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Chapter 29: Problem 1

Suppose that you draw a card from a standard deck of 52 cards. What is the probability of drawing a. an ace of any suit? b. the ace of spades? c. How would your answers to parts (a) and (b) change if you were allowed to draw three times, replacing the card drawn back into the deck after each draw?

Short Answer

Expert verified
The probabilities for drawing a card from a standard deck of 52 cards are as follows: a. P(Ace of any suit) = \(\frac{1}{13}\) b. P(Ace of spades) = \(\frac{1}{52}\) When drawing three times with replacement: c. P(Ace of any suit in all three draws) = \(\frac{1}{2197}\) P(Ace of spades in all three draws) = \(\frac{1}{140608}\)

Step by step solution

01

(a) Probability of drawing an ace of any suit

In a standard deck, there are 4 aces (one for each suit: hearts, diamonds, clubs, and spades). The total number of cards in the deck is 52. The probability of drawing an ace of any suit is the ratio of the number of aces to the total number of cards. Therefore, the probability is: P(Ace of any suit) = \(\frac{Number\ of\ Aces}{Total\ Number\ of\ Cards}\) P(Ace of any suit) = \(\frac{4}{52} = \frac{1}{13}\)
02

(b) Probability of drawing the ace of spades

There is only 1 ace of spades in the deck. The probability of drawing the ace of spades is the ratio of the single ace of spades to the total number of cards. Therefore, the probability is: P(Ace of spades) = \(\frac{Number\ of\ Ace\ of\ Spades}{Total\ Number\ of\ Cards}\) P(Ace of spades) = \(\frac{1}{52}\)
03

(c) Probabilities when drawing three times with replacement

When drawing three times with replacement, each draw will be independent and have the same probabilities as in parts (a) and (b) because the deck will be the same after each draw. For each draw: P(Ace of any suit) = \(\frac{1}{13}\) P(Ace of spades) = \(\frac{1}{52}\) Since the draws are independent, we can multiply the probabilities to find the probability of getting the desired outcome in all three draws. P(Ace of any suit in all three draws) = \(\left(\frac{1}{13}\right)^3 = \frac{1}{2197}\) P(Ace of spades in all three draws) = \(\left(\frac{1}{52}\right)^3 = \frac{1}{140608}\) Thus, the probabilities for parts (a) and (b) when drawing three times with replacement are: P(Ace of any suit in all three draws) = \(\frac{1}{2197}\) P(Ace of spades in all three draws) = \(\frac{1}{140608}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
Understanding independent events is crucial when calculating probabilities in scenarios such as card games. An independent event is one where the outcome of one event does not affect the outcome of another. For instance, when you draw a card from a deck and then return it before drawing another, each draw is considered an independent event because the composition of the deck is restored before each draw, meaning the probabilities remain constant.

In the provided exercise, the act of drawing a card three times with replacement exemplifies independent events. Each time a card is drawn and replaced, the outcome of the next draw is not influenced by the previous one. Therefore, the probability of drawing an ace on the second or third draw is the same as on the first draw. This concept allows us to multiply the probabilities of each draw to find the overall probability when multiple independent events occur sequentially. For example, the probability of drawing an ace of any suit in all three draws is the cube of the probability of drawing an ace in one draw: \(\left(\frac{1}{13}\right)^3 = \frac{1}{2197}\).
Probability with Replacement
Probability with replacement comes into play when we perform repeated trials of an experiment and ensure that the conditions of each trial remain unchanged. In card games, this means returning a drawn card back to the deck and reshuffling it to restore the original conditions before the next draw.

Type of Visible Mathematics
For instance, after drawing an ace and placing it back into the deck, the probability of drawing an ace again stays the same, unlike without replacement, where the probability would change. This approach can lead to outcomes that seem counterintuitive, such as being able to draw the ace of spades three times in a row, however unlikely that may be. The calculations for such events are made simpler by the fact that each trial's outcome does not depend on the previous ones, hence the use of the multiplication rule as seen in the exercise: \(P(Ace of spades in all three draws) = \left(\frac{1}{52}\right)^3 = \frac{1}{140608}\).
Standard Deck of 52 Cards
A standard deck of 52 cards is a common starting point for calculating probabilities in card games. This deck consists of 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, ranking from ace to king. It is essential to understand the composition of the deck to accurately determine probabilities of drawing specific cards.

For example, knowing that there are four aces in a standard deck allows us to calculate the probability of drawing an ace as \(\frac{4}{52} = \frac{1}{13}\). Likewise, being aware that there is only one ace of spades in the deck leads to the probability of drawing that specific card being \(\frac{1}{52}\). Familiarity with the deck's structure aids in solving problems and understanding the likelihood of different outcomes, which is vital for card players and enthusiasts alike.

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Most popular questions from this chapter

In nonlinear optical switching devices based on dye-doped polymer systems, the spatial orientation of the dye molecules in the polymer is an important parameter. These devices are generally constructed by orienting dye molecules with a large dipole moment using an electric field. Imagine placing a vector along the molecular dipole moment such that the molecular orientation can be described by the orientation of this vector in space relative to the applied field ( \(z\) direction) as illustrated here: For random molecular orientation about the \(z\) axis, the probability distribution describing molecular orientation along the \(z\) axis is given by \(P(\theta)=\sin \theta d \theta / \int_{0}^{\pi} \sin \theta d \theta .\) Orientation is quantified using moments of \(\cos \theta\). a. Determine \(\langle\cos \theta\rangle\) for this probability distribution. b. Determine \(\left\langle\cos ^{2} \theta\right\rangle\) for this probability distribution.

You are dealt a hand consisting of 5 cards from a standard deck of 52 cards. Determine the probability of obtaining the following hands: a. flush (five cards of the same suit) b. a king, queen, jack, ten, and ace of the same suit (a "royal flush")

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Determine the numerical values for the following: a. the number of configurations employing all objects in a six-object set b. the number of configurations employing four objects from a six-object set c. the number of configurations employing no objects from a six-object set d. \(C(50,10)\)

Radio station call letters consist of four letters (for example, KUOW ). a. How many different station call letters are possible using the 26 letters in the English alphabet? b. Stations west of the Mississippi River must use the letter \(\mathrm{K}\) as the first call letter. Given this requirement, how many different station call letters are possible if repetition is allowed for any of the remaining letters? c. How many different station call letters are possible if repetition is not allowed for any of the letters?
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