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You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3\) on 1 st card and 10 on 2 nd). (c) Find \(P(10\) on 1 st card and 3 on 2 nd \()\). (d) Find the probability of drawing a 10 and a 3 in either order.

Short Answer

Expert verified
(a) No, the outcomes are not independent. (b) \( \frac{4}{663} \). (c) \( \frac{4}{663} \). (d) \( \frac{8}{663} \).

Step by step solution

01

Understanding Independence

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. In drawing cards without replacement, the outcome of the first draw affects the probability of the second draw because it changes the total number of cards remaining in the deck. Thus, the outcomes on the two cards are not independent.
02

Calculating Probability for Specific Order (3 then 10)

To find the probability of drawing a 3 on the first card and a 10 on the second: 1. Probability of drawing a 3 on the first card is the number of 3s in the deck divided by the total number of cards, which is \(\frac{4}{52}\).2. After drawing a 3, there are 51 cards left, and the probability of drawing a 10 from these cards is \(\frac{4}{51}\).3. Multiply these probabilities: \[ P(3 \text{ on 1st and 10 on 2nd}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} \approx \frac{4}{663} \]
03

Calculating Probability for Reverse Order (10 then 3)

To find the probability of drawing a 10 on the first card and a 3 on the second: 1. Probability of drawing a 10 on the first card is \(\frac{4}{52}\).2. After drawing a 10, there are 51 cards left, and the probability of drawing a 3 is \(\frac{4}{51}\).3. Multiply these probabilities: \[ P(10 \text{ on 1st and 3 on 2nd}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} \approx \frac{4}{663} \]
04

Calculating Probability for Either Order

To find the probability of drawing a 10 and a 3 in either order, sum the probabilities from Steps 2 and 3:\[ P(3 \text{ and } 10 \text{ in either order}) = \frac{4}{663} + \frac{4}{663} = \frac{8}{663} \]

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Key Concepts

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

Dependent Events
In probability theory, it's important to recognize the concept of dependent events. Dependent events occur when the outcome of one event affects the likelihood of the other. This concept is evident in scenarios involving card draws without replacement from a deck.
If you draw a card from a deck and do not place it back before drawing the next card, the number of available cards changes, thereby changing the probability for the next draw.
  • The total count of remaining cards decreases by one.
  • The specific types of cards remaining also change, depending on which card was drawn first.
In our exercise, since we're drawing cards without replacing them, the events are certainly dependent. This makes calculating probabilities a little more complex, as opposed to independent events where each card draw would have no impact on the other. Understanding dependent events is crucial for tackling many real-world probability problems involving sequential actions.
Card Probability
The concept of card probability is based on determining the likelihood of drawing specific cards from a deck. A standard deck has 52 cards, with 4 suits (hearts, diamonds, clubs, and spades), each containing an equal number of cards.
When calculating card probability, these considerations are key:
  • Each suit contains one 3 and one 10, so there are four 3s and four 10s in the total deck.
  • The initial probability of drawing any specific card (like a 3 or a 10) from the full deck is calculated by dividing the number of such cards by 52.
For example, to calculate the chance of drawing a 3 as the first card, you consider the four 3s spread across the suits, resulting in a probability of \( \frac{4}{52} \). As cards are removed without replacement, you will need to adjust your calculations and consider the reduced number of cards and corresponding probabilities for subsequent draws.
Order of Events
Order of events is a critical factor when it comes to determining probabilities where the sequence affects the outcome.
In card drawing scenarios, the sequence in which cards are drawn can change the probabilities significantly.
  • For instance, the probability of drawing a 3 first, then a 10, is calculated separately from drawing a 10 first, then a 3.
  • Each sequence has its own set of calculations based on the card drawn first influencing the deck's composition for the second draw.
When considering both sequences in our problem:
  • Drawing a 3 first and a 10 second requires multiplying their respective probabilities, resulting in \( \frac{4}{52} \times \frac{4}{51} \).
  • Conversely, drawing a 10 first and a 3 second follows the same process with adjusted probabilities upon card removal.
Because these are dependent events, the order impacts the probabilities, underlining the importance of considering all possible orders when solving similar exercises.
Multiplication Rule for Probability
The multiplication rule for probability helps determine the likelihood of two or more events happening in sequence, especially for dependent events.
This rule is crucial when calculating probabilities of drawing cards sequentially from a deck without replacement.
  • The rule states that the probability of multiple dependent events occurring is the product of their individual probabilities adjusted by each previous outcome.
  • In our exercise, multiplying the probability of drawing a 3 on the first card and a 10 on the second involves the calculation \( \frac{4}{52} \times \frac{4}{51} \).
  • Similar multiplication applies in reverse order: drawing a 10 first, then a 3.
By understanding and applying the multiplication rule, you can effectively calculate complex probabilities involving sequences of dependent events, crucial in many statistical and real-world applications. Whether dealing with games or predictions, mastering this rule is essential for anyone aiming to work competently with probabilities.

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Most popular questions from this chapter

(a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? (b) Assign probabilities to the outcomes of the sample space of part (a). Do the probabilities add up to 1 ? Should they add up to \(1 ?\) Explain. (c) What is the probability of getting a number less than 5 on a single throw? (d) What is the probability of getting 5 or 6 on a single throw?

At Litchfield College of Nursing, \(85 \%\) of incoming freshmen nursing students are female and \(15 \%\) are male. Recent records indicate that \(70 \%\) of the entering female students will graduate with a BSN degree, while \(90 \%\) of the male students will obtain a BSN degree. If an incoming freshman nursing student is selected at random, find (a) \(P(\) student will graduate \(\mid\) student is female). (b) \(P\) (student will graduate and student is female). (c) \(P\) (student will graduate I student is male). (d) \(P(\) student will graduate and student is male). (e) \(P\) (student will graduate). Note that those who will graduate are either males who will graduate or females who will graduate. (f) The events described by the phrases "will graduate and is female" and "will graduate, given female" seem to be describing the same students. Why are the probabilities \(P(\) will graduate and is female ) and \(P\) (will graduate female) different?

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Arches National Park is located in southern Utah. The park is famous for its beautiful desert landscape and its many natural sandstone arches. Park Ranger Edward McCarrick started an inventory (not yet complete) of natural arches within the park that have an opening of at least 3 feet. The following table is based on information taken from the book Canyon Country Arches and Bridges, by F. A. Barnes. The height of the arch opening is rounded to the nearest foot. $$\begin{array}{l|lllll} \hline \text { Height of arch, feet } & 3-9 & 10-29 & 30-49 & 50-74 & 75 \text { and higher } \\ \hline \begin{array}{l} \text { Number of arches } \\ \text { in park } \end{array} & 111 & 96 & 30 & 33 & 18 \\ \hline \end{array}$$ For an arch chosen at random in Arches National Park, use the preceding information to estimate the probability that the height of the arch opening is (a) 3 to 9 feet tall (d) 10 to 74 feet tall (b) 30 feet or taller (e) 75 feet or taller (c) 3 to 49 feet tall

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