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Chapter 3: Problem 8

You ask your neighbor to water a sickly plant while you are on vacation. Without water it will die with probability .8; with water it will die with probability .15. You are 90 percent certain that your neighbor will remember to water the plant. (a) What is the probability that the plant will be alive when you return? (b) If 'it is dead, what is the probability your neighbor forgot to water it?

Short Answer

Expert verified
(a) The probability that the plant will be alive when you return is \(0.785\). (b) If the plant is dead, the probability that your neighbor forgot to water it is approximately \(0.3721\).

Step by step solution

01

Define the probabilities

Let's assign some variables to the probabilities given in the problem: - P(A): Probability that the plant is alive - P(D): Probability that the plant is dead - P(W): Probability that the neighbor waters the plant - P(F): Probability that the neighbor forgets to water the plant - P(D|W): Probability that the plant is dead, given that the neighbor waters it - P(D|F): Probability that the plant is dead, given that the neighbor forgets to water it We are given: - P(D|W) = 0.15 - P(D|F) = 0.8 - P(W) = 0.9 Since the neighbor either waters the plant or forgets, P(F) = 1 - P(W) = 1 - 0.9 = 0.1.
02

Find the probability of the plant being alive

We want to find P(A). We can find P(A) by using the Law of Total Probability, considering both scenarios (neighbor waters the plant or forgets). P(A) = P(A|W)P(W) + P(A|F)P(F) We can find P(A|W) and P(A|F) by noting that since the plant is either alive or dead, P(A|W) = 1 - P(D|W) and P(A|F) = 1 - P(D|F). P(A|W) = 1 - P(D|W) = 1 - 0.15 = 0.85 P(A|F) = 1 - P(D|F) = 1 - 0.8 = 0.2 Now, we can calculate P(A): P(A) = P(A|W)P(W) + P(A|F)P(F) = 0.85 * 0.9 + 0.2 * 0.1 = 0.765 + 0.02 = 0.785 So, the probability of the plant being alive is 0.785.
03

Find the probability that the neighbor forgot, given the plant is dead

We want to find P(F|D), which is the probability that the neighbor forgot to water the plant, given that it is dead. We can use Bayes' theorem for this calculation: P(F|D) = P(D|F)P(F) / P(D) First, we need to find P(D), which can be found since P(D) = 1 - P(A) = 1 - 0.785 = 0.215. Now, we can calculate P(F|D): P(F|D) = P(D|F)P(F) / P(D) = (0.8 * 0.1) / 0.215 鈮� 0.3721 The probability that the neighbor forgot to water the plant, given that the plant is dead, is approximately 0.3721. To summarize our results: (a) The probability that the plant will be alive when you return is 0.785. (b) If the plant is dead, the probability that your neighbor forgot to water it is approximately 0.3721.

