/*! 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 29 Three cards are drawn without re... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Three cards are drawn without replacement from a wellshuffled deck of 52 playing cards. What is the probability that the third card drawn is a diamond?

Short Answer

Expert verified
The probability that the third card drawn is a diamond is \(\frac{13}{50}\) or 26%.

Step by step solution

01

Understanding the problem

There are 52 cards in a well-shuffled deck, which includes 13 diamond cards and 39 non-diamond cards. We will be drawing three cards without replacement.
02

Calculate the probability of drawing a diamond on the third draw

To find the probability of drawing a diamond on the third draw, we first need to analyze different scenarios. There are three scenarios: 1. The first and second cards drawn are not diamonds: In this scenario, we have 39 non-diamond cards and 13 diamond cards remaining. 2. The first card is a diamond and the second card is not: In this scenario, we have 38 non-diamond cards and 12 diamond cards remaining. 3. The first card is not a diamond, but the second card is: In this scenario, we have 38 non-diamond cards and 12 diamond cards remaining. Notice that scenario 2 and 3 are identical; hence, we can consider the joint scenario and calculate the probability of drawing a diamond on the third draw.
03

Calculate probabilities for each scenario

1. Probability of drawing non-diamond, non-diamond cards, and then a diamond card: First two non-diamond cards: \( \frac{39}{52} \times \frac{38}{51} \) The third card is a diamond: \( \frac{13}{50} \) Total probability for this scenario: \( \frac{39}{52} \times \frac{38}{51} \times \frac{13}{50} \) 2. Probability of (drawing a diamond and then a non-diamond) or (drawing a non-diamond and then a diamond), and then drawing a diamond card: First card: diamond; second card: non-diamond: \( \frac{13}{52} \times \frac{39}{51} \) First card: non-diamond; second card: diamond: \( \frac{39}{52} \times \frac{13}{51} \) Total probability for this scenario: We need to multiply each case probability by \(\frac{12}{50}\), which is the probability of drawing a diamond card as the third card. \( \frac{12}{50} \times \left(\frac{13}{52} \times \frac{39}{51} + \frac{39}{52} \times \frac{13}{51} \right) \)
04

Calculate the total probability

Add the probabilities from step 3 to find the total probability. \( \frac{39}{52} \times \frac{38}{51} \times \frac{13}{50} + \frac{12}{50} \times \left(\frac{13}{52} \times \frac{39}{51} + \frac{39}{52} \times \frac{13}{51} \right) \)
05

Simplify the total probability

Simplify the expression to get the final probability. \(=\frac{13}{50}\) The probability that the third card drawn is a diamond is \(\frac{13}{50}\) or 26%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Copykwik has four photocopy machines: \(A, B, C\), and \(D .\) The probability that a given machine will break down on a particular day is \(P(A)=\frac{1}{50} \quad P(B)=\frac{1}{60} \quad P(C)=\frac{1}{75} \quad P(D)=\frac{1}{40}\) Assuming independence, what is the probability on a particular day that a. All four machines will break down? b. None of the machines will break down?

Suppose the probability that an event will occur in one trial is \(p\). Show that the probability that the event will occur at least once in \(n\) independent trials is \(1-(1-p)^{n}\).

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Let \(S=\left\\{s_{1}, s_{2}, \ldots, s_{n}\right\\}\) be a uniform sample space for an experiment. If \(n \geq 5\) and \(E=\left\\{s_{1}, s_{2}, s_{5}\right\\}\), then \(P(E)=3 / n\).

A study of the faculty at U.S. medical schools in 2006 revealed that \(32 \%\) of the faculty were women and \(68 \%\) were men. Of the female faculty, \(31 \%\) were full/ associate professors, \(47 \%\) were assistant professors, and \(22 \%\) were instructors. Of the male faculty, \(51 \%\) were full/associate professors, \(37 \%\) were assistant professors, and \(12 \%\) were instructors. If a faculty member at a U.S. medical school selected at random holds the rank of full/associate professor, what is the probability that she is female?

A halogen desk lamp produced by Luminar was found to be defective. The company has three factories where the lamps are manufactured. The percentage of the total number of halogen desk lamps produced by each factory and the probability that a lamp manufactured by that factory is defective are shown in the accompanying table. What is the probability that the defective lamp was manufactured in factory III? $$ \begin{array}{ccc} \hline & & \text { Probability of } \\ \text { Factory } & \text { Percent of } & \text { Defective } \\ \text { Total Production } & \text { Component } \\ \hline \text { I } & 35 & .015 \\ \hline \text { II } & 35 & .01 \\ \hline \text { III } & 30 & .02 \\ \hline \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.