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You are playing with an ordinary deck of 52 cards by drawing cards at random and looking at them. a) Find the probability that the card you draw is (i) the ace of hearts (ii) the ace of hearts or any spade (iii) an ace or any heart (iv) not a face card. b) Now you draw the ten of diamonds, put it on the table and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card? c) Now you draw the ten of diamonds, return it to the deck and draw a second card. What is the probability that the second card is (i) the ace of hearts? (ii) not a face card?

Short Answer

Expert verified
a(i) \( \frac{1}{52} \), (ii) \( \frac{7}{26} \), (iii) \( \frac{4}{13} \), (iv) \( \frac{10}{13} \); b(i) \( \frac{1}{51} \), (ii) \( \frac{40}{51} \); c(i) \( \frac{1}{52} \), (ii) \( \frac{10}{13} \).

Step by step solution

01

Probability of Drawing the Ace of Hearts

To find the probability of drawing the ace of hearts from a standard 52-card deck, recognize that there is only one ace of hearts in the deck. Therefore, the probability is the ratio of favorable outcomes to the total number of outcomes: \[ P(Ace\ of\ Hearts) = \frac{1}{52} \]
02

Probability of Drawing the Ace of Hearts or Any Spade

To calculate this probability, note that there are 13 spades in the deck plus one additional ace of hearts that is not a spade. So, the total number of favorable outcomes is 14. The probability is: \[ P(Ace\ of\ Hearts\ or\ Spade) = \frac{14}{52} = \frac{7}{26} \]
03

Probability of Drawing an Ace or Any Heart

There are 4 aces in the deck and 13 hearts. However, the ace of hearts is counted twice in these sets, so we must subtract it once. The total number of favorable outcomes is therefore \( 4 + 13 - 1 = 16 \). Hence, the probability is: \[ P(Ace\ or\ Heart) = \frac{16}{52} = \frac{4}{13} \]
04

Probability of Drawing a Non-Face Card

A deck has 52 cards, where face cards (Jacks, Queens, Kings) total 12 (3 per suit). The non-face cards total \( 52 - 12 = 40 \). Therefore, the probability is: \[ P(Not\ a\ Face\ Card) = \frac{40}{52} = \frac{10}{13} \]
05

Probability of Drawing the Ace of Hearts After the First Draw

After drawing and keeping the ten of diamonds, the deck has 51 cards remaining. The number of ace of hearts remains 1. Thus, the probability of drawing the ace of hearts is: \[ P(Ace\ of\ Hearts\ |\ After\ First\ Draw) = \frac{1}{51} \]
06

Probability of Drawing a Non-Face Card After the First Draw

With 51 cards left and still 40 being non-face cards, the probability of drawing a non-face card is: \[ P(Not\ a\ Face\ Card\ |\ After\ First\ Draw) = \frac{40}{51} \]
07

Probability of Drawing the Ace of Hearts After Returning First Draw

Return the ten of diamonds to the deck, which restores the deck to 52 cards with one ace of hearts. So, the probability of drawing the ace of hearts is then: \[ P(Ace\ of\ Hearts\ |\ Deck\ Restored) = \frac{1}{52} \]
08

Probability of Drawing a Non-Face Card After Returning First Draw

Restoring the deck means there are 52 cards again with 40 non-face cards. The probability remains: \[ P(Not\ a\ Face\ Card\ |\ Deck\ Restored) = \frac{40}{52} = \frac{10}{13} \]

