/*! 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 18 \- Arrows An Olympic archer is a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\- Arrows An Olympic archer is able to hit the bull's-eye \(80 \%\) of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what's the probability of each of the following results? a) Her first bull's-eye comes on the third arrow. b) She misses the bull's-eye at least once. c) Her first bull's-eye comes on the fourth or fifth arrow. d) She gets exactly 4 bull's-eyes. e) She gets at least 4 bull's-eyes. f) She gets at most 4 bull's-eyes.

Short Answer

Expert verified
a) 0.032 b) 0.737856 c) 0.00768 d) 0.24576 e) 0.90112 f) 0.34464

Step by step solution

01

Understanding the Geometric Distribution

The event in part (a) is that the first success (bull's-eye) occurs on the third arrow. This is a geometric distribution scenario where you want 2 failures and then 1 success. The probability for success (hitting the bull's-eye) is 0.8, and the probability of failure (missing) is 0.2 after 2 trials then a success on the third. Hence, the probability is given by the formula:\[ P(X = 3) = (1-p)^{k-1} \cdot p = 0.2^2 \cdot 0.8 = 0.032 \]
02

Calculating Complement Probability for At Least One Miss

b) To find the probability she misses at least once, calculate the complement of her hitting all bull's-eyes. The probability she hits all of them: \( 0.8^6 = 0.262144 \). Thus, the probability she misses at least once: \( 1 - 0.262144 = 0.737856 \)
03

First Success on Specific Trials

c) We need the first bull's-eye on the 4th or 5th trial. For the 4th: \( P(X=4) = 0.2^3 \times 0.8 = 0.0064 \). For the 5th: \( P(X=5) = 0.2^4 \times 0.8 = 0.00128 \). Sum these to find the answer: \( 0.0064 + 0.00128 = 0.00768 \).
04

Using Binomial Distribution for Exact Count

d) For exactly 4 bull's-eyes, use the binomial formula: \( P(X = x) = \binom{6}{4} \cdot 0.8^4 \cdot 0.2^2 \). Calculate:\[ \binom{6}{4} = 15 \]\[ P(X=4)= 15 \cdot 0.8^4 \cdot 0.2^2 = 0.24576 \]
05

Accumulating Binomial Probabilities for At Least

e) For at least 4 bull's-eyes, calculate probabilities for getting 4, 5, or 6 bull's-eyes. Use the binomial distribution:\[ P(X=5)= \binom{6}{5} \cdot 0.8^5 \cdot 0.2^1 = 0.393216 \]\[ P(X=6)= \binom{6}{6} \cdot 0.8^6 = 0.262144 \]Add probabilities for X=4, X=5, and X=6:\[ 0.24576 + 0.393216 + 0.262144 = 0.90112 \]
06

Using Complement Rule for At Most

f) For at most 4 bull's-eyes, calculate using complement: at least 5 bull's-eyes, calculated from part e:\[ 1 - P(X \geq 5) = 1 - (0.393216 + 0.262144) = 0.34464 \]

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.

Binomial Distribution
The Binomial Distribution is a powerful statistical tool used to model events with two possible outcomes. For example, hitting or missing a bull's-eye. It's commonly used when you want to find out the number of successes in a series of trials. Each trial is independent, meaning the outcome of one doesn't influence another. The probability of success remains constant across trials. For an Olympic archer hitting a bull's-eye 80% of the time, the binomial distribution helps calculate probabilities like in parts d), e), and f) of the original exercise.

The binomial distribution formula is: \[P(X = x) = \binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}\]where:
  • \( n \) is the number of trials (arrows shot)
  • \( x \) is the number of successful outcomes (bull's-eyes)
  • \( p \) is the probability of success on a single trial
For the archer, she shoots 6 arrows. To find probabilities for achieving exactly 4 bull's-eyes, or at least 4, you use this formula and adjust \( x \) to match the desired outcome.
Probability Calculations
Calculating probability involves understanding the likelihood of an event occurring. In the context of the original exercise, probability helps in determining how often the archer hits the target or misses it. Important steps include using complementary probabilities and summing up probabilities for different scenarios.

For instance, to find the probability of missing at least once (part b), you calculate the probability of hitting all targets and then subtract it from one: \[1 - 0.8^6 = 0.737856\]This approach utilizes the fact that the chances of hitting every time is one event, and missing at least once is its complement.

Another calculation involves knowing the probability for her first bull's-eye being on a specific arrow, like in parts a) and c). Here, the geometric distribution can be applied if interested specifically in the occurrence of the first successful trial.
Independent Trials
Independent trials are a core concept in probability and statistics. In simple terms, whether or not the archer hits the bull's-eye in any shot doesn't affect her chances in the next shot. This independence is critical when applying both geometric and binomial distributions in probability exercises.

The independence assumption simplifies calculations. For the Olympic archer, each of her shots is treated as an independent trial with an 80% success rate.
  • In Section 1 (Binomial Distribution), the independence of trials ensures the validity of the binomial probability model over multiple shots.
  • Similarly, when using the geometric distribution as discussed in the probability of her first success, it assumes each shot is taken independently.
When trial outcomes are independent, the chance of success in the entire series of shots remains consistent, providing a clearer outlook for modeling real-world scenarios where repeated actions are involved.

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

\- Simulation II You are one space short of winning a child's board game and must roll a l on a die to claim victory. You want to know how many rolls it might take. a) Describe how you would simulate rolling the die until you get a 1 b) Run at least 30 trials. c) Based on your simulation, estimate the probabilities that you might win on the first roll, the second, the third, etc. d) Calculate the actual probability model. e) Compare the distribution of outcomes in your simulation to the probability model.

\- Colorblindness About \(8 \%\) of males are colorblind. A researcher needs some colorblind subjects for an experiment and begins checking potential subjects. a) On average, how many men should the researcher expect to check to find one who is colorblind? b) What's the probability that she won't find anyone colorblind among the first 4 men she checks? c) What's the probability that the first colorblind man found will be the sixth person checked? d) What's the probability that she finds someone who is colorblind before checking the 10th man?

Tennis, anyone? A certain tennis player makes a successful first serve \(70 \%\) of the time. Assume that each serve is independent of the others. If she serves 6 times, what's the probability she gets a) all 6 serves in? b) exactly 4 serves in? c) at least 4 serves in? d) no more than 4 serves in?

Chips Suppose a computer chip manufacturer rejects \(2 \%\) of the chips produced because they fail presale testing. a) What's the probability that the fifth chip you test is the first bad one you find? b) What's the probability you find a bad one within the first 10 you examine?

Annoying phone calls A newly hired telemarketer is told he will probably make a sale on about \(12 \%\) of his phone calls. The first week he called 200 people, but only made 10 sales. Should he suspect he was misled about the true success rate? Explain.

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.