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A bowl contains 10 chips numbered \(1,2, \ldots, 10\), respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability that two even-numbered chips are drawn and they occur on even- numbered draws?

Short Answer

Expert verified
The probability that two even-numbered chips are drawn and they occur on even-numbered draws is approximately 0.4.

Step by step solution

01

Determine the total number of outcomes

When 5 chips are drawn from a pool of 10 without replacement, the total number of outcomes can be calculated using combination notation: \( \binom{10}{5} = 252 \).
02

Determine the favorable outcomes

For favorable outcomes, the two even-numbered chips must be drawn on draws 2 and 4. So, there are \( \binom{5}{2} = 10 \) ways to select 2 even-numbered chips, and \( \binom{5}{3} = 10 \) ways to select the remaining 3 odd-numbered chips.
03

Calculate the probability

The probability is the ratio of favorable outcomes to total outcomes. Therefore, \( P = \frac{10 \times 10}{252} \) simplifying this equation results in \( P = \frac{100}{252} = 0.3968 \) or approximately 0.4 when rounded to one decimal place.

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Key Concepts

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

Permutations and Combinations
Understanding the fundamental difference between permutations and combinations is crucial when solving probability problems that involve the arrangement or selection of objects.

Permutations refer to arrangements where the order matters, while combinations are selections where the order is irrelevant. For instance, if you're drawing chips from a bowl and the sequence in which they're drawn is important, you're dealing with permutations. On the other hand, if only the selected chips matter, regardless of the order, combinations come into play.

In the given exercise, we're looking at combinations. Specifically, we calculate the total number of ways to draw 5 chips from 10 without regard to order, which is denoted as \( \binom{10}{5} \). It's a classic example of a combination, as the sequence of drawing is not considered here—only the fact that 5 chips are drawn out of 10.
Odds and Evens in Probability
Odds and evens often come up in probability problems, particularly in situations with a finite and well-defined set of outcomes. In our exercise, we focus on the probability of drawing even-numbered chips on even-numbered draws. This is a subset of a larger probability space and requires careful calculations.

The concept applies as we need to ensure that our selections meet specific criteria, which complicates the problem. The presence of conditions like drawing even numbers during even draws (second and fourth) has a profound impact on the resulting probability. The calculation combines understanding of both conditional probability and the basic principles of odds and evens within a probability framework.

To solve such problems effectively, it’s important to break down the conditions and understand how they interact with one another. Then, use permutations or combinations to calculate the desired probability accurately. It's also crucial to realize this impacts the nature of the calculation, as we switch from combinations to permutations once sequence-based conditions are introduced.
Mathematical Statistics
Mathematical statistics involves collecting, analyzing, interpreting, and presenting data. It underlies the principles of probability and helps us make sense of what the numbers actually tell us.

In the context of our problem, the calculation of probability is a form of statistical analysis. We gather all possible outcomes (the complete set of all 5-chip draws from the bowl) and then examine a specific event (drawing two even chips on the second and fourth draws). Mathematical statistics includes tools such as combination and permutation calculations, which are essential for solving our example problem.

After determining the number of favorable and possible outcomes, we calculate the probability—a statistical measure that gives us insight into the likelihood of an event occurring. Expressing the probability as a simplified fraction, decimal, or percentage aids clarity and supports better understanding, alignment with standard practices in mathematical statistics. Importantly, the interpretation of the result (in this case, approximately 0.4) is just as vital as the calculation itself, guiding us on the odds of our conditioned event occurring within a defined set of possibilities.

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Most popular questions from this chapter

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