Math Problem Statement

Suppose one card is drawn at random from a standard deck. What is the probability that the card drawn is (a) a face card, (b) a club, (c) a face card or a club?

Solution

The image shows a probability problem related to a deck of 52 playing cards. Let’s solve the three parts step-by-step:

(a) What is the probability that the card drawn is a face card?

Face cards are the Jack, Queen, and King in each suit. Each suit (Hearts, Diamonds, Spades, Clubs) has 3 face cards, so there are: 3 face cards×4 suits=12 face cards in total.3 \text{ face cards} \times 4 \text{ suits} = 12 \text{ face cards in total}. The total number of cards is 52, so the probability of drawing a face card is: P(Face card)=1252=313.P(\text{Face card}) = \frac{12}{52} = \frac{3}{13}.

(b) What is the probability that the card drawn is a club?

Each suit (Hearts, Diamonds, Spades, Clubs) has 13 cards. Therefore, the number of club cards is 13. The probability of drawing a club is: P(Club)=1352=14.P(\text{Club}) = \frac{13}{52} = \frac{1}{4}.

(c) What is the probability that the card drawn is a face card or a club?

To find this probability, we need to use the principle of inclusion and exclusion: P(Face card or Club)=P(Face card)+P(Club)P(Face card and Club).P(\text{Face card or Club}) = P(\text{Face card}) + P(\text{Club}) - P(\text{Face card and Club}).

  • P(Face card)=313P(\text{Face card}) = \frac{3}{13},
  • P(Club)=14=1352P(\text{Club}) = \frac{1}{4} = \frac{13}{52},
  • P(Face card and Club)P(\text{Face card and Club}): There are 3 face cards in the club suit (Jack, Queen, King of Clubs), so P(Face card and Club)=352P(\text{Face card and Club}) = \frac{3}{52}.

Now, calculating the combined probability: P(Face card or Club)=313+1352352.P(\text{Face card or Club}) = \frac{3}{13} + \frac{13}{52} - \frac{3}{52}.

First, express 313\frac{3}{13} with denominator 52: 313=1252.\frac{3}{13} = \frac{12}{52}. Now, substitute: P(Face card or Club)=1252+1352352=2252=1126.P(\text{Face card or Club}) = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}.

So, the probability of drawing a face card or a club is: P(Face card or Club)=1126.P(\text{Face card or Club}) = \frac{11}{26}.


Would you like more details or have any questions about these solutions?

Here are 5 related questions:

  1. What is the probability of drawing a red card from a standard deck?
  2. How many face cards are in a red suit?
  3. What is the probability of drawing a numbered card (2 through 10)?
  4. What is the probability of drawing a non-face card that is a spade?
  5. How would the probabilities change if jokers were included in the deck?

Tip: The principle of inclusion and exclusion helps to avoid double-counting in probability when events overlap.

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

P(Event) = Number of favorable outcomes / Total number of outcomes
P(A or B) = P(A) + P(B) - P(A and B)

Theorems

Inclusion-Exclusion Principle

Suitable Grade Level

Grades 6-8