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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) 0.475 (d) 0.5385

Step by step solution

01

Probability of Wind and Hit

The event of interest is where there is a gust of wind, and she hits the target. Given:- Probability of it being windy: \( P(W) = 0.3 \)- Probability of hitting the target when windy: \( P(H|W) = 0.4 \) We find the joint probability: \[ P(W \cap H) = P(H|W) \cdot P(W) = 0.4 \times 0.3 = 0.12 \]
02

Probability of Hitting on First Shot

To find the probability she hits the target with the first shot, considering whether it is windy or not:- Probability it is not windy: \( P(NW) = 0.7 \)- Probability of hitting target without wind: \( P(H|NW) = 0.7 \)The total probability she hits the target is:\[ P(H) = P(W \cap H) + P(NW \cap H) = (0.3 \times 0.4) + (0.7 \times 0.7) = 0.12 + 0.49 = 0.61 \]
03

Probability of Hitting Exactly Once in Two Shots

Let's define two scenarios where she hits exactly one target:1. Hits on the first shot and misses on the second.2. Misses on the first shot and hits on the second.Calculate the probability she hits the first and misses the second:\[ P(H_1 \cap M_2) = P(H) \times P(M) = 0.61 \times (1 - 0.61) = 0.61 \times 0.39 \approx 0.2379 \]Now calculate the probability she misses the first and hits the second:\[ P(M_1 \cap H_2) = P(M) \times P(H) = 0.39 \times 0.61 = 0.2379 \]Add the two probabilities: \[ P(\text{Hit exactly once}) = P(H_1 \cap M_2) + P(M_1 \cap H_2) = 2 \times 0.2379 \approx 0.4758 \]
04

Probability of No Gust when Missed

Find the probability of no wind given she misses the target:- Probability of missing with wind: \( P(M|W) = 1 - P(H|W) = 0.6 \)- Probability of missing without wind: \( P(M|NW) = 1 - P(H|NW) = 0.3 \)Use Bayes’ theorem to determine: \[ P(NW|M) = \frac{P(NW \cap M)}{P(M)} \] Find missing total probability first:\[ P(M) = P(W \cap M) + P(NW \cap M) = (0.3 \times 0.6) + (0.7 \times 0.3) = 0.18 + 0.21 = 0.39 \]Compute probability:\[ P(NW|M) = \frac{0.21}{0.39} \approx 0.5385 \]

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Key Concepts

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

Conditional Probability
When dealing with probability, understanding conditional probability is crucial. Conditional probability is focused on finding the probability of an event occurring, given that another event has already occurred. In simpler terms, it investigates how the chance of one event is influenced by the occurrence of a different event.

In the archer problem, we are interested in finding how the probability of hitting the target changes when it is windy compared to when it is not. Mathematically, conditional probability is expressed as:
  • \( P(A|B) \): the probability of event \( A \) occurring given that \( B \) has occurred.
  • This can be calculated using the formula: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
In the solution provided, the conditional probabilities were defined as \( P(H|W) = 0.4 \) for hitting the target in windy conditions and \( P(H|NW) = 0.7 \) for non-windy conditions. Each probability tells us how likely it is to hit the target considering if there is supposed wind or not.
Bayes' Theorem
Bayes' theorem is a powerful tool in probability theory that allows one to update the probability measures of an event based on new evidence. It helps us reason backwards from observed events to deduce the probabilistic causes.

In the given problem, we used Bayes’ theorem to find the probability that there was no wind given that the target was missed, represented mathematically as \( P(NW|M) \). This theorem is pivotal when we need to "invert" the conditional probability using the known likelihoods:
  • The general form is: \[ P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)} \]
For the archer scenario:
  • \( P(NW|M) \) uses the precomputed probability of missing and whether that miss occurred during no wind. It provided insights into the likelihood of conditions based on the outcome.
Joint Probability
The concept of joint probability refers to the likelihood of two events occurring simultaneously. This tells us about the intersection of two events, often denoted as \( P(A \cap B) \).

In the archer’s case, step one sought the probability of both windy conditions occurring and hitting the target at the same time, expressed as \( P(W \cap H) \). This joint probability is calculated by multiplying the conditional probability of hitting the target in wind by the occurrence probability of wind:
  • \( P(W \cap H) = P(H|W) \cdot P(W) = 0.4 \times 0.3 = 0.12 \)
Understanding joint probability helps in finding situations where multiple criteria must be met simultaneously, which is crucial in probabilistic models.
Total Probability
The rule of total probability is an essential concept that provides a way to calculate the probability of an event by considering all possible scenarios or paths leading up to that event.

This rule is incredibly useful when you are dealing with a partitioned sample space and want to find the overall probability of an event across all these partitions.

In this exercise, calculating the probability that the archer hits the target on her first shot, considering both possibilities of wind presence and absence, involves applying the total probability concept:
  • The two separate scenarios (windy and not windy) are considered.
  • The individual probabilities from each scenario are summed to find the total probability of the target being hit.
  • The calculation was: \( P(H) = P(W \cap H) + P(NW \cap H) = 0.12 + 0.49 = 0.61 \)
By dissecting the total possibility into these distinct yet exhaustive segments, one can arrive at an overall measure of outcome effectively.

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Most popular questions from this chapter

Weather Days can be sunny or cloudy. The weather tomorrow is the same as the weather today with probability \(p\), or it is different with probability \(q\), where \(p+q=1\). If it is sunny today, show that the probability \(s_{n}\) that it will be sunny \(n\) days from today satisfies $$ s_{n}=(p-q) s_{n-1}+q ; \quad n \geq 1, $$ where \(s_{0}=1\). Deduce that $$ s_{n}=\frac{1}{2}\left(1+(p-q)^{n}\right) ; \quad n \geq 1 $$

A man has five coins in his pocket. Two are double-headed, one is double- tailed, and two are normal. They can be distinguished only by looking at them. (a) The man shuts his eyes, chooses a coin at random, and tosses it. What is the probability that the lower face of the coin is a head? (b) He opens his eyes and sees that the upper face is a head. What is the probability that the lower face is a head? (c) He shuts his eyes again, picks up the coin, and tosses it again. What is the probability that the lower face is a head? (d) He opens his eyes and sees that the upper face is a head. What is the probability that the lower face is a head?

A 12 -sided die \(A\) has 9 green faces and 3 white faces, whereas another 12 -sided die \(B\) has 3 green faces and 9 white faces. A fair coin is tossed once. If it falls heads, a series of throws is made with die \(A\) alone; if it falls tails then only the die \(B\) is used. (a) Show that the probability that green turns up at the first throw is \(\frac{1}{2}\). (b) If green turns up at the first throw, what is the probability that die \(A\) is being used? (c) Given that green turns up at the first two throws, what is the probability that green turns up at the third throw?

Candidates are allowed at most three attempts at a given test. Given \(j-1\) previous failures, the probability that a candidate fails at his \(j\) th attempt is \(p_{j}\). If \(p_{1}=0.6, p_{2}=0.4\), and \(p_{3}=0.75\), find the probability that a candidate: (a) Passes at the second attempt: (b) Passes at the third attempt: (c) Passes given that he failed at the first attempt; (d) Passes at the second attempt given that he passes.

Let \(A, B\) be two events with \(\mathbf{P}(B)>0\). Show that (a) If \(B \subset A\), then \(\mathbf{P}(A \mid B)=1\), (b) If \(A \subset B\), then \(\mathbf{P}(A \mid B)=\mathbf{P}(A) / \mathbf{P}(B)\).

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