/*! 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 151 Six tissues are extracted from a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Identify the Probability of Infestation

The probability that a single tissue sample is taken from an infested area is given as 20%, or 0.2. This will be our 'success' probability, denoted as \( p = 0.2 \).
02

Determine the Number of Trials

Since four successive samples are considered, we define this as four independent trials, where each trial represents a sample. Thus, the number of trials \( n \) is equal to 4.
03

Probability of Four Infested Samples

To find the probability that all four samples show signs of infestation, we calculate \( P(X = 4) \), where \( X \) is the number of infested samples. This is given by the binomial formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Substituting for \( n = 4 \), \( k = 4 \), and \( p = 0.2 \), we have: \[ P(X = 4) = \binom{4}{4} (0.2)^4 (0.8)^0 = 1 \cdot 0.0016 \cdot 1 = 0.0016 \]
04

Probability of Three Infested Samples

Next, we calculate the probability that three out of four samples show infestation, \( P(X = 3) \). Using the binomial formula, with \( k = 3 \): \[ P(X = 3) = \binom{4}{3} (0.2)^3 (0.8)^1 \] Calculating each part, we get: \[ \binom{4}{3} = 4, \ (0.2)^3 = 0.008, \ (0.8)^1 = 0.8 \] Thus, \[ P(X = 3) = 4 \times 0.008 \times 0.8 = 0.0256 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that studies the likelihood of events happening. It helps us to quantify the uncertainty of various outcomes. In the context of the ivy plant exercise, probability theory allows us to measure the likelihood that a certain number of tissue samples are infested by spider mites.
A key concept in probability is the distinction between independent and dependent events. Independent events are those whose outcomes do not affect each other. This concept is crucial in this exercise because the infestation of one tissue sample does not affect the infestation status of another, making each sampling an independent event.
In our scenario, the probability of one sample being infested is given as 20%, or 0.2. This value is used to calculate other probabilities related to multiple samples, as demonstrated in the provided solutions. Our calculations need to factor in both the probability of success (an infested sample) and the probability of failure (a non-infested sample), which is 0.8 in this case.
Statistical Analysis
Statistical analysis involves interpreting data to make informed decisions or predictions. In this exercise, we use a statistical tool called the Binomial Probability formula. This formula allows us to calculate the probability of a specific number of successes in a series of trials. A 'success' is defined in context — here, it refers to a tissue sample being infested.
The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:
  • \( n \) is the total number of trials (or samples),
  • \( k \) is the number of successful trials (or infested samples),
  • \( p \) is the probability of success in a single trial.
By applying statistical analysis and this formula, we calculated the probability for four infested samples as 0.0016 and for three as 0.0256. These calculations help in understanding the expected frequency of different outcomes.
Independent Trials
The concept of independent trials is fundamental in understanding and applying probability in real-world scenarios. Independent trials refer to a series of experiments or observations where the outcome of one does not affect the others. This idea is essential to apply binomial probability effectively.
In the ivy plant example, each tissue sample taken is an independent trial. Whether one sample is infested does not influence whether another will also be infested. Because these samples are considered independently, we can accurately use the binomial probability formula without introducing errors from dependence.
Independent trials assure us that the collected data truly reflect the randomness of the processes we're observing. In statistical terms, this means that each trial has precisely the same probability of success — in this case, 20% — and allows us to confidently predict and analyze patterns in outcomes, like the probabilities we calculated for different numbers of infested samples in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the manufacturing of a chemical adhesive, \(3 \%\) of all batches have raw materials from two different lots. This occurs when holding tanks are replenished and the remaining portion of a lot is insufficient to fill the tanks. Only \(5 \%\) of batches with material from a single lot require reprocessing. However, the viscosity of batches consisting of two or more lots of material is more difficult to control, and \(40 \%\) of such batches require additional processing to achieve the required viscosity. Let \(A\) denote the event that a batch is formed from two dif- ferent lots, and let \(B\) denote the event that a lot requires additional processing. Determine the following probabilities: (a) \(P(A)\) (b) \(P\left(A^{\prime}\right)\) (c) \(P(B \mid A)\) (d) \(P\left(B \mid A^{\prime}\right)\) (e) \(P(A \cap B)\) (f) \(P\left(A \cap B^{\prime}\right)\) (g) \(P(B)\)

Go Tutorial An optical storage device uses an error recovery procedure that requires an immediate satisfactory readback of any written data. If the readback is not successful after three writing operations, that sector of the disk is eliminated as unacceptable for data storage. On an acceptable portion of the disk, the probability of a satisfactory readback is \(0.98 .\) Assume the readbacks are independent. What is the probability that an acceptable portion of the disk is eliminated as unacceptable for data storage?

A lot contains 15 castings from a local supplier and 25 castings from a supplier in the next state. Two castings are selected randomly, without replacement, from the lot of \(40 .\) Let \(A\) be the event that the first casting selected is from the local supplier, and let \(B\) denote the event that the second casting is selected from the local supplier. Determine: (a) \(P(A)\) (b) \(P(B \mid A)\) (c) \(P(A \cap B)\) (d) \(P(A \cup B)\) Suppose that 3 castings are selected at random, without replacement, from the lot of \(40 .\) In addition to the definitions of events \(A\) and \(B,\) let \(C\) denote the event that the third casting selected is from the local supplier. Determine: (e) \(P(A \cap B \cap C)\) (f) \(P\left(A \cap B \cap C^{\prime}\right)\)

\(+\) If \(A, B,\) and \(C\) are mutually exclusive events with \(P(A)=0.2, P(B)=0.3,\) and \(P(C)=0.4,\) determine the follow. ing probabilities: (a) \(P(A \cup B \cup C)\) (b) \(P(A \cap B \cap C)\) (c) \(P(A \cap B)\) (d) \(P[(A \cup B) \cap C]\) (e) \(P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)\)

Heart failures are due to either natural occurrences \((87 \%)\) or outside factors \((13 \%) .\) Outside factors are related to induced substances \((73 \%)\) or foreign objects \((27 \%) .\) Natural occurrences are caused by arterial blockage \((56 \%),\) disease \((27 \%),\) and infection (e.g., staph infection) (17\%). (a) Determine the probability that a failure is due to an induced substance (b) Determine the probability that a failure is due to disease or infection.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.