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ViOLATIONS OF THE BuILDING CoDE Suppose that one third of the new buildings in a town are in violation of the building code. If a building inspector inspects five of the buildings chosen at random, find the probability that a. The first three buildings will pass the inspection and the remaining two will fail the inspection. b. Exactly three of the buildings will pass inspection.

Short Answer

Expert verified
a. The probability that the first three buildings will pass inspection and the remaining two will fail is approximately \(0.027\). b. The probability that exactly three of the buildings will pass inspection is approximately \(0.329\).

Step by step solution

01

Part a: First three buildings pass, remaining two fail

For this specific arrangement, we need to find the probability of the first three inspections resulting in passing and the remaining two inspections resulting in buildings violating the building code. In this case, we can just multiply the probabilities for each of the five events since they are independent: P(first three pass, last two fail) = P(1st passes) * P(2nd passes) * P(3rd passes) * P(4th fails) * P(5th fails) Substituting the probabilities for each event: P(first three pass, last two fail) = \((\frac{2}{3})^3 * (\frac{1}{3})^2\) Now calculate the probability: P(first three pass, last two fail) ≈ 0.027
02

Part b: Exactly three of the buildings will pass inspection

Now we need to calculate the probability of exactly three buildings passing inspection, regardless of their order. For this, we will be using the binomial probability formula that we discussed earlier. In this case, we have: - n = 5 (total number of buildings inspected) - k = 3 (number of buildings that pass inspection) - p = 2/3 (probability of a building passing inspection) Now plug these values into the binomial probability formula: P(3 buildings pass) = C(5, 3) * \((\frac{2}{3})^3 * (\frac{1}{3})^2\) To find the number of combinations (C(5, 3)): C(5, 3) = \(\frac{5!}{3!(5-3)!}\) = 10 Substitute this value back into the formula: P(3 buildings pass) = 10 * \((\frac{2}{3})^3 * (\frac{1}{3})^2\) Now calculate the probability: P(3 buildings pass) ≈ 0.329 So the probability of having exactly three buildings pass the inspection is approximately 0.329.

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Key Concepts

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

Binomial Probability Formula
Understanding the binomial probability formula is essential to solving problems involving a fixed number of independent trials, each with two possible outcomes - success or failure. This formula is invaluable when you need to find the probability of a certain number of successes in a given number of trials.

To apply the binomial probability formula, you must know three key parameters: the number of trials (), the probability of success in a single trial (p), and the number of successes(k) you're interested in. The formula is written as:
[P(k successes in n trials) = C(n, k) * p^k * (1-p)^(n-k)]

where C(n, k) is the number of combinations of n items taken k at a time. This component of the formula ensures that we're considering all possible arrangements of successes and failures.

For instance, in the building code inspection scenario, to find the probability of exactly three buildings passing (which we label as successes), we use the formula with p = 2/3, n = 5, and k = 3. Calculating this gives us the probability that out of five inspections, three will be passes and two will be fails, regardless of the order of these outcomes.
Combinatorial Mathematics
Combinatorial mathematics is the field of mathematics that deals with counting, arranging, and combining objects. This field is the backbone of determining the number of ways a specific event can occur, which is essential in calculating probabilities.

When we speak about combinations in the context of probability, we refer to the number of ways in which we can choose k items from a larger set of n items, without regard to the order of selection. The formula to calculate combinations is:
[C(n, k) = frac{n!}{k!(n-k)!}]

Here, the symbol ! denotes the factorial operation, which involves multiplying a series of descending natural numbers. For the building inspection problem, to determine the number of combinations for three passes out of five buildings (C(5, 3)), we calculated it to be 10. This signifies there are 10 different ways in which any three of the five buildings can pass the inspection.
Probability Calculation
Probability calculation is the quantitative measurement of the likelihood of an event’s occurrence. The fundamental rule in probability is that the probability of all possible outcomes must equal 1.

The general approach to calculating probability is to divide the number of favorable outcomes by the total number of possible outcomes. In certain scenarios, like the building code inspection, we multiply the probabilities of independent events to get the combined probability of a series of outcomes.

Continuing with our building inspection example, the calculations involve multiplying the individual probabilities of each independent building inspection. The calculation for the first part of the scenario, where the first three buildings pass and the last two fail, is simply the multiplication of each independent probability: (frac{2}{3})^3 * (frac{1}{3})^2, which makes the probability calculation quite direct and straightforward.
Independent Events
Independent events are pivotal in the realm of probability, as they refer to events where the outcome of one does not affect the outcome of another. In simpler terms, the probability of one event occurring has no bearing on the probability of the other event occurring.

When dealing with independent events, we apply the multiplication rule. This rule states that the probability of two or more independent events occurring together is the product of their individual probabilities.
[P(Event A and Event B) = P(Event A) * P(Event B)]

In the context of our exercise, the inspections of individual buildings are considered independent events. The probability of each building passing or failing the inspection is independent of the results from other buildings. The multiplication rule allows us to find the probability that the first three buildings inspected will pass and the last two will fail by simply multiplying their respective probabilities together.

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