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A box contains four \(40-\mathrm{W}\) bulbs, five \(60-\mathrm{W}\) bulbs, and six 75 -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 \(75 \mathrm{~W}\) ?

Short Answer

Expert verified
The probability is \( \frac{9}{35} \).

Step by step solution

01

Understand the Total Number of Bulbs

First, we need to calculate the total number of bulbs. The box contains 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

Probability of Selecting a Non-75W Bulb First

To find the probability that the first bulb selected is not 75W, we need to sum the non-75W bulbs, which are the 40W and 60W bulbs. Together, they make:\[ 4 + 5 = 9 \text{ bulbs} \]Hence, the probability of selecting a non-75W bulb first is:\[ \frac{9}{15} = \frac{3}{5} \]
03

Probability of Selecting a 75W Bulb Second

If the first bulb is not a 75W, we then calculate the probability of selecting a 75W bulb on the next draw. After one non-75W bulb is removed, there are 14 bulbs left:The 75W bulbs remain as 6, so the probability is:\[ \frac{6}{14} = \frac{3}{7} \]
04

Calculate Probability of the Event

Since the problem asks that at least two bulbs are selected to obtain a 75W bulb, we need to multiply the probabilities from Step 2 and Step 3:\[ \frac{3}{5} \times \frac{3}{7} = \frac{9}{35} \]
05

Interpret the Result

The calculation from Step 4 gives us the probability of needing to select at least two bulbs to get a 75W bulb. Thus, the probability is \( \frac{9}{35} \).

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Key Concepts

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

Random Selection
In this exercise, the concept of random selection is crucial. It refers to the idea that each bulb picked from the box is chosen entirely by chance, with no bias towards any specific type. When we randomly select, it implies that every bulb has an equal opportunity to be chosen during each draw. This is important as it allows us to calculate probabilities using the law of probability.

Random selection ensures fairness and unpredictability in the selection process, which is a critical aspect of probability exercises. In our scenario, each draw that starts with a total of 15 bulbs will have the same initial probability distribution, regardless of which bulb is picked first or second. This unbiased approach forms the foundation for calculating the likelihood of picking a non-75W bulb first and a 75W bulb on subsequent draws.
Step by Step Solution
The step-by-step solution approach ensures that we break down complex problems into manageable parts, making it easier for students to follow and understand.
  • First, by identifying and summing up the total number of bulbs, we establish the sample space of our probability problem, which is 15 in this case.
  • Next, calculating the probability of first picking a non-75W bulb narrows down our focus by initially considering only the possible non-target outcomes.
  • The subsequent step involves identifying the outcome we want on the second draw: picking a 75W bulb, and then determining the new probability.
  • Finally, multiplying these probabilities enables us to find the likelihood of the combined events (picking a non-75W first, followed by a 75W).
Each of these steps simplifies the process of solving the problem by narrowing our focus and using probability calculations to answer the exercise's question.
Cognitive Understanding
Cognitive understanding refers to the mental process of comprehending, processing, and applying information. When students approach a probability exercise, it's important for them to understand rather than memorize solutions.

By working through logical steps, students reinforce their understanding of probability laws and their application. For example, when students comprehend why each bulb has a chance of being selected and how outcomes influence future possibilities, they enhance their conceptual grasp of randomness and probability.

The step-by-step breakdown aids cognitive understanding by making each part of the solution clear and connectable. This clarity in smaller chunks supports building a strong foundation in solving not only this specific problem but also more complex probability exercises they might encounter.
Problem-Solving Skills
Developing problem-solving skills is essential for mathematical tasks such as this one. These skills involve analytical thinking, making connections between concepts, and applying learned techniques to new situations.

Through this exercise, students practice:
  • Identifying necessary information (like total and relevant numbers of bulbs).
  • Isolating parts of the problem (finding probability by focusing on one bulb at a time).
  • Connecting the solutions of each small step to form the final answer.
These skills are not only valuable in math but also in scientific inquiries, everyday tasks, and future learning scenarios. By engaging in exercises like these, students learn to think critically and solve problems systematically, emphasizing the process rather than just the outcome.

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Most popular questions from this chapter

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent. [Hint: First establish a relationship between \(P\left(A^{\prime} \cap B\right), P(B)\), and \(P(A \cap B)\).]

A college library has five copies of a certain text on reserve. Two copies ( 1 and 2 ) are first printings, and the other three \((3,4\), and 5\()\) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5 , and another is \(213 .\) a. List the outcomes in \(\mathcal{S}\). b. Let \(A\) denote the event that exactly one book must be examined. What outcomes are in \(A\) ? c. Let \(B\) be the event that book 5 is the one selected. What outcomes are in \(B\) ? d. Let \(C\) be the event that book 1 is not examined. What outcomes are in \(C\) ?

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+\) \(P\left(A^{\prime} \mid B\right)=1\).

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, cassette deck). A purchaser is offered a choice of manufacturer for each component: Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood Compact disc player: Onkyo, Pioneer, Sony, Technics Speakers: Boston, Infinity, Polk Cassette deck: Onkyo, Sony, Teac, Technics 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?

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