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Chapter 3: Problem 41

A multiple-choice examination has 15 questions, each with five possible answers, only one of which is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent random guess. What is the probability that he answers at least ten questions correctly?

Short Answer

Expert verified
Use binomial distribution: very low probability of at least ten correct answers by random guessing.

Step by step solution

01

Understand the Problem

The student answers each question by guessing, which means there's a fixed probability of getting a question correct. We need to find the probability that the student gets at least 10 questions correct out of 15 by guessing randomly.
02

Determine the Probability of Success

Since there are 5 possible answers and only 1 is correct, the probability of correctly guessing one answer is \( \frac{1}{5} = 0.2 \).
03

Apply Binomial Distribution

This is a binomial probability problem because each question is an independent trial with two possible outcomes: correct guess or incorrect guess. The number of trials is 15, the probability of success on each trial is 0.2, and the random variable X is the number of correct answers.
04

Define the Binomial Random Variable

Let \( X \) be the random variable representing the number of questions answered correctly. \( X \) follows a binomial distribution with parameters \( n = 15 \) and \( p = 0.2 \), so \( X \sim B(15, 0.2) \).
05

Calculate the Complementary Probability

We want \( P(X \geq 10) \). It is easier to find \( P(X < 10) \) and then subtract from 1: \[ P(X \geq 10) = 1 - P(X < 10) \].
06

Use Binomial Probability Formula

The probability for exactly \( k \) successes is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]. Use this formula to calculate probabilities for \( k = 0, 1, 2, \, ..., 9 \).
07

Calculate Total Probability for X < 10

Compute \( P(X = k) \) for \( k = 0 \) to \( k = 9 \) using a calculator or statistical software and sum these probabilities to get \( P(X < 10) \).
08

Find P(X 鈮� 10)

Subtract the sum of probabilities from Step 7 from 1 to get \( P(X \geq 10) \).
09

Conclusion

Compute the final probability from the calculations and form your conclusion about \( P(X \geq 10) \).

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Key Concepts

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

Probability of Success
In a binomial distribution, "probability of success" is a key concept. It refers to the probability that a single trial results in a success.
For example, in our multiple-choice exam scenario, a success is when the student guesses the correct answer. Since there are 5 possible answers for each question, and only 1 is correct, the probability of success is calculated as:
  • Probability of Success, \( p = \frac{1}{5} = 0.2 \)
This probability remains constant throughout the trials, as each guess has the same chance of being correct. Understanding this concept is fundamental because it is used to model the binomial distribution, which assesses the likelihood of a given number of successes in a set of trials.
Complementary Probability
The idea of complementary probability is useful when calculating the chance of something happening indirectly.
In the binomial distribution context, sometimes it is easier to find the probability of the complement of an event and subtract from 1.
This approach can be seen in our problem where we need to find the probability of getting at least 10 questions correct.
  • Complementary Probability: \( P(X \geq 10) = 1 - P(X < 10) \)
We first find the probability of getting fewer than 10 questions right, then subtract this result from 1 to find the target probability. This technique often simplifies calculations.
Random Variable
A random variable is a numerical outcome of a random phenomenon.
In the exercise, the random variable \( X \) represents the number of questions the student answers correctly.
  • \( X \) is a discrete random variable, as it only takes integer values from 0 to 15.
  • The value of \( X \) depends on the random guesses made by the student.
The understanding of random variables is fundamental, as they help to quantify uncertain outcomes and form the basis for calculating probabilities in statistics.
Independent Trials
When we refer to independent trials, we mean that the outcome of one trial does not affect the outcome of another.
In the context of our problem, every guess the student makes on each question is independent.
The probability of guessing correctly remains constant at 0.2, regardless of previous answers.
  • Independent trials ensure that the total probability across all trials adds up to 1, maintaining the integrity of the binomial model.
  • Independence helps simplify calculations, as each event's probability stands alone, unaffected by other events.
Mastering the concept of independence is crucial for working with probability distributions like the binomial distribution, where each trial is independent of others.

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Most popular questions from this chapter

Suppose that \(30 \%\) of the applicants for a certain industrial job possess advanced training in computer programming. Applicants are interviewed sequentially and are selected at random from the pool. Find the probability that the first applicant with advanced training in programming is found on the fifth interview.

Of a population of consumers, \(60 \%\) are reputed to prefer a particular brand, \(A\), of toothpaste. If a group of randomly selected consumers is interviewed, what is the probability that exactly five people have to be interviewed to encounter the first consumer who prefers brand \(A\) ? At least five people?

a. We observe a sequence of independent identical trials with two possible outcomes on each trial, \(S\) and \(F,\) and with \(P(S)=p .\) The number of the trial on which we observe the fifth success, \(Y\), has a negative binomial distribution with parameters \(r=5\) and \(p\). Suppose that we observe the fifth success on the eleventh trial. Find the value of \(p\) that maximizes \(P(Y=11\) ). b. Generalize the result from part (a) to find the value of \(p\) that maximizes \(P\left(Y=y_{0}\right)\) when \(Y\) has a negative binomial distribution with parameters \(r\) (known) and \(p\)

An experiment consists of tossing a fair die until a 6 occurs four times. What is the probability that the process ends after exactly ten tosses with a 6 occurring on the ninth and tenth tosses?

The number of people entering the intensive care unit at a hospital on any single day possesses a Poisson distribution with a mean equal to five persons per day. a. What is the probability that the number of people entering the intensive care unit on a particular day is equal to 2 ? Is less than or equal to 2 ? b. Is it likely that \(Y\) will exceed \(10 ?\) Explain.
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