Problem 30 - Algebra and Trigonometry

Question

Selecting Cards Three cards are randomly selected from a standard 52 -card deck, one at a time, with each card re- placed in the deck before the next one is picked. Find the probability of each event. (a) All three cards are hearts. (b) Exactly two of the cards are spades. (d) None of the cards is a diamond. (d) At least one of the cards is a club.

Step-by-Step Solution

Answer

a) \( \frac{1}{64} \), b) \( \frac{9}{64} \), c) \( 0.4226 \), d) \( 0.5774 \)

Step 1: Understand the Problem

We have a deck of 52 cards, and we are selecting three cards one at a time with replacement. This means we have the same probabilities for each card selection, as the deck returns to its full number after each draw.

Step 1: Probability of Three Hearts

To find the probability that all three cards are hearts, we calculate the probability of drawing a heart on each selection. Since there are 13 hearts in a 52-card deck, the probability of drawing a heart is \( \frac{13}{52} = \frac{1}{4} \). The probability that all three cards are hearts is \( \left( \frac{1}{4} \right)^3 = \frac{1}{64} \).

Step 2: Probability of Exactly Two Spades

We use the binomial probability formula. The probability of drawing a spade is \( \frac{13}{52} = \frac{1}{4} \). For exactly two spades (and one non-spade), we calculate \( \binom{3}{2} \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^1 = 3 \times \frac{1}{16} \times \frac{3}{4} = \frac{9}{64} \).

Step 3: Probability of No Diamonds

To find the probability that none are diamonds, first calculate the probability of not drawing a diamond in one draw. Since there are 39 non-diamonds, the probability of a non-diamond is \( \frac{39}{52} \). For three non-diamonds: \( \left( \frac{39}{52} \right)^3 = \frac{59319}{140608} \approx 0.4226 \).

Step 4: Probability of At Least One Club

To find the probability of at least one club, calculate the complement: the probability that none are clubs. The probability of not drawing a club is \( \frac{39}{52} \), and for three cards: \( \left( \frac{39}{52} \right)^3 \approx 0.4226 \). Therefore, the probability of at least one club is \( 1 - 0.4226 = 0.5774 \).

Key Concepts

Binomial Probability

Binomial probability deals with situations where we perform a series of identical experiments, known as Bernoulli trials, each with two possible outcomes: success or failure. It's commonly used when calculating probabilities like getting a certain number of spades in our card selection exercise.
When we learned about finding the probability of exactly two spades in three draws, this was an application of binomial probability. Each draw is a Bernoulli trial where "success" means drawing a spade, which happens with probability \( \frac{1}{4} \) in our card example. The number of successes (drawing spades) had to be two out of three draws.
The formula for binomial probability is:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Where:

• \( n \) is the number of trials (draws)
• \( k \) is the number of successful trials (spades drawn)
• \( p \) is the probability of success (probability of drawing a spade)
• \( \binom{n}{k} \) is the binomial coefficient, which calculates the number of different combinations possible

In our example, we calculated it as \( 3 \cdot \left( \frac{1}{4} \right)^2 \cdot \frac{3}{4} = \frac{9}{64} \), showing us how to work with repeated, independent trials.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations, permutations, and counting techniques. It helps in determining the number of possible ways an event can occur, as seen with our example of picking cards.
When you calculated the probability of exactly two spades among three cards, combinatorics came into play through the binomial coefficient \( \binom{3}{2} \). This term represents how many different combinations of two spades out of three cards can occur.
To understand this better, remember:

• The binomial coefficient \( \binom{n}{k} \), read as "n choose k," calculates the number of ways to choose \( k \) successes (like spades) out of \( n \) trials (cards drawn).
• Mathematically, it's represented as: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
• The factorial notation \( n! \) means \( n \) multiplied by every whole number down to 1.

Understanding combinatorics is crucial for accurately solving problems that involve selecting or arranging items under certain restrictions, like our card probabilities.

Card Probability

Card probability is a fascinating topic that explores the likelihood of drawing particular cards from a standard deck. Our exercise involved determining the chances of selecting specific types of cards, such as hearts, spades, diamonds, and clubs.
In a standard deck, there are 52 cards, sorted equally among four suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards. This uniform distribution allows us to calculate simple probabilities, like drawing a heart or a non-diamond.
Here are some key points to remember:

• The probability of drawing a heart from the deck on any single draw is \( \frac{13}{52} = \frac{1}{4} \).
• Since cards are replaced after each draw, drawing remains independent and the probability stays consistent across draws.
• For events like "at least one club," it’s often useful to calculate the probability of the complement event (no clubs) and subtract from 1.
• For example, finding the probability of no clubs among three cards is \( \left( \frac{39}{52} \right)^3 \), and therefore the opposite, at least one club, is about \( 0.5774 \).

Understanding these concepts allows you to tackle more complex probability problems involving cards, enhancing your statistical reasoning capabilities.