcomes of \( Y \): \( \{E, F, G\} \) The union \( X \cup Y \) is: - Unique letters: \( B, E, A, D, F, G \) Thus, the outcome for event " \( X \) or \( Y \) " is: \[ \{B, E, A, D, F, G\} \] ### (c) The complement of the event \( Y \) The complement of event \( Y \) includes all letters that are not in \( Y \). - Outcomes of \( Y \): \( \{E, F, G\} \) - The complement of \( Y \) will include the remaining letters: \( A, B, C, D \) Thus, the outcome for the complement of event \( Y \) is: \[ \{A, B, C, D\} \] ### Summary of Outcomes: - (a) Event " \( X \) and \( Y \) ": \(\{E\}\) - (b) Event " \( X \) or \( Y \) ": \(\{B, E, A, D, F, G\}\) - (c) The complement of the event \( Y \): \(\{A, B, C, D\}\)

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Beyond the Answer

Let's break down the events based on the letters available and the definitions given. (a) Event \( X \) specifies letters found in "BEAD," which are \( B, E, A, D \). Event \( Y \) includes letters that come after "D," which are \( E, F, G \). The intersection of these two sets (where both events happen) is simply the letters that are both in "BEAD" and come after "D." The only letter that satisfies this condition is: **\{E\}** (b) The event “not \( X \)” corresponds to the letters that are *not* in "BEAD" from the set of letters \( \{A, B, C, D, E, F, G\} \). Therefore, the letters in \( { }^{n} X \) include \( C, F, G \). Adding the letters in \( Y \), which are \( E, F, G \), the union of these events results in: **\{C, E, F, G\}** (c) The complement of the event \( Y \) includes the letters that do not come after “D.” From our set \( \{A, B, C, D, E, F, G\} \), the letters before and including \( D \) are: **\{A, B, C, D\}**

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