/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A forest contains 20 elk, of whi... [FREE SOLUTION] | 91Ó°ÊÓ

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A forest contains 20 elk, of which 5 are captured, tagged, and then released. A certain time later 4 of the 20 elk are captured. What is the probability that 2 of these 4 have been tagged? What assumptions are you making?

Short Answer

Expert verified
The probability of capturing exactly 2 tagged elk out of 4 is \(\frac{70}{323}\). The assumptions made are that the elk are being captured randomly and independently, and that tagging an elk does not affect its likelihood of being captured again.

Step by step solution

01

The total number of ways to capture 4 elk out of 20 is given by the combinations formula, which is: \[C(n, r) = \frac{n!}{r!(n - r)!}\] For our problem, n = 20 and r = 4, so: \[C(20, 4) = \frac{20!}{4!(20 - 4)!}\] #Step 2: Calculate the total possible combinations#

Plugging the values for n and r into the formula, we get: \[C(20, 4) = \frac{20!}{4!(16)!}\] \[C(20, 4) = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1}\] \[C(20, 4) = 4845\] There are 4845 possible ways to choose 4 elk out of 20. #Step 3: Determine the successful outcomes#
02

We want to find the probability of capturing exactly 2 tagged elk out of 4. There are 5 tagged elk and 15 untagged elk in the forest. It means we must choose 2 elk from the tagged elk and 2 elk from the untagged elk. The number of ways to choose 2 tagged elk out of 5 is: \[C(5, 2) = \frac{5!}{2!(5 - 2)!}\] \[C(5, 2) = 10\] The number of ways to choose 2 untagged elk out of 15 is: \[C(15, 2) = \frac{15!}{2!(15 - 2)!}\] \[C(15, 2) = 105\] #Step 4: Calculate the total successful outcomes#

We multiply the number of ways to choose 2 tagged and 2 untagged elk: \(10 \times 105 = 1050\) There are 1050 successful outcomes (exactly 2 tagged elk and 2 untagged elk). #Step 5: Calculate the probability#
03

We can now determine the probability of capturing exactly 2 tagged elk. The probability is given by the ratio of successful outcomes to total possible outcomes: \[P(\text{2 tagged}) = \frac{\text{successful outcomes}}{\text{total outcomes}}\] \[P(\text{2 tagged}) = \frac{1050}{4845}\] #Step 6: Simplify the probability#

Both the numerator and the denominator can be divided by 15, so we simplify the fraction: \[P(\text{2 tagged}) = \frac{70}{323}\] Thus, the probability of capturing exactly 2 tagged elk out of 4 is \(\frac{70}{323}\).

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Key Concepts

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

Combinatorics
Combinatorics deals with counting, arrangement, and combination of objects. This field is fundamental in probability theory, as it helps us calculate possibilities and probabilities. It becomes especially useful when determining how many ways a certain event, like capturing elk, can happen.
In the context of these elk, we use combinations to find out how many different groups of 4 elk can be created out of 20. Combinations answer the question: "How many ways can I choose r objects from a set of n objects without regard to the order of selection?"
The notation used for combinations is often \( C(n, r) \) or read as "n choose r." The formula for calculating combinations is:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
Where \( n! \) (factorial of n) is the product of all positive integers up to n. This formula derives from the fact that when choosing r objects out of n, there are n choices for the first object, n-1 for the second, all the way down to n-r+1 choices, and so on.
In our elk example, the reason we calculate \( C(20, 4) \) is to understand the total possible combinations from which the tagged ones can be drawn.
Binomial Coefficient
The binomial coefficient is a central concept in probability and combinatorics. It is represented by the combinations symbol \( C(n, r) \). This coefficient gives the number of ways to choose r items from n items without considering the order. Hence, it is exactly the number of ways some event can happen in random situations.
The binomial coefficient formula, \( C(n, r) = \frac{n!}{r!(n-r)!} \), tells us about these combinations. In our problem, calculating \( C(5, 2) \) gives us the number of ways to choose 2 tagged elk from a group of 5. This same approach calculates \( C(15, 2) \) for the 2 untagged elk out of 15.
These coefficients help differentiate between looking at all ways to choose items (all possibilities) versus successful specific attributes (like specifically tagged elk), thus drilling down into targeted probabilities. They simplify calculations, turning complex probability scenarios into solvable mathematical problems.
Tagged Population Estimation
Estimating a tagged population can relate to ecological studies, such as tracking animal populations, and involves probability to predict or understand certain quantities based on limited data. In this scenario, after tagging some elk, we estimate how many of the newly captured elk are from the previously tagged set.
In a more general sense, this concept models real-world cases where only a subset of a population is known or modified, like tagging elk. Key assumptions made usually include:
  • Tagged individuals mix back into the population randomly and thoroughly.
  • There is no loss of tags, and tagged individuals have the same likelihood of capture.

Using combinations determines how likely it is to randomly select a certain number of tagged individuals from a previously captured and released group. It highlights how well these assumptions hold by reflecting a realistic probability under those modeled conditions.
This estimation approach allows ecologists and researchers to make scientifically sound predictions about animal behaviors and population dynamics, even in complex ecosystems with limited direct observation.

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Most popular questions from this chapter

A customer visiting the suit department of a certain store will purchase a suit with probability .22, a shirt with probability .30, and a tie with probability 28. The customer will purchase both a suit and a shirt with probability .11, both a suit and a tie with probability .14, and both a shirt and a tie with probability ,10. A customer will purchase all 3 items with probability \(.06 .\) What is the probability that a customer purchases (a) none of these items: (b) exactly 1 of these items?

For any sequence of events \(E_{1}, E_{2}, \ldots\), define a new sequence \(F_{1}, F_{2}, \ldots\) of disjoint events (that is, events such that \(F_{i} F_{j}=\varnothing\) whenever \(i \neq j\) ) such that for all \(n \geq 1\), $$ \bigcup_{1}^{n} F_{i}=\bigcup_{1}^{n} E_{i} $$

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