Question

question 3 how many ways can you select a face card, or a queen from a deck of 52 cards?

Explanation:

Step1: Define sets and their counts

Let \( A \) be the set of face cards, and \( B \) be the set of queens. In a standard deck, face cards are Jacks, Queens, and Kings. There are 3 face card types (Jack, Queen, King) and 4 suits, so \( n(A)=3\times4 = 12 \). The set of queens: there are 4 queens (one in each suit), so \( n(B) = 4 \). But the intersection of \( A \) and \( B \) (queens that are face cards) is just the set of queens, so \( n(A\cap B)=4 \) (since all queens are face cards).

Step2: Apply the principle of inclusion - exclusion

The formula for \( n(A\cup B) \) (the number of ways to select a face card or a queen) is \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \). Substituting the values: \( n(A\cup B)=12 + 4-4 \).

Step3: Calculate the result

Simplify the expression: \( 12+4 - 4=12 \). Wait, that seems off? Wait, no: Wait, face cards include queens. So the set of face cards is 12 (J, Q, K in 4 suits: 3*4=12). The set of queens is 4. But when we take "face card or queen", since all queens are face cards, the union is just the set of face cards. Wait, maybe my initial set definition was wrong. Let's re - examine:

Face cards: Jack (J), Queen (Q), King (K). So in each suit (hearts, diamonds, clubs, spades), there are 3 face cards. So total face cards: \( 3\times4 = 12 \). Queens: 4 (one per suit). But the queens are already included in the face cards. So the number of elements in \( A\cup B \) is the number of face cards, because \( B\subseteq A \). So \( n(A\cup B)=n(A)=12 \)? Wait, no, that can't be. Wait, the problem says "select a face card, or a queen". But since every queen is a face card, the "or" here (in set theory, inclusive or) means that we are counting all face cards (because queens are already face cards). Wait, but maybe the problem is using "or" in the sense of "face card (non - queen) or queen"? No, the standard inclusive or. Let's re - do the inclusion - exclusion correctly.

\( A \): face cards (J, Q, K: 12 cards)

\( B \): queens (4 cards)

\( A\cap B \): cards that are both face cards and queens, which is the set of queens (4 cards)

So \( n(A\cup B)=n(A)+n(B)-n(A\cap B)=12 + 4-4 = 12 \). Wait, but that's the same as the number of face cards. Because when you take face card or queen, since queens are face cards, you are just taking all face cards. But let's think again: how many cards are face cards or queens? A face card is J, Q, K. A queen is Q. So the union is J, Q, K (all face cards). So there are 12 such cards.

Wait, maybe I made a mistake in the problem interpretation. Let's check the number of face cards: in a deck, face cards are Jack, Queen, King. So 3 ranks, 4 suits: 3*4 = 12. Queens: 4. But the overlap is 4 (queens are face cards). So by inclusion - exclusion, the number of distinct cards that are face cards or queens is 12 + 4-4 = 12.

Answer:

\( \boldsymbol{12} \)