Problem 46
Question
Four cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability of exactly one pair?
Step-by-Step Solution
Verified Answer
The probability is approximately 58.51%.
1Step 1: Understand the Problem
We need to find the probability of drawing exactly one pair from a deck of 52 cards when drawing four cards in total.
2Step 2: Calculate the Total Number of Ways to Draw 4 Cards
First, calculate the total number of ways to draw 4 cards from a deck of 52 cards, which can be given by the combination formula \( \binom{52}{4} \). This results in 270725.
3Step 3: Calculate the Ways to Draw Exactly One Pair
To have exactly one pair, choose a rank for the pair (one of 13 possible ranks), choose 2 out of the 4 available cards of that rank, choose 2 ranks for the non-paired cards (12 choose 2), and then choose one card from each of these ranks (4 times 4). The calculation involves the product \( \binom{13}{1} \times \binom{4}{2} \times \binom{12}{2} \times \binom{4}{1}^2 = 13 \times 6 \times 66 \times 4 \times 4 = 158544 \).
4Step 4: Calculate the Probability
Divide the number of favorable outcomes (one pair) by the total number of possible outcomes. This is \( \frac{158544}{270725} \approx 0.5851 \), or approximately 58.51%.
Key Concepts
CombinationsPermutationsDeck of 52 Cards
Combinations
Understanding the concept of combinations is essential when solving problems related to probability, especially when dealing with scenarios like drawing cards from a deck. Combinations refer to the selection of items from a larger pool, where the order does not matter. This means that choosing cards A, B, and C is the same as choosing C, B, and A.A combination is mathematically represented by the binomial coefficient, expressed using the formula:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]where:- \(n\) is the total number of items to choose from,- \(r\) is the number of items to select, and- \(!\) denotes factorial, which is the product of all positive integers up to that number.In our exercise on drawing cards, we apply combinations to determine the total number of ways to select 4 cards out of 52. The formula \(\binom{52}{4}\) helps us capture all possible selections without worrying about the order. This foundational understanding is essential as it sets the stage to calculate more specific probabilities, like having exactly one pair.
Permutations
Permutations provide a way to understand scenarios where order or sequence matters. Unlike combinations, permutations consider different orderings of the selected items, making it critical for situations where the sequence affects the outcome.The general permutation formula is:\[P(n, r) = \frac{n!}{(n-r)!}\]Although permutations themselves aren't directly used in the given card-drawing problem (because we don't care about card order), understanding the distinction is crucial. Permutations would come into play, for example, when determining the number of ways to order a hand of cards once specific cards have been chosen.Knowing when to use combinations versus permutations can save you from making mistakes in complex probability problems. Always ask: does the order of selection matter in this scenario? For our card problem, the order is irrelevant for selection purposes, hence, the use of combinations.
Deck of 52 Cards
The standard deck of 52 cards is a staple in probability and combinatorics exercises, providing a familiar context for applying mathematical principles. A typical deck consists of four suits: Hearts, Diamonds, Clubs, and Spades. Each suit contains 13 ranks, ranging from Ace up to King.
Understanding the deck's composition is pivotal as it allows you to calculate probabilities related to drawing certain types of cards, like pairs, straights, or flushes. In the problem of drawing exactly one pair, this knowledge helps to determine:
- The number of ranks available (13 different ranks).
- The fact that each rank consists of 4 cards, allowing for the pairing logic to apply.
- The remaining ranks and cards available for selection after a pair is chosen.
Other exercises in this chapter
Problem 45
In how many ways can four red and five black cards be selected from a standard deck of cards if cards are drawn without replacement?
View solution Problem 46
A true-false exam has 20 questions. Find the expected number of correct answers if a student guesses the answers at random.
View solution Problem 46
In how many ways can two aces and three kings be selected from a standard deck of cards if cards are drawn without replacement?
View solution Problem 47
An urn contains 12 green and 24 blue balls. (a) You take 10 balls out of the urn. Find the probability that 6 of the 10 balls are blue. (b) You take a ball out
View solution