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Chapter 7: Problem 43

An experiment consists of selecting a card at random from a well-shuffled 52 -card deck. Let \(E\) denote the event that an ace is drawn and let \(F\) denote the event that a spade is drawn. Show that \(n(E \cup F)=n(E)+n(F)-n(E \cap F)\).

Short Answer

Expert verified
In a well-shuffled 52-card deck, there are 16 favorable outcomes for drawing either an Ace or a Spade, as calculated using the formula \(n(E \cup F) = n(E) + n(F) - n(E \cap F) = 4 + 13 - 1 = 16\).

Step by step solution

01

Identify the number of favorable outcomes for drawing an Ace

There are 4 Aces in a standard 52-card deck (one Ace per suit). Thus, \(n(E) = 4\).
02

Identify the number of favorable outcomes for drawing a Spade

There are 13 Spades in a standard 52-card deck (one for each rank). Thus, \(n(F) = 13\).
03

Identify the number of favorable outcomes for drawing an Ace of Spades

There is only 1 Ace of Spades in a standard 52-card deck. Thus, \(n(E \cap F) = 1\).
04

Use the formula to calculate the number of favorable outcomes for drawing either an Ace or a Spade

Plug the values of \(n(E)\), \(n(F)\), and \(n(E \cap F)\) into the formula: \(n(E \cup F) = n(E) + n(F) - n(E \cap F) = 4 + 13 - 1 = 16\) Hence, there are 16 favorable outcomes for drawing either an Ace or a Spade.

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