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Chapter 5: Problem 15

Assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are \(n=8\) trials, each with probability of success (correct) given by \(p=0.20 .\) Find the indicated probability for the number of correct answers. Find the probability that the number \(x\) of correct answers is exactly 7 .

Short Answer

Expert verified
The probability of getting exactly 7 correct answers is 0.000008192.

Step by step solution

01

- Understand the Binomial Formula

The problem involves finding the probability of getting a specific number of successes (correct answers) in a series of trials (questions). Here, we need the Binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where - \( n \) is the number of trials, - \( k \) is the number of successes, - \( p \) is the probability of success, - \( \binom{n}{k} \) is the binomial coefficient.
02

- Define Given Variables

Identify the given variables from the problem: - Number of trials, \( n = 8 \)- Probability of success, \( p = 0.20 \)- Number of successes, \( k = 7 \)
03

- Calculate the Binomial Coefficient

Calculate the binomial coefficient \( \binom{n}{k} \): \[ \binom{8}{7} = \frac{8!}{7!(8-7)!} = \frac{8!}{7! \times 1!} = 8 \]
04

- Apply the Binomial Probability Formula

Substitute the values into the binomial probability formula: \[ P(X = 7) = \binom{8}{7} (0.20)^7 (1-0.20)^{8-7} \]\[ P(X = 7) = 8 \times (0.20)^7 \times (0.80)^1 \]\[ P(X = 7) = 8 \times 0.00000128 \times 0.80 \]
05

- Perform Final Calculation

Calculate the final value: \[ P(X = 7) = 8 \times 0.00000128 \times 0.80 = 0.000008192 \]

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Key Concepts

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

Probability of Success
In probability, the 'probability of success' is a key concept when dealing with Binomial Distributions. Let's break it down further: Probability of success, denoted by \( p \), is the likelihood of a single event resulting in a success. In our exercise, a success is defined as correctly answering a multiple-choice question.

Given that each question has 5 potential answers with only one correct answer, if a student guesses randomly, they have a probability of success of \( p = 0.20 \). This means on any given question, the chance of guessing correctly is 20%.

It's crucial to understand that 'success' doesn't necessarily mean a positive outcome; it merely refers to the event of interest. For a Binomial Distribution, this success probability \( p \) remains constant across all trials.
Binomial Coefficient
The Binomial Coefficient is central to calculating probabilities using the Binomial Distribution. It's represented by \( \binom{n}{k} \) and helps determine the number of ways to achieve \( k \) successes in \( n \) trials. Essentially, it tells us how many different sequences of successes and failures can occur.

In our case, we need to calculate \( \binom{8}{7} \). This coefficient is computed as:

\[ \binom{8}{7} = \frac{8!}{7!(8-7)!} \]
This simplifies to 8 because:
\[ \frac{8!}{7!1!} = \frac{8 \times 7!}{7!} = 8 \]

Understanding this helps you grasp why Binomial Coefficient is important: it captures the different possible ways outcomes can occur across multiple trials.
Multiple Choice Questions
Multiple Choice Questions (MCQs) are commonly used in exams. Each question typically has several possible answers, but only one correct one. For our problem, the multiple-choice scenario means random guessing can be modelled using a Binomial Distribution.

Notice we dealt with 8 multiple choice questions, with a probability of guessing correctly set at \( p = 0.20 \). The challenge was to find the probability of exactly 7 correct guesses, which is a classic application of the binomial probability formula.

In summary, when facing a series of MCQs where answers are guessed, each question's outcome becomes a trial in a Binomial experiment. This allows us to use Binomial Distribution to calculate the probability of various results, like getting a certain number of correct answers purely by chance.

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Most popular questions from this chapter

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World War II Bombs In Exercise 1 "Notation" we noted that in analyzing hits by \(\mathrm{V}-1\) buzz bombs in World War II, South London was partitioned into 576 regions, each with an area of \(0.25 \mathrm{~km}^{2}\). A total of 535 bombs hit the combined area of 576 regions. a. Find the probability that a randomly selected region had exactly 2 hits. b. Among the 576 regions, find the expected number of regions with exactly 2 hits. c. How does the result from part (b) compare to this actual result: There were 93 regions that had exactly 2 hits?

Powerball: Poisson Approximation to Binomial There is a \(1 / 292,201,338\) probability of winning the Powerball lottery jackpot with a single ticket. Assume that you purchase a ticket in each of the next 5200 different Powerball games that are run over the next 50 years. Find the probability of winning the jackpot with at least one of those tickets. Is there a good chance that you would win the jackpot at least once in 50 years?

Hybrids One of Mendel's famous experiments with peas resulted in 580 offspring, and 152 of them were yellow peas. Mendel claimed that under the same conditions, \(25 \%\) of offspring peas would be yellow. Assume that Mendel's claim of \(25 \%\) is true, and assume that a sample consists of 580 offspring peas. a. Use the range rule of thumb to identify the limits separating values that are significantly low and those that are significantly high. Based on the results, is the result of 152 yellow peas either significantly low or significantly high? b. Find the probability of exactly 152 yellow peas: c. Find the probability of 152 or more yellow peas. d. Which probability is relevant for determining whether 152 peas is significantly high: the probability from part (b) or part (c)? Based on the relevant probability, is the result of 152 yellow peas significantly high? e. What do the results suggest about Mendel's claim of \(25 \%\) ?
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