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The probability that an archer hits the target when it is windy is \(0.4\); when it is not windy, her probability of hitting the target is \(0.7\). On any shot, the probability of a gust of wind is \(0.3\). Find the probability that: (a) On a given shot, there is a gust of wind and she hits the target. (b) She hits the target with her first shot. (c) She hits the target exactly once in two shots. (d) There was no gust of wind on an occasion when she missed.

Short Answer

Expert verified
(a) 0.12, (b) 0.61, (c) 0.1134, (d) 0.5385

Step by step solution

01

Understand the Problem

We are given that the probability the archer hits the target is conditional on whether it is windy. We need to calculate probabilities for scenarios involving being windy and not being windy. Let A be the event she hits the target, and W be the event it is windy.
02

Define the Probabilities

- Probability it is windy, \( P(W) = 0.3 \).- Probability it is not windy, \( P(W^c) = 0.7 \).- Probability she hits the target when it is windy, \( P(A|W) = 0.4 \).- Probability she hits the target when it is not windy, \( P(A|W^c) = 0.7 \).
03

Calculate Probability for Part (a)

We need to find \( P(W \cap A) \). Using the formula for conditional probability, \( P(W \cap A) = P(A|W) \cdot P(W) = 0.4 \cdot 0.3 = 0.12 \).
04

Calculate Probability for Part (b)

We use the law of total probability: \[ P(A) = P(A|W) \cdot P(W) + P(A|W^c) \cdot P(W^c) \]Substituting the known values, \[ P(A) = 0.4 \cdot 0.3 + 0.7 \cdot 0.7 = 0.12 + 0.49 = 0.61 \]
05

Calculate Probability for Part (c)

Find the probability of hitting exactly once out of two shots. Define the events:- Hits once out of two if either hits first and misses second or misses first and hits second.- Probability of missing when it is windy, \( P(A^c|W) = 1 - 0.4 = 0.6 \).- Probability of missing when it is not windy, \( P(A^c|W^c) = 1 - 0.7 = 0.3 \).- Probability of hitting within windy conditions is \( P(W \cap A) = 0.12 \) (from Step 3).- \( P(W^c \cap A) = P(A|W^c) \cdot P(W^c) = 0.7 \cdot 0.7 = 0.49 \).Calculate:\[ P( ext{hits exactly once}) = P( ext{hit first, miss second}) + P( ext{miss first, hit second}) \]The detailed calculation involves:- Hit first, miss second: - Wind first then no wind: \( P(W \cap A) \cdot P(W^c \cap A^c) = 0.12 \cdot 0.21 = 0.0252 \) - No wind first then wind: \( P(W^c \cap A) \cdot P(W \cap A^c) = 0.49 \cdot 0.18 = 0.0882 \)Sum these probabilities to get: \( 0.0252 + 0.0882 = 0.1134 \)
06

Calculate Probability for Part (d)

We want \( P(W^c|A^c) \). Using Bayes' theorem:\[ P(W|A^c) = \frac{P(A^c|W) \cdot P(W)}{P(A^c)} \]First calculate \( P(A^c) = 1 - P(A) = 1 - 0.61 = 0.39 \).Then, find \( P(W^c \cap A^c) = P(A^c|W^c) \cdot P(W^c) = 0.3 \cdot 0.7 = 0.21 \).Finally, compute:\[ P(W^c|A^c) = \frac{P(W^c \cap A^c)}{P(A^c)} = \frac{0.21}{0.39} = 0.5385 \]

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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 deals with predicting the likelihood of events occurring. It uses a scale of 0 to 1, where 0 represents an event that cannot happen, and 1 represents an event that is certain to happen. In probability problems, various events and their outcomes are analyzed to determine how probable they are.
For example, in the archer exercise, the probability of the weather being windy or not affects the likelihood of hitting a target.
By identifying the probability of each event (windy, not windy) and the subsequent effect on hitting the target, we apply probability theory to predict outcomes.
Law of Total Probability
The law of total probability is a fundamental concept in probability theory that helps compute the probability of an event by considering all possible scenarios. In essence, it breaks down the overall probability of an event into sub-events that contribute to it. These sub-events are generally mutually exclusive and cover all possibilities.
In our exercise, we apply this law to calculate the probability that the archer hits the target. We consider separate events: when it is windy and when it is not windy.
  • Probability of hitting the target when windy: \( P(A|W) \cdot P(W) = 0.4 \cdot 0.3 \)
  • Probability of hitting the target when not windy: \( P(A|W^c) \cdot P(W^c) = 0.7 \cdot 0.7 \)
Adding these probabilities gives the total probability of hitting the target. This method simplifies complex probability problems by considering different possible outcomes and their probabilities.
Bayes' Theorem
Bayes' theorem is a powerful mathematical tool used to update the probability of an event given new evidence. It relates the probability of event A given B to the reverse: the probability of B given A. This theorem helps to revise predictions or expectations as more data becomes available.
Within our exercise, Bayes’ theorem is applied to find the probability that there was no gust of wind given that the archer missed the target.
First, we find the probability that she missed the target (no hit), using our previous calculations:
\( P(A^c) = 1 - P(A) \).
We then calculate \( P(W^c|A^c) \), which is the probability that it wasn’t windy given the miss. This requires using Bayes' theorem to first determine \( P(W^c \cap A^c) \), and subsequently \( P(W^c|A^c) = \frac{P(W^c \cap A^c)}{P(A^c)} \).
Bayes' theorem is especially useful in real-world situations where evidence influences the likelihood of outcomes.
Independent and Dependent Events
Understanding the distinction between independent and dependent events is crucial in probability theory. Independent events are those whose occurrence or non-occurrence does not affect each other. The probability of one event happening has no impact on the probability of the other event.
Dependent events, on the other hand, are related such that the probability of one event changes because of the occurrence of another event.
In the exercise scenario, whether it is windy is an influencing (dependent) factor on whether the archer hits the target. The probability she hits the target changes based on whether the day is windy or not. This differs from independent events where the probability of one event would remain constant irrespective of another event.
Recognizing this dependency allows us to apply conditional probability and tools like Bayes' theorem to accurately calculate compound probabilities.

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