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Key Concepts

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

Understanding a Deck of Cards
A standard deck of cards consists of 52 cards divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards: Ace through 10, and three face cards: Jack, Queen, and King. This setup creates a fascinating way to explore probability as each card has an equal chance of being drawn. This means every individual card in the deck possesses a probability of \(\frac{1}{52}\) to be drawn on any given attempt. The organization into suits and values allows for different types of pairings and probabilities, such as finding the chance of drawing a specific value or a card from a particular suit.
Identifying Face Cards
Face cards in a deck are notable for their uniquely illustrated characters, commonly known as the King, Queen, and Jack. Each suit contains one of each of these face cards, resulting in 3 face cards per suit and 12 face cards in total for the entire deck. Since there are 52 cards in a deck, the probability of drawing a face card in one draw is the number of face cards divided by the total number of cards. Hence, the probability is \( \frac{12}{52} \) or simplified, \( \frac{3}{13} \). Understanding face cards is crucial as they often constitute specific scenarios in probability exercises such as avoiding face cards or calculating chances involving face cards.
Analyzing Card Drawing Probabilities
Card drawing involves selecting cards from a deck at random. When we talk about the probability of drawing a particular card or combination of cards, it's essential to consider how the deck changes with each draw. Initially, the deck has a known layout but drawing a card without replacement alters the total number of cards, thus changing the probabilities. For example, drawing and keeping a card changes the deck to contain 51 cards instead of 52, affecting calculations like \( P(Ace\ of\ Hearts\ \mid\ After\ First\ Draw) = \frac{1}{51} \). If you draw a card and return it, the probabilities revert to the original, unchanged state. This dynamic nature of card drawing is a fundamental aspect of understanding probabilities in card games.

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Most popular questions from this chapter

An experiment involves rolling a pair of dice, 1 white and 1 red, and recording the numbers that come up. Find the probability a) that the sum is greater than 8 b) that a number greater than 4 appears on the white die c) that at most a total of 5 appears.

Some young people do not like to wear glasses. A survey considered a large number of teenage students as to whether they needed glasses to correct their vision and whether they used the glasses when they needed to. Here are the results. $$\begin{array}{|l|c|c|c|} \hline \multirow{2}{*} & \multirow{2}{*}\multicolumn{2}{|c|}\text { Used glasses when needed }\text { Used glasses when needed } \\ \hline & & \text { Yes } & \text { No } \\ \hline \multirow{2}{*}\begin{array}{l} \text { Need glasses for } \\ \text { correct vision } \end{array} & \text { Yes } & 0.41 & 0.15 \\ \hline & \text { No } & 0.04 & 0.40 \\ \hline \end{array}$$ a) Find the probability that a randomly chosen young person from this group (i) is judged to need glasses (ii) needs to use glasses but does not use them. b) From those who are judged to need glasses, what is the probability that he she does not use them? c) Are the events of using and needing glasses independent?

Events \(A\) and \(B\) are given such that \(P(A)=\frac{7}{10} P(A \cup B)=\frac{9}{10}\) and \(P(A \cap B)=\frac{3}{10}\) Find a) \(P(B)\) b) \(P\left(B^{\prime} \cap A\right)\) c) \(P\left(B \cap A^{\prime}\right)\) d) \(P\left(B^{\prime} \cap A^{\prime}\right)\) e) \(P\left(B | A^{\prime}\right)\)

A class consists of 10 girls and 12 boys. A team of 6 members is to be chosen at random. What is the probability that the team contains a) one boy? b) more boys than girls?

In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate. Give reasons for your answer. a) A die is loaded such that the probability of each face is according to the following assignment ( \(x\) is the number of spots on the upper face and \(P(x)\) is its probability.) $$\begin{array}{c|cccccc} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & 0 & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6} & 0 \\\ \hline \end{array}$$ b) A student at your school categorized in terms of gender and whether they are diploma candidates or not. P(female, diploma candidate) \(=0.57,\) P(female, not a diploma candidate) \(=0.23\) \(\mathrm{P}\) (male, diploma candidate) \(=0.43, \mathrm{P}\) (male, not a diploma candidate) \(=0.18\) c) Draw a card from a deck of 52 cards ( \(x\) is the suit of the card and \(P(x)\) is its probability). $$\begin{array}{|c|c|c|c|c|} \hline x & \text { Hearts } & \text { Spades } & \text { Diamonds } & \text { Clubs } \\ \hline P(x) & \frac{12}{52} & \frac{15}{52} & \frac{12}{52} & \frac{13}{52} \\\ \hline \end{array}$$

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