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Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a fascinating concept in probability theory that allows us to update our beliefs based on new evidence. In our plant-watering problem, we wanted to find the probability that the neighbor forgot to water the plant, given that the plant ends up dead. This situation is a perfect candidate for using Bayes' Theorem. The formula for Bayes' Theorem is:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
In our context:
  • \( P(F|D) \) is the probability that the neighbor forgot given the plant is dead.
  • \( P(D|F) \) is the probability of the plant being dead given the neighbor forgot.
  • \( P(F) \) is the overall probability that the neighbor forgets.
  • \( P(D) \) is the total probability of the plant ending up dead.
Bayes' Theorem enabled us to bridge the gap between what we knew about the individual probabilities and what we wanted to find out about our neighbor's forgetfulness.
Law of Total Probability
The Law of Total Probability is a crucial concept when finding out the probability of an event that can happen in different ways. For the problem of the plant being alive, the Law of Total Probability helped us consider both scenarios: if the neighbor waters the plant or forgets.
This is expressed as:
  • \( P(A) = P(A|W)P(W) + P(A|F)P(F) \)
This breaks down the overall probability into probabilities conditioned on different scenarios (watering and forgetting). We calculate:
  • \( P(A|W) \) as the probability of the plant being alive if watered, which is given by \( 1 - P(D|W) \).
  • \( P(A|F) \) as the probability of the plant being alive if forgotten, which is \( 1 - P(D|F) \).
Thus, the Law of Total Probability allowed us to calculate that the plant will be alive with probability 0.785 by considering each potential outcome separately.
Probability Theory
Probability Theory is the foundation of analyzing and understanding random events. In problems like our plant situation, it provides a structured way to assign likely outcomes to real-world situations based on given data.
Key principles include:
  • Assigning probabilities to outcomes (e.g., probability of plant survival with or without water).
  • Understanding conditional probabilities, such as the chance of the plant dying or living based on whether it was watered.
  • Calculating total probabilities by using knowledge from different scenarios, as seen with the Law of Total Probability.
Probability theory forms the backbone that allows us to calculate, reason, and predict in uncertain conditions.
Problem Solving in Probability
Problem solving in probability often involves breaking down complex real-world situations into simpler, analyzable components. For example, in our plant-water problem, we identified different possible outcomes (plant living or dying) and actions (neighbor watering or forgetting) and used given probabilities to solve the problem.
Effective problem solving in probability includes several steps:
  • Clearly defining all events and their respective probabilities.
  • Using relevant theorems such as Bayes鈥� and the Law of Total Probability to find unknown probabilities.
  • Re-evaluating given known values to calculate unknown values, i.e., transforming the problem in manageable steps.
This structured approach makes probability questions less daunting by breaking them into solvable parts.

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Most popular questions from this chapter

Show that $$ \frac{P(H \mid E)}{P(G \mid E)}=\frac{P(H)}{P(G)} \frac{P(E \mid H)}{P(E \mid G)} $$ Suppose that before observing new evidence the hypothesis \(H\) is three times as likely to be true as is the hypothesis \(G\). If the new evidence is twice as likely when \(G\) is true than it is when \(H\) is true, which hypothesis is more likely after the evidence has been observed?

\(A\) and \(B\) are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of \(A\) will hit \(B\) with probability \(p_{A}\), and each shot of \(B\) will hit \(A\) with probability \(p_{B}\). What is (a) the probability that \(A\) is not hit; (b) the probability that both duelists are hit; (c) the probability that the duel ends after the \(n\)th round of shots; (d) the conditional probability that the duel ends after the \(n\)th round of shots given that \(A\) is not hit; (e) the conditional probability that the duel ends after the \(n\)th round of shots given that both duelists are hit?

Independent trials that result in a success with probability \(p\) are successively performed until a total of \(r\) successes is obtained. Show that the probability that exactly \(n\) trials are required is $$ \left(\begin{array}{l} n-1 \\ r-1 \end{array}\right) p^{r}(1-p)^{n-r} $$ Use this result to solve the problem of the points (Example 4i). HINT. In order for it to take \(n\) trials to obtain \(r\) successes, how many successes must occur in the first \(n-1\) trials?

An event \(F\) is said to carry negative information about an event \(E\), and we write \(F \searrow E\) if $$ P(E \mid F) \leq P(E) $$ Prove or give counterexamples to the following assertions: (a) If \(F \searrow E\), then \(E \searrow F\). (b) If \(F \searrow E\) and \(E \searrow G\), then \(F \searrow G\). (c) If \(F \searrow E\) and \(G \searrow E\), then \(F G \searrow E\). Repeat parts (a), (b), and (c) when \(\searrow\) is replaced by \(\lambda\), where we say that \(F\) carries positive information about \(E\), written \(F \nearrow E\), when \(P(E \mid F) \geq P(E)\)

A parallel system functions whenever at least one of its components works. Consider a parallel system of \(n\) components and suppose that each component independently works with probability \(\frac{1}{2}\). Find the conditional probability that component 1 works given that the system is functioning.
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