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Chapter 2: Problem 40

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 -W bulb is found. What is the probability that it is necessary to examine at least \(\operatorname{six}\) bulbs?

Short Answer

Expert verified
a) Probability for exactly two 75-W bulbs is approximately 0.264. b) Probability for all same rating is approximately 0.051. c) Probability for one of each type is approximately 0.264. d) Probability for at least six selections is approximately 0.023.

Step by step solution

01

Determine Total Bulbs and Probability Basics

First, count the total number of bulbs: there are four 40-W bulbs, five 60-W bulbs, and six 75-W bulbs. This gives us a total of \(4 + 5 + 6 = 15\) bulbs. For any probability problem, the number of ways to choose a certain outcome divided by the total number of possible outcomes gives us the probability. For choosing 3 bulbs from 15, the number of ways is given by the binomial coefficient \(\binom{15}{3}\).
02

Calculate Probability for Exactly Two 75-W Bulbs

To have exactly two 75-W bulbs, we must choose 2 out of the 6 available 75-W bulbs, and 1 out of the remaining 9 bulbs (4 40-W + 5 60-W). The number of ways to do this is \(\binom{6}{2} \times \binom{9}{1}\). The probability is then given by \( \frac{\binom{6}{2} \times \binom{9}{1}}{\binom{15}{3}} \). Calculate the values of these binomial coefficients and find the probability.
03

Calculate Probability for All Bulbs Having Same Rating

We have three scenarios: choosing all 40-W bulbs, all 60-W bulbs, or all 75-W bulbs. For each, we calculate the number of ways to choose 3 bulbs from the respective quantity using the binomial coefficient \(\binom{4}{3}, \binom{5}{3}, \binom{6}{3}\). Sum these and divide by \(\binom{15}{3}\) for the total probability.
04

Calculate Probability for One Bulb of Each Type

This requires choosing 1 bulb from each of the three groups (40-W, 60-W, 75-W). The number of combinations for this scenario is the product of choosing 1 bulb from each wattage class: \(\binom{4}{1} \times \binom{5}{1} \times \binom{6}{1}\). Divide this product by \(\binom{15}{3}\) to find the probability.
05

Calculate Probability of Checking at Least Six Bulbs to Find a 75-W

This means in the first 5 selections, no 75-W bulb is found. Thus, all selected bulbs are either 40-W or 60-W. Calculate the probability of selecting 5 bulbs from the 9 bulbs (non 75-W) using \(\binom{9}{5}\). Divide this by \(\binom{15}{5}\) (ways to choose any 5 bulbs) to find the probability.

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Key Concepts

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

Combinatorics
Combinatorics involves counting and arranging elements within a set. In probability, combinatorics helps determine the number of possible outcomes for an event.
For instance, when selecting lightbulbs from a box, combinatorics can tell us how many ways we can pick our desired bulbs.
In the context of our exercise, we need to count:
  • The total number of ways to choose bulbs
  • The number of ways to achieve specific outcomes, such as selecting exactly two 75-W bulbs or ensuring all selected bulbs have the same wattage
Understanding combinatorics allows us to methodically calculate all possible selections or group arrangements, vital for finding probabilities.
Binomial Coefficient
The binomial coefficient, denoted as \( \binom{n}{k} \), represents the number of ways to choose \( k \) items from a set of \( n \) items, without considering the order.
It's crucial when calculating probabilities involving combinations.
For example, when choosing 3 bulbs from a box with 15 different bulbs, the binomial coefficient \( \binom{15}{3} \) tells us the number of ways to make that choice.
Use the formula:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Where \( n! \) (n factorial) is the product of all positive integers up to \( n \).
This concept aids in solving problems where order does not matter, such as finding the particular distributions of lightbulbs.
Random Sampling
Random sampling refers to selecting elements from a population in such a way that each element has an equal chance of being chosen. This ensures fairness and unpredictability in the sample selection.
In our exercise, when lightbulbs are randomly selected, it means each bulb, regardless of wattage, initially has the same probability of being picked.
This unpredictability is what makes the selection "random".
Random sampling is crucial for applying probability theory since accurate probability calculations must assume randomness in selection to be valid.
Every mild shift from randomness can significantly affect the probability outcomes and introduce biases in real-world scenarios.
Probability Distributions
A probability distribution describes how probabilities are distributed over the values of a random variable. In our context, it refers to the likelihood of drawing different combinations of lightbulbs.
To understand this concept:
  • Imagine different scenarios like pulling specific wattage combinations.
  • Each scenario is associated with a probability derived from the total possible combinations.
The problems solved involve finding these probabilities, illustrating common probability distributions.
For instance, determining the probability that all three selected bulbs have the same rating is a part of understanding how probabilities distribute across possible bulb combinations.
Probability distributions are key to interpreting the results of random sampling, allowing predictions about an overall population based on sample data.

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Most popular questions from this chapter

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