Problem 7 - Algebra 2

Question

Three cards are drawn from a standard deck of cards. Find each probability. \(P(3 \text { hearts }),\) if no replacement occurs

Step-by-Step Solution

Answer

The probability is \( \frac{11}{850} \).

Step 1: Understand the Event

In this problem, we need to calculate the probability of drawing three hearts consecutively from a standard deck of 52 playing cards without replacement.

Step 2: Calculate Probability of First Heart

The probability of drawing the first heart from the deck is \[ P(\text{First heart}) = \frac{13}{52} = \frac{1}{4} \] since there are 13 hearts in a deck of 52 cards.

Step 3: Calculate Probability of Second Heart

Once the first heart has been drawn, there are now 51 cards left, of which 12 are hearts. Therefore, the probability of drawing a second heart is \[ P(\text{Second heart | First heart}) = \frac{12}{51} \]

Step 4: Calculate Probability of Third Heart

After drawing the second heart, there are 50 cards remaining and 11 hearts left. So, the probability for drawing a third heart is \[ P(\text{Third heart | First and Second hearts}) = \frac{11}{50} \]

Step 5: Calculate the Combined Probability

The probability of all three events occurring in sequence is the product of their individual probabilities. Therefore: \[ P(3 \text{ hearts}) = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \]This can be simplified to: \[ P(3 \text{ hearts}) = \frac{13 \times 12 \times 11}{52 \times 51 \times 50} \] \[ = \frac{1716}{132600} \]\[ = \frac{11}{850} \] Therefore, the probability of drawing three hearts consecutively without replacement is \( \frac{11}{850} \).

Key Concepts

Deck of Cards

A standard deck of playing cards is a great way to understand probability because it is both complex and structured at the same time. Each deck consists of 52 cards, which are evenly divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, including one ace, numbered cards from 2 to 10, and three face cards (jack, queen, and king). In this exercise, we're focusing on one suit, the hearts.

• Suits: Four distinct suits — hearts, diamonds, clubs, and spades.
• Number of Cards per Suit: 13 cards each.
• Card Types: Numeric cards from 2 to 10, along with face cards and an ace.

Knowing how the deck is organized is key to understanding the probability of drawing specific cards.

Without Replacement

In probability, doing something 'without replacement' means that once an object (or card, in this case) is used, it isn't put back into the initial group. This changes the odds of certain events occurring, as the total number of objects decreases and thus, options become limited.
For example, in our card scenario:

• After drawing one heart, only 51 cards remain in the deck.
• The number of hearts lessens with each heart drawn, which shifts the probability for the subsequent card draws.

"Without replacement" introduces a dependency between events, leading us to the concept of conditional probability – where the probability of future outcomes depends on previous results.

Conditional Probability

Conditional probability is a fundamental concept in probability that deals with the likelihood of an event occurring given that another event has already happened.This is particularly important in scenarios like drawing cards "without replacement".

Calculating Conditional Probability

The formula for conditional probability is expressed by: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]Where:

• \(P(A|B)\) stands for the probability of event A occurring given that B has already occurred.
• \(P(A \cap B)\) is the probability that both events A and B occur.
• \(P(B)\) is the probability of event B.

For example, if the first card drawn is a heart, the probability changes for drawing another heart because there are now fewer hearts left, and fewer cards overall.Therefore:

• Probability of second heart: \(\frac{12}{51}\)
• Probability of third heart after two draws: \(\frac{11}{50}\)

In compound events, conditional probability allows us to track changes in likelihood across multiple steps.