Problem 32 - Algebra and Trigonometry

Question

Selecting Cards Three cards are randomly selected from a standard 52 -card deck, one at a time, with each card replaced 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. (c) 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) \( \frac{27}{64} \); (d) \( \frac{37}{64} \).

Step 1: Understand the Problem

We have a standard 52-card deck and we are selecting cards with replacement. This means that after selecting a card, we put it back into the deck, maintaining 52 cards for each selection. We want to calculate the probability of specific events given this scenario.

Step 2: Determine Relevant Probabilities

For a standard deck, there are 13 cards of each suit (hearts, spades, diamonds, clubs). The probability of drawing a specific suit in one draw is \( \frac{13}{52} = \frac{1}{4} \), and the probability of drawing a card that is not of one suit (e.g., not a heart) is \( \frac{39}{52} = \frac{3}{4} \). These will be used to calculate the probabilities of the events described.

Step 3: Calculate Probability for Part (a)

The probability that all three cards are hearts is the probability of drawing a heart three times in a row. Each draw is independent, so the probability is \[ P(\text{3 hearts}) = \left( \frac{1}{4} \right)^3 = \frac{1}{64}. \]

Step 4: Calculate Probability for Part (b)

For exactly two cards to be spades, two out of the three cards must be spades, and the other must not be. There are \( \binom{3}{2} = 3 \) ways to choose which two cards are spades. The probability for one specific arrangement (two spades and one non-spade) is \[ \left( \frac{1}{4} \right)^2 \times \frac{3}{4} = \frac{3}{64}. \] Therefore, the total probability is \[ 3 \times \frac{3}{64} = \frac{9}{64}. \]

Step 5: Calculate Probability for Part (c)

For none of the cards to be a diamond, each card drawn must be from the other three suits. The probability of one card not being a diamond is \( \frac{3}{4} \). The probability that none of the three cards is a diamond is \[ \left( \frac{3}{4} \right)^3 = \frac{27}{64}. \]

Step 6: Calculate Probability for Part (d)

To find the probability of at least one club, we use the complement rule. First, find the probability that none of the cards are clubs (all are other suits): \( \left( \frac{3}{4} \right)^3 = \frac{27}{64} \). The probability of at least one club is then \[ 1 - \frac{27}{64} = \frac{37}{64}. \]

Key Concepts

Independent Events

In probability, independent events are those whose outcomes do not affect each other. When you draw a card from a deck with replacement, this action creates independent events. Since the card is replaced, the outcome of one draw does not change the outcomes of the subsequent draws. Each draw has the same probability, regardless of the past ones.
This is crucial when calculating probabilities in sequences. For example, when calculating the chance of drawing three hearts consecutively with replacement, each draw remains independent. The probability of drawing a heart remains at \( \frac{1}{4} \) per draw. This is why we multiply \( \frac{1}{4} \) by itself three times, giving us \[ P(3\, \text{hearts}) = \left( \frac{1}{4} \right)^3 = \frac{1}{64}. \]

Complement Rule

The complement rule is a useful tool in probability. It allows us to calculate the probability of an event by finding the probability of its opposite and subtracting it from 1. This is particularly handy when finding probabilities like "at least one".
For example, if we wish to find the probability of getting at least one club in three draws, it's simpler to calculate the complement: the probability of getting no clubs at all. This probability is \( \left( \frac{3}{4} \right)^3 \), since each draw has a \( \frac{3}{4} \) chance of not being a club. Thus, \( \left( \frac{3}{4} \right)^3 = \frac{27}{64} \).
Finally, subtract that from 1 to get the probability of at least one club: \[ 1 - \frac{27}{64} = \frac{37}{64}. \]

Deck of Cards

A standard deck of cards is an essential concept in probability exercises related to card games. Understanding the structure of the deck helps in calculating probabilities. A typical deck contains 52 cards, divided equally into four suits: hearts, diamonds, clubs, and spades.
Each suit holds 13 cards, numbered from 2 to 10, and includes the face cards Jack, Queen, King, and the Ace. This consistent structure means each card has the same chance of being drawn from the full deck. These characteristics help solve probability problems, such as the likelihood of drawing particular suits or combinations.

Suits in a Deck

Suits play a crucial role in card probability problems. The four suits—hearts, diamonds, clubs, and spades—each have an equal share of the deck, with 13 cards per suit. This balance is foundational in calculating specific suit-based probabilities.
Knowing that each suit accounts for \( \frac{13}{52} \), or \( \frac{1}{4} \) of the deck, you can determine the probability of drawing a card from any suit with replacement simply and directly. For instance, if calculating the chance of drawing two spades in three draws, knowing each draw is independent, you multiply the probability of drawing a spade twice and a non-spade once, adjusting the arrangement appropriately.