A circular garden bed of radius \(1 \mathrm{~m}\) is to be planted so that \(N\) seeds are uniformly distributed oser the circular area. Then we can talk about the number \(n\) of seeds in some particular area \(A\), or we can call \(n / N\) the probability for any one particular seed to be in the area \(A\). Find the probability \(F(r)\) that a seed (that is, some particular seed) is within \(r\) of the center.

Short Answer

Expert verified
The probability \(F(r)\) that a seed is within \(r\) meters of the center is \(r^2\).

Step by step solution

01

Calculate the total area of the circular garden bed

The total area, \(A_{\text{total}}\), of a circle is given by the formula \(A = \pi r^2\). Substituting the radius of the garden bed, which is 1 meter, we find \(A_{\text{total}} = \pi (1)^2 = \pi \).
02

Calculate the area of a circle of radius r within the garden

The area, \(A_r\), of a circle with radius \(r\) is given by \(A = \pi r^2\). This represents the area within which we want to find the probability of a seed falling.
03

Use the probability interpretation

The probability \( F(r) \) that a seed lies within a radius \( r \) is given by the ratio of the area \(A_r \) to the total area \( A_{\text{total}} \). Thus, \( F(r) = \frac{A_r}{A_{\text{total}}} = \frac{\bf{(\bf{\frac{\pi r^2}{\bf{\pi}}}\)}}{\}= }\r^2 \ }.

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Key Concepts

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

circular garden bed
A circular garden bed is essentially a garden patch shaped like a circle. In this problem, the garden bed has a radius of 1 meter. This implies that any point on the boundary of this garden is exactly 1 meter away from its center.

Understanding the structure of the garden bed is crucial. All calculations and probabilities will be based on this circular layout. A circle's geometry simplifies many assumptions and calculations due to its symmetrical properties.

The fundamental attributes of the circle, like its radius and area, will form the basis of our discussion.
uniform distribution of seeds
When we say that seeds are uniformly distributed in the circular garden bed, it means every part of the garden has an equal chance of having seeds.

Imagine spreading seeds evenly across the garden. No matter where you look within the circle, the density of seeds remains consistent.

This uniform spread ensures that the probability calculations remain straightforward. Each section of the garden, regardless of its size, will have a proportionate share of seeds relative to its area.
area calculation
The area is an essential property when dealing with circles. The total area of our circular garden bed can be calculated using the formula for the area of a circle: \[\text{Area} = \pi r^2\].

Given that the radius \((r)\) of the garden is 1 meter:

\[A_{\text{total}} = \pi (1)^2 = \pi \text{ square meters}\]

This area forms the foundation for understanding how seeds are spread across the garden.

Similarly, for a smaller circle within the garden with a radius \((r)\), the area can again be computed using the same formula:

\[A_r = \pi r^2\].
probability within a radius
Probability helps us determine the chance that a particular event will occur. In this case, we are interested in finding the probability \(F(r)\) that a seed lies within a specific radius \((r)\) from the center of the garden.

The method involves comparing the area of the smaller circle (of radius \((r)\)) to the total area of the garden:

\[F(r) = \frac{A_r}{A_{\text{total}}} = \frac{\pi r^2}{\pi} = r^2\]

This result indicates a simple relationship: the probability \(F(r)\) that a seed falls within a radius \((r)\) from the center is directly proportional to \(r^2\). For example, if the radius \(r = 0.5\) meters, then:

\[F(0.5) = (0.5)^2 = 0.25 \(or 25%\)\]

This straightforward formula makes it easy to calculate the probability for any given radius within the circular garden.

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Most popular questions from this chapter

Using the normal approximation to the binomial distribution, and tables [or calculator for \(\phi(t)\) ], find the approximate probability of each of the following: Between 195 and 205 tails in 400 tosses of a coin.

(a) Find the probability that in two tosses of a coin, one is heads and one tails. That in six tosses of a die, all six of the faces show up. That in 12 tosses of a 12-sided die, all 12 faces show up. That in \(n\) tosses of an \(n\)-sided die, all \(n\) faces show up. (b) The last problem in (a) is equivalent to finding the probability that, when \(n\) balls are distributed at random into \(n\) boxes, each box contains exacty one ball. Show that for large \(n\), this is approximateiy \(e^{-n} \sqrt{2 \pi n}\)

You are trying to find instrument \(A\) in a laboratory. Unfortunately, someone has put both instruments \(A\) and another kind (which we shall call \(B\) ) away in identical unmarked boxes mixed at random on a shelf. You know that the laboratory has \(3 \mathrm{~A} \mathrm{~s}\) and \(7 \mathrm{~B}^{\prime} \mathrm{s}\). If you take down one box, what is the probability that you get an \(A\) ? If it is a \(B\) and you put it on the table and take down another box, what is the probability that you get an \(A\) this time?

Given a family of two children (assume boys and girls equally likely, that is, probability \(\frac{1}{2}\) for each), what is the probability that both are boys? That at least one is a girl? Given that at least one is a girl, what is the probability that both are girls? Given that the first two are giris, what is the probability that an expected third child will be a boy?

A student claims in Problem \(1.5\) that if one child is a girl, the probability that both are girls is \(\frac{1}{2}\). Lise appropriate sample spaces to show what is wrong with the following argument: It doesn't matter whether the girl is the older child or the younger; in either case the probabitity is \(\frac{1}{2}\) that the other child is a girl.

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