/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 You draw two cards from a standa... [FREE SOLUTION] | 91Ó°ÊÓ

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You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (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) Yes, the outcomes are independent. (b) \(\frac{1}{169}\) (c) \(\frac{1}{169}\) (d) \(\frac{2}{169}\)

Step by step solution

01

- Define Card Independence

Two events are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. Since after drawing the first card, we put it back and reshuffle the deck, the composition of the deck remains unchanged for the second draw. Therefore, the outcome of the first card does not influence the outcome of the second card.
02

- Calculate Probability of 3 on 1st and 10 on 2nd

The probability of drawing a 3 on the first card is \(\frac{4}{52}\) since there are 4 threes in the deck. The deck is reshuffled, so the probability of drawing a 10 on the second card is also \(\frac{4}{52}\). Using the formula for independent events, the probability is: \[P(3 \text{ on 1st and } 10 \text{ on 2nd}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169}.\]
03

- Calculate Probability of 10 on 1st and 3 on 2nd

Similar to Step 2, the probability of drawing a 10 on the first card is \(\frac{4}{52}\) and, after reshuffling, the probability of drawing a 3 on the second card is \(\frac{4}{52}\). Thus, the probability is: \[P(10 \text{ on 1st and } 3 \text{ on 2nd}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169}.\]
04

- Calculate Total 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: \[P(\text{3 and 10 in any order}) = \frac{1}{169} + \frac{1}{169} = \frac{2}{169}.\]

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Key Concepts

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

Independent Events
In probability theory, understanding independent events is crucial for accurate calculations. Two events are deemed independent when the occurrence of one event does not alter the probability of the other event occurring.
In simpler terms, knowing the outcome of the first event does not provide any information about the second event.
When dealing with card games, like in our exercise with a standard deck of cards, independence is critical. Here, since the first drawn card is put back into the deck and the deck is reshuffled, drawing the second card becomes an independent event from drawing the first card. This reshuffling maintains the equal probability for any card to be drawn each time. Thus, the drawing of cards fits the definition of independent events.
Probability Calculation
Calculating the probability of independent events involves using the multiplication rule. For two independent events, say Event A and Event B, the probability of both A and B occurring is given by:
  • \( P(A \text{ and } B) = P(A) \times P(B) \)
This means you multiply the individual probabilities of each event. In our exercise with the deck of cards, when calculating the probability of drawing specific cards in sequence, this rule applies.
For instance, both drawing a '3' on the first draw and a '10' on the second involves multiplying their individual probabilities. The same logic applies to calculating the reverse order.
By using this formula, you ensure that the independence of events is considered in your calculations, giving you a more precise probability for scenarios like the ones involving card draws.
Standard Deck of Cards
A standard deck of cards is a fundamental tool when learning probability due to its familiar structure and limited number of outcomes. Let's explore its composition to understand probability calculations better.
A standard deck consists of 52 cards, with each card belonging to one of four suits: hearts, diamonds, clubs, and spades.
Each suit contains 13 cards, ranging from Ace through to King, which includes numbers 2 through 10 and the face cards Jack, Queen, and King.
This consistent structure means that for any given card, like a '3' or a '10', there are always 4 cards of that kind within the deck, assuming no card has been removed. Thus, for any specific card, the probability of drawing it is
  • \( \frac{4}{52} \)
By thoroughly understanding the setup of a standard deck, you can more effectively apply probability concepts, especially for exercises that involve sequential draws with replacement, as every draw is independent and probabilities remain the same.

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Most popular questions from this chapter

A recent Harris Poll survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

What is the probability that a day of the week selected at random will be a Wednesday?

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Expand Your Knowledge: Odds Against Betting odds are usually stated against the event happening (against winning). The odds against event \(W\) are the ratio \(\frac{P(n o t W)}{P(W)}=\frac{P\left(W^{c}\right)}{P(W)}\). In horse racing, the betting odds are based on the probability that the horse does not win. (a) Show that if we are given the odds against an event \(W\) as \(a: b\), the probability of not \(W\) is \(P\left(W^{c}\right)=\frac{a}{a+b} .\) Hint \(:\) Solve the equation \(\frac{a}{b}=\frac{P\left(W^{c}\right)}{1-P\left(W^{c}\right)}\) for \(P\left(W^{c}\right)\) (b) In a recent Kentucky Derby, the betting odds for the favorite horse, Point Given, were 9 to \(5 .\) Use these odds to compute the probability that Point Given would lose the race. What is the probability that Point Given would win the race? (c) In the same race, the betting odds for the horse Monarchos were 6 to 1 . Use these odds to estimate the probability that Monarchos would lose the race. What is the probability that Monarchos would win the race? (d) Invisible Ink was a long shot, with betting odds of 30 to \(1 .\) Use these odds to estimate the probability that Invisible Ink would lose the race. What is the probability the horse would win the race? For further information on the Kentucky Derby, visit the Brase/Brase statistics site at http://www.cengage .com/statistics/brase and find the link to the Kentucky Derby.

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