91Ó°ÊÓ

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. 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 \(P(3 \text { on } 1 \text { st card and } 10 \text { on } 2 \text { nd })\) (c) Find \(P(10 \text { on } 1 \text { st card and } 3 \text { on } 2 \text { nd })\) (d) Find the probability of drawing a 10 and a 3 in either order.

Step by step solution

01

Determine Independence

Two events are independent if the outcome of one event does not affect the outcome of another. In this case, drawing a card from the deck changes the number of cards for the second draw, affecting probabilities. Therefore, the outcomes on the two drawn cards are not independent.

02

Calculate P(3 on 1st, 10 on 2nd)

Initially, there are 52 cards in total, and four of these are 3s. Thus, the probability of drawing a 3 first is \( \frac{4}{52} \). After drawing a 3, 51 cards remain, with four 10s to possibly draw. Hence, the probability of then drawing a 10 is \( \frac{4}{51} \). Therefore, the joint probability \( P(3 \text{ on 1st and } 10 \text{ on 2nd}) \) is calculated as:\[P(3 \text{ on 1st}) \times P(10 \text{ on 2nd | 3 on 1st}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}\]

03

Calculate P(10 on 1st, 3 on 2nd)

Initially, there are 52 cards in total, and four of these are 10s. Thus, the probability of drawing a 10 first is \( \frac{4}{52} \). After drawing a 10, 51 cards remain, with four 3s to possibly draw. Hence, the probability of then drawing a 3 is \( \frac{4}{51} \). Therefore, the joint probability \( P(10 \text{ on 1st and } 3 \text{ on 2nd}) \) is calculated as:\[P(10 \text{ on 1st}) \times P(3 \text{ on 2nd | 10 on 1st}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663}\]

04

Calculate Probability of Drawing a 10 and a 3 in Any Order

The probability of drawing a 10 and a 3 in any order combines the two mutually exclusive scenarios calculated in Steps 2 and 3: \( P(3 \text{ on 1st, 10 on 2nd}) \) and \( P(10 \text{ on 1st, 3 on 2nd}) \). Thus,\[P(3 \, \text{and} \, 10 \, \text{in any order}) = \frac{4}{663} + \frac{4}{663} = \frac{8}{663}\]

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Key Concepts

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

Deck of Cards

A standard deck of playing cards consists of 52 cards, divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards ranked from 2 up to Ace, including numbered cards and the three face cards, which are Jack, Queen, and King. The hearts and diamonds suits are red, while the clubs and spades suits are black.

• There are four cards of each rank, meaning there are four Aces, four Kings, and so on, down to four 2s.
• This structured distribution allows us to calculate probabilities for card draws when considering the deck as a whole.

Understanding the composition of a card deck is crucial for calculating probabilities related to card games and other similar exercises in probability.

Independent Events

Two events are considered independent if the occurrence of one event does not impact the probability of the other event occurring. In other words, the outcome of the first event has no bearing on the likelihood of the second.
In the context of drawing cards without replacement, the events are **not** independent. Drawing one card changes the number of remaining cards, consequently affecting the probabilities of any subsequent draws.
This shift highlights the importance of recognizing dependent versus independent events when calculating probabilities in exercises like these.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already happened. In card draws, it often comes into play when considering the sequential drawing of cards without replacement.
For instance, after drawing a 3 from the deck initially, the probability of drawing a 10 with the remaining 51 cards must consider that one card is now missing from the deck. This affects the denominator in the probability calculation.

• The probability of drawing a 3 first is \( \frac{4}{52} \).
• Provided that a 3 was drawn, the probability of a 10 next becomes \( \frac{4}{51} \).

This reveals how drives the need for adjusting probabilities based on prior outcomes in events.

Combinatorics

Combinatorics is a branch of mathematics concerned with counting, arranging, and finding probabilities in a finite structure. When approaching problems involving card draws, combinatorics helps determine the total number of possible outcomes and the likelihood of specific events occurring.
By calculating combinations and permutations, we find the number of successful outcomes relative to the total possibilities. For example, calculating the probability of drawing a 10 and a 3 in any order involves combining the probabilities of different sequences:

• Drawing a 3 then a 10: \( \frac{4}{663} \)
• Drawing a 10 then a 3: \( \frac{4}{663} \)

We add these probabilities to find the overall probability of drawing a 10 and a 3 in any sequence, showing how combinatorial principles support probability problem-solving.