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A box in a certain supply room contains four 40 -W lightbulbs, five \(60-\mathrm{W}\) bulbs, and six \(75-\mathrm{W}\) bulbs. Suppose that three bulbs are randomly selected. a. What is the probability that exactly two of the selected bulbs are rated \(75-\) W? b. What is the probability that all three of the selected bulbs have the same rating? c. What is the probability that one bulb of each type is selected? d. Suppose now that bulbs are to be selected one by one until a \(75-\mathrm{W}\) bulb is found. What is the probability that it is necessary to examine at least six bulbs?

Short Answer

Expert verified
a) \(\frac{27}{91}\); b) \(\frac{34}{455}\); c) \(\frac{24}{91}\); d) \(\approx 0.04195\).

Step by step solution

01

Understand the Problem

We have a box with 4 40-W bulbs, 5 60-W bulbs, and 6 75-W bulbs. We need to calculate probabilities based on selecting 3 bulbs randomly without replacement.
02

Calculate Total Possibilities

The total number of ways to select 3 bulbs out of 15 is given by the combination formula \( \binom{15}{3} \). Compute this value:\[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \]
03

Calculate Probability for Part (a)

We want exactly two 75-W bulbs, so we choose 2 out of 6 75-W bulbs and 1 bulb from the remaining 9 (4 40-W and 5 60-W). Calculate selections and probability:Number of ways to choose 2 75-W bulbs: \( \binom{6}{2} = 15 \)Number of ways to choose 1 non-75-W bulb: \( \binom{9}{1} = 9 \)Total favorable outcomes: \( 15 \times 9 = 135 \)Therefore, the probability is:\[ P(\text{Exactly two 75-W}) = \frac{135}{455} = \frac{27}{91} \]
04

Calculate Probability for Part (b)

We want all three bulbs to be of the same wattage. Calculate the probabilities separately:- All 40-W bulbs: \( \binom{4}{3} = 4 \)- All 60-W bulbs: \( \binom{5}{3} = 10 \)- All 75-W bulbs: \( \binom{6}{3} = 20 \)Total favorable outcomes: \( 4 + 10 + 20 = 34 \)Therefore, the probability is:\[ P(\text{All three same}) = \frac{34}{455} \]
05

Calculate Probability for Part (c)

We want one bulb of each type. Calculate the number of ways to choose 1 bulb each from 40-W, 60-W, and 75-W:- 40-W: \( \binom{4}{1} = 4 \)- 60-W: \( \binom{5}{1} = 5 \)- 75-W: \( \binom{6}{1} = 6 \)Total favorable outcomes: \( 4 \times 5 \times 6 = 120 \)Therefore, the probability is:\[ P(\text{One of each}) = \frac{120}{455} = \frac{24}{91} \]
06

Calculate Probability for Part (d)

We need the probability that at least 6 bulbs are selected before finding the first 75-W bulb.To achieve this, all first 5 bulbs must be either 40-W or 60-W:- Total 40-W and 60-W bulbs = 9- Ways to select 5 non-75-W from 9:\[ \binom{9}{5} = 126 \]Thus the probability is:\[ P(\text{At least 6 bulbs examined}) = \frac{126}{\binom{15}{5}} = \frac{126}{3003} \approx 0.04195 \]

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Key Concepts

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

Combinatorics
Combinatorics is a fascinating branch of mathematics that deals with counting, arrangement, and combination of objects. It's the fundamental tool we use to solve problems involving selections, permutations, and combinations. In the exercise above, combinatorics helps us determine the number of ways to select certain numbers of bulbs from the total available. For instance, when we calculate how many ways we can pick 3 bulbs from a total of 15, we use the combination formula, denoted as \( \binom{n}{k} \) where \( n \) is the total number of items and \( k \) is the number of selections. This mathematical tool is essential for calculating probabilities, as it simplifies the process of considering all possible outcomes in a structured manner. It takes a complex scenario and translates it into a straightforward calculation.
Random Selection
Random selection ensures that each item in a given set has an equal chance of being chosen. In the context of the problem, selecting the light bulbs randomly means that any bulb could potentially be picked no matter its wattage. This principle of randomness ensures fairness and unpredictability in selection processes. It's essential in probability theory because it forms the base criteria for evaluating the likelihood of an event. In our exercise, when we select bulbs randomly, we need to consider all possible combinations and their respective probabilities correctly. It helps in understanding that through random selection, we're looking at an unbiased scenario which is critical for fair and accurate probability assessments.
Binomial Coefficients
Binomial coefficients are vital in probability calculations, particularly when dealing with selections. They are part of what we refer to as combinations, found using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). In the problem, these coefficients show up frequently to determine how many ways we can choose specific sets of bulbs. For example, choosing 2 bulbs rated 75-W out of 6 and 1 bulb from the remaining, involves calculating \( \binom{6}{2} \) and \( \binom{9}{1} \) respectively. This coefficient indicates how many different groups of items can be picked, regardless of the order, which is crucial when evaluating random scenarios where the order does not matter, just like picking bulbs from a box.
Replacement in Probability
Replacement in probability refers to whether or not items are returned to the original set after being chosen. This affects the total number of outcomes in a selection process. In our exercise, the bulbs are selected **without replacement**, which means each selection changes the number of total bulbs left in the box. Without replacement drastically affects calculations, as the probability of picking a bulb changes with each draw. It makes the problem more complex because it narrows down choices with each selection, thereby influencing the independent probabilities. Understanding whether a problem involves replacement is key to setting up accurate probability equations and solving problems correctly. In scenarios like this, each choice must adapt based on previous selections, reflecting the real-world constraints where items are removed from the pool once selected.

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Most popular questions from this chapter

A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company. a. If a night crew consists of 3 machinists, how many different crews are possible? b. If the machinists are ranked \(1,2, \ldots, 20\) in order of competence, how many of these crews would not have the best machinist? c. How many of the crews would have at least 1 of the 10 best machinists? d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

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