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Irena throws at a target. After each throw she moves further away so that the probability of a hit is two-thirds of the probability of a hit on the previous throw. The probability of a hit on the first throw is \(\frac{1}{4}\). Find the probability of a hit on the \(n\)th throw. Deduce that the probability of never hitting the target is greater than \(\frac{1}{4}\).

Short Answer

Expert verified
\( P_n = \frac{1}{4} \left(\frac{2}{3}\right)^{n-1} \). Probability of never hitting: \( \frac{1}{4} \).

Step by step solution

01

Understand the Problem

The probability changes with each throw based on its relationship to the previous throw. We start with a given probability for the first throw and reduce it by two-thirds for each subsequent throw.
02

Define the Sequence

Let \( P_n \) be the probability of hitting the target on the \( n \)th throw. We know \( P_1 = \frac{1}{4} \) and \( P_{n+1} = \frac{2}{3}P_n \). This defines a geometric sequence.
03

Derive the General Formula

To find \( P_n \), we recognize that it follows the pattern of a geometric sequence: \( P_n = \frac{1}{4} \left(\frac{2}{3}\right)^{n-1} \). This represents the probability of each subsequent throw being two-thirds of the previous one.
04

Calculate Probability of Never Hitting the Target

The probability of never hitting the target is the complement of at least one hit. Using the formula for an infinite geometric series, \( S = \sum_{n=1}^{\infty} P_n = \frac{\frac{1}{4}}{1 - \frac{2}{3}} = \frac{3}{4} \). Thus, the probability of never hitting the target is \( 1 - S = 1 - \frac{3}{4} = \frac{1}{4} \).
05

Conclusion on the Probability

It was previously believed that the probability of never hitting the target should be greater than \( \frac{1}{4} \), however, on calculating, it exactly equals \( \frac{1}{4} \). Therefore, the given condition that it should be greater than \( \frac{1}{4} \) seems incorrect based on calculation.

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Key Concepts

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

Probability Calculation
Probability calculation involves determining the likelihood of an event occurring. To calculate a probability, you generally divide the number of favorable outcomes by the total number of possible outcomes. In the context of Irena's target practice, calculating the probability for each throw involves understanding that because she moves further back after every throw, the probability of hitting the target decreases exponentially.

For Irena's case, we use her initial probability of hitting the target, which is \( \frac{1}{4} \), and then multiply it by \( \frac{2}{3} \) for each subsequent throw. This decrease follows a geometric progression, making each \( n^{th} \) throw's probability given by the geometric sequence formula: \( P_n = \frac{1}{4} \left( \frac{2}{3} \right)^{n-1} \).

- **Key Points**:
  • Understand how probabilities change with consecutive trials and distances.
  • Recognize the geometric sequence pattern used to determine subsequent probabilities.
Infinite Series
An infinite series involves summing a sequence of numbers indefinitely. In mathematics, particularly in probability and calculus, infinite series are crucial when you want to calculate the entirety of an event where each occurrence has a diminishing probability.

In the problem involving Irena, the challenge is finding the probability that she never hits the target. The infinite series sums up all possible probabilities of each throw missing the target. The formula for the sum of an infinite geometric series is \( S = a_1 \left( \frac{1}{1-r} \right) \), where \( a_1 \) is the first term and \( r \) is the common ratio. In Irena's case, this becomes \( \frac{1}{4} \left( \frac{1}{1-\frac{2}{3}} \right) = \frac{3}{4} \).

- **Wisdom Bits**:
  • Understand how to apply series sums to probability problems.
  • Recognize the significance of convergence (the series sum reaching a limit).
Complementary Probability
Complementary probability is a concept that states the probability of an event occurring and not occurring must add up to 1. This is handy when determining the probability of all possible outcomes when one outcome's probability is easier to calculate.

For Irena’s target exercises, calculating the probability of never hitting the target involved first assessing the probability of her hitting it at least once. We found that the sum of all probabilities (probability of at least one hit) is \( \frac{3}{4} \) using the infinite series. Therefore, the complementary probability, which is never hitting the target, was found to be \( 1 - \frac{3}{4} = \frac{1}{4} \).

- **Key Idea**:
  • Complementing probabilities simplifies the calculations of coupled events.
  • Always ensure sum of complementary probabilities equals 1.

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Most popular questions from this chapter

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.

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Prisoners' Paradox Three prisoners are informed by their warder that one of them is to be released and the other two shipped to Devil's Island, but the warder cannot inform any prisoner of that prisoner's fate. Prisoner \(A\) thus knows his chance of release to be \(\frac{1}{3}\). He asks the warder to name some one of the other two who is destined for Devil's Island, and the warder names \(B\). Can \(A\) now calculate the conditional probability of his release?

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?

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?

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