/*! 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 20 A box contains four \(40-W\) bul... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A box contains four \(40-W\) bulbs, five \(60-W\) bulbs, and six \(75-\mathrm{W}\) bulbs. If bulbs are selected one by one in random order, what is the probability that at least two bulbs must be selected to obtain one that is rated 75 W?

Short Answer

Expert verified
The probability is \(\frac{3}{5}\) that at least two bulbs must be selected to get a 75-W bulb.

Step by step solution

01

Determine Total Number of Bulbs

Count the total number of bulbs in the box. We have 4 bulbs of 40-W, 5 bulbs of 60-W, and 6 bulbs of 75-W. Therefore, the total number of bulbs is \(4 + 5 + 6 = 15\).
02

Identify the Favorable Outcome for 'At Least Two Selections'

To find the probability that at least two bulbs are needed to obtain a 75-W bulb, we consider the case where the first bulb selected is not 75-W (i.e., it must be either 40-W or 60-W).
03

Calculate the Probability of the First Selection Not 75-W

The probability that the first bulb selected is not 75-W is the probability of selecting either a 40-W or a 60-W bulb.There are \(4\) 40-W bulbs and \(5\) 60-W bulbs, so the total number of bulbs that are not 75-W is \(4 + 5 = 9\). Thus, the probability of selecting a bulb that is not 75-W in the first selection is:\[\frac{9}{15} = \frac{3}{5}\].
04

Confirm Initial Assumption Is Correct

Since we've calculated the probability of selecting a non-75-W bulb first, if at least two selections are required, we are guaranteed to pick a 75-W bulb by at least the second attempt or further after initially picking a non-75-W bulb.

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!

Key Concepts

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

Random Selection
When dealing with random selection, it's all about chance and equal opportunities. But what does this mean? Imagine you have a box with different types of bulbs, just like in our exercise. When you choose a bulb without looking, every bulb in the box has an equal chance of being picked. This process is random selection.

Here, we're concerned with making a random choice from a collection of 15 bulbs, each type of bulb being mixed together. Think of it like a fair coin toss, where each side has a 50% chance, except in our scenario, some bulbs might be more plentiful than others, modifying their likelihood of being selected first. This randomness adds an element of probability into the analysis, which we need to unlock by calculating different scenarios. Such a random act often requires a mathematical approach to truly understand the outcomes.

By approaching the exercise with randomness in mind, it becomes easier to see why multiple selections might be needed to achieve a particular result—specifically, picking a 75-W bulb. Remember, each pick is affected by pure chance, and every possibility is just one mystery of the random selection world.
Probability Calculation
Calculating probability is like decoding chance into numbers. Understanding probability involves evaluating how likely an event is to occur among all possible outcomes. In the exercise, we're calculating the probability of picking a bulb that is not 75-W on the first try.

This is how we broke it down:
  • We have 9 bulbs out of the total 15, which are either 40-W or 60-W.
  • The probability of the first pick being a non-75-W is thus \( \frac{9}{15} \).
By simplifying that fraction, we get \( \frac{3}{5} \). This means on a first random trial, there is a 60% probability of selecting either a 40-W or a 60-W bulb, implying one would need another selection to land a 75-W bulb.

Probability helps us prepare for the likely and unlikely, based on given data. Here, calculating the probability of non-selection of a 75-W bulb first provides insight into "at least two selections", which relies on confirming that our initial non-selection was not a 75-W bulb, aligning perfectly with the solution aimed for.
Mathematical Problem Solving
Mathematical problem solving unveils hidden truths within puzzles. Every step in a solution is crucial to unraveling the final answer. In our bulb selection problem, we employed logical reasoning paired with probability rules.

An important reign to problem-solving is to break the problem down. We began by identifying all elements—counting bulbs of different wattages, determining favorable conditions, and computing probabilities. This structured approach provides clarity and guides us to an accurate result.

The formulation: "The probability that at least two bulbs are selected to get one 75-W," requires understanding what happens in different scenarios. Here's how we concluded:
  • First selection isn’t 75-W, calculated to be \( \frac{3}{5} \).
  • Thus, you proceed to a second pick, anticipating at least one 75-W bulb.
It's the interplay between calculated probabilities and logical assumption-checks, dictating that mathematical solving is both an art of reasoning and science of calculations. The resolution, therefore, becomes not just a number or a probability but an understanding transferred into steps, illustrating both elegance and complexity within mathematical problems.

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

A particular state has elected both a govemor and a senator. Let \(A\) be the event that a randomly selected voter has a favorable view of a certain party's senatorial candidate, and let \(B\) be the corresponding event for that party's gubernatorial candidate. Suppose that \(P\left(A^{\prime}\right)=.44, P\left(B^{\prime}\right)=\) \(.57\), and \(P(A \cup B)=.68\) (these figures are suggested by the 2010 general election in California). a. What is the probability that a randomly selected voter has a favorable view of both candidates? b. What is the probability that a randomly selected voter has a favorable view of exactly one of these candidates? c. What is the probability that a randomly selected voter has an unfavorable view of at least one of these candidates.

A certain company sends \(40 \%\) of its overnight mail parcels via express mail service \(E_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (denote the event "late delivery" by \(L\) ). If a record of an overnight mailing is randomly selected from the company's file, what is the probability that the parcel went via \(E_{1}\) and was late?

The three major options on a car model are an automatic transmission \((A)\), a sunroof \((B)\), and an upgraded stereo \((C)\). If \(70 \%\) of all purchasers request \(A, 80 \%\) request \(B, 75 \%\) request \(C, 85 \%\) request \(A\) or \(B, 90 \%\) request \(A\) or \(C, 95 \%\) request \(B\) or \(C\), and \(98 \%\) request \(A\) or \(B\) or \(C\), compute the probabilities of the following events. [Hint: "A or \(B^{\prime \prime}\) is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and neither of the other two options. d. The next purchaser will select exactly one of these three options.

1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the extra remaining pump is … # A system consists of two identical pumps, #1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the extra remaining pump is now more likely to fail than was originally the case. That is, \(r=P(\\# 2\) fails \(\mid\) # 1 fails \()>P(\\# 2\) fails \()=q\). If at least one pump fails by the end of the pump design life in \(7 \%\) of all systems and both pumps fail during that period in only \(1 \%\), what is the probability that pump #1 will fail during the pump design life?

A chain of stereo stores is offering a special price on a complete set of components (receiver, compact disc player, speakers). A purchaser is offered a choice of manufacturer for each component: A switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions: a. In how many ways can one component of each type be selected? b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony? c. In how many ways can components be selected if none is to be Sony? d. In how many ways can a selection be made if at least one Sony component is to be included? e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?

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.