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Chapter 2: Problem 131

Six tissues are extracted from an ivy plant infested by spider mites. The plant in infested in \(20 \%\) of its area. Each tissue is chosen from a randomly selected area on the ivy plant. (a) What is the probability that four successive samples show the signs of infestation? (b) What is the probability that three out of four successive samples show the signs of infestation?

Short Answer

Expert verified
(a) 0.0016 (b) 0.0256

Step by step solution

01

Understand the problem

We are given an infestation rate of 20% across an ivy plant. We're selecting tissues randomly, so each tissue selection represents a random trial with a probability of infestation of 20% (or 0.2). We need to determine the probability of finding a certain number of infested tissues in given sample sizes.
02

Determine the probability of a sample showing infestation

For each tissue sample, the probability of showing signs of infestation (success) is 0.2. The probability of not showing signs of infestation (failure) is 0.8. We treat this scenario as a binomial distribution where each sample is a Bernoulli trial.
03

Calculate (a): Probability that four successive samples are infested

We need the probability of 4 out of 4 successive samples being infested. Using the binomial probability formula where success probability (peq 0.2 ext{ and number of trials }n=4 ext{)}, we find \( P(X=4) = \binom{4}{4} (0.2)^4 (0.8)^0 = (0.2)^4 \).
04

Compute (a)

\( (0.2)^4 \) computes to 0.0016. Thus, the probability that all four samples are infested is 0.0016.
05

Calculate (b): Probability that three out of four samples are infested

Here, we again use the binomial probability formula for success with parameters: \(n = 4\), \(k = 3\), and \(p = 0.2\). \( P(X=3) = \binom{4}{3} (0.2)^3 (0.8)^1 = 4 \times (0.2)^3 \times 0.8 \).
06

Compute (b)

Compute \(4 \times (0.2)^3 \times 0.8\): \(4 \times 0.008 \times 0.8 = 0.0256\). Therefore, the probability is 0.0256.

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Key Concepts

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

Probability Calculation
Calculating probability involves determining the likelihood of a particular event occurring out of the total number of possible outcomes. In the context of this exercise, it refers to understanding how likely it is to find infested tissues when samples are taken from a plant. Each tissue selection can be viewed as a separate event, where either you find signs of infestation (success) or you don鈥檛 (failure).

To begin, we need to understand the basic terms:
  • Success Probability (p): This is the probability of a single trial resulting in a success, which in this case is an infested tissue. For the ivy plant, this probability is given as 0.2 or 20%.
  • Failure Probability (q): This is the probability of a single trial resulting in a failure, calculated as 1 - p. Therefore, the probability that a tissue is not infested is 0.8 or 80%.
  • Trials (n): This is the number of times the event/trial is repeated. Here, each tissue sample is one trial, so if we sample four tissues, we have four trials.
Understanding these probabilities helps us setup and solve problems using the binomial probability formula, a key aspect when dealing with multiple trials.
Binomial Probability Formula
The binomial probability formula is used to find the probability of having exactly \(k\) successes in \(n\) independent Bernoulli trials, with success probability \(p\). The formula is given by:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Here, \[\binom{n}{k}\] denotes the binomial coefficient, commonly calculated as \[\frac{n!}{k!(n-k)!}\], representing the number of ways to choose \(k\) successes from \(n\) trials.

For our scenario, this means calculating:
  • Four Infestations in Four Samples: Using \(n = 4\), \(k = 4\), and \(p = 0.2\), we find \(P(X=4)\). This is computed as \( (0.2)^4 \), leading to a probability of 0.0016.
  • Three Infestations in Four Samples: With \(n = 4\), \(k = 3\), and \(p = 0.2\), we calculate \(P(X=3)\). Utilizing the formula, the result is\(4 \times (0.2)^3 \times 0.8 = 0.0256\).
These calculations demonstrate how we use the binomial probability formula in practice, providing insights on the likelihood of different outcomes when sampling from our infested plant.
Bernoulli Trials
A Bernoulli trial is a simple random experiment that can have only two possible outcomes: success or failure. Each trial is independent of the others, meaning the outcome of one trial does not affect the others.

In the context of the ivy plant scenario:
  • Each tissue sample is a Bernoulli trial, where finding signs of infestation is regarded as a "success," and not finding them is a "failure."
  • The probability of success (infestation) is constant at 0.2 for each trial, reflecting the uniform infestation rate across the plant.
  • When combined, these trials can be analyzed using the binomial distribution to determine the probability of different numbers of successes (i.e., infested tissues in this case) over multiple trials.
Bernoulli trials underpin the calculations involving the binomial distribution, simplifying the study of sequences of random events like sampling plant tissues for infestation.

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