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Chapter 4: Problem 4

It is estimated that two out of five high school students would fall victim to a phishing e-mail (an online scam asking for sensitive information) if it appears to originate from their high school main office. In a random sample of five high school students, what is the probability that exactly two fall victim to a phishing e-mail that appears to originate from their high school main office? (A) 0.4 (B) 1.0 (C) \(\left(\begin{array}{l}5 \\ 2\end{array}\right)(0.4)^{2}(0.6)^{3}\) (D) \(\left(\begin{array}{l}5 \\ 2\end{array}\right)(0.4)^{2}(0.6)^{3}\) (E) \((0.4)^{2}(0.6)^{3}\)

Short Answer

Expert verified
The correct answer is (C) or (D): \( \binom{5}{2} (0.4)^2 (0.6)^3 \)

Step by step solution

01

Identifying the problem type

Recognize that this problem involves finding the probability of a specific number of successes (victims) in a fixed number of trials (students), with a given probability of success. This indicates it is a binomial probability problem.
02

Define the binomial probability formula

The binomial probability formula is given by \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where- \(n\) is the number of trials (5 students)- \(k\) is the number of successes (2 victims)- \(p\) is the probability of success (0.4)- \(\binom{n}{k}\) is the binomial coefficient
03

Calculate the binomial coefficient

Calculate the binomial coefficient \(\binom{5}{2}\).\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
04

Substitute values into the formula

Substitute \(n = 5\), \(k = 2\), and \(p = 0.4\) into the binomial probability formula.\[ P(X = 2) = \binom{5}{2} (0.4)^2 (0.6)^{5-2} \]Which simplifies to\[ P(X = 2) = 10 (0.4)^2 (0.6)^3 \]
05

Calculate the probability

Calculate the value of the simplified expression.\[ P(X = 2) = 10 (0.4)^2 (0.6)^3 = 10 (0.16) (0.216) = 10 \times 0.03456 = 0.3456 \]
06

Identify the correct answer choice

From the answer choices given, identify which matches the calculated expression.Both choices (C) and (D) match the expression, but they both represent the same calculation. Therefore, either (C) or (D) is correct.

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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 probability problem, the 'probability of success' is a key element. This term refers to the likelihood of a particular outcome occurring in one trial of a given experiment.
In our exercise, the probability of a high school student falling victim to a phishing email is given as 0.4.
To fully understand this concept, think about flipping a fair coin. The 'probability of success' for landing heads is 0.5, or 50%.
Similarly, in our problem, there is a 40% chance or 0.4 probability that a student will fall for the phishing email.
This probability will be used in the binomial formula to determine how likely it is that exactly two out of five students will be victims.
Binomial Coefficient
The binomial coefficient, represented as \(\binom{n}{k}\), is another vital part of binomial probability.
It's used to determine the number of ways to choose k successes in n trials. In our exercise, \(\binom{5}{2}\) calculates how many ways we can have exactly two students fall victim out of the five sampled.
The formula for the binomial coefficient is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Applying this formula to our problem involves calculating \(\binom{5}{2}\). We get: \[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Thus, there are 10 different ways two out of five students can fall victim to the phishing email.
Random Sample
A 'random sample' is a subset of a population that is selected randomly to represent the entire group.
In our exercise, the random sample consists of five high school students chosen to evaluate the probability of falling victim to a phishing email.
Random sampling ensures that every student has an equal chance of being selected and that the results are not biased.
This method is commonly used in statistics to make inferences about a larger population. It is crucial because it allows for generalization from the sample to the larger group.
By using a random sample in this exercise, we can apply the results to get an idea of how phishing might affect the broader group of high school students.

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Most popular questions from this chapter

In a 1974 "Dear Abby" letter, a woman lamented that she had just given birth to her eighth child and all were girls! Her doctor had assured her that the chance of the eighth child being a girl was less than 1 in 100 . What was the real probability that the eighth child would be a girl? (A) 0.0039 (B) 0.5 (C) \((0.5)^{7}\) (D) \((0.5)^{8}\) (E) \(\frac{(0.5)^{7}+(0.5)^{8}}{2}\)

A city water supply system involves three pumps, the failure of any one of which crashes the system. The probabilities of failure for each pump in a given year are \(0.025,0.034,\) and o.02, respectively. Assuming the pumps operate independently of each other, what is the probability that the system does crash during the year? (A) \(0.025+0.034+0.02\) (B) \(1-(0.025+0.034+0.02)\) (C) \(1-(0.025)(0.034)(0.02)\) (D) \((1-0.025)(1-0.034)(1-0.02)\) (E) \(1-(1-0.025)(1-0.034)(1-0.02)\)

Sandy Koufax, one of the greatest baseball pitchers ever, struck out approximately one out of every four batters he faced. Assuming independence, what is the probability that the first batter Koufax strikes out is before the third batter he faces? (A) \((0.25)^{2}\) (B) \((0.75)^{2}\) (C) \(1-(0.25)^{2}\) (D) \(1-(0.75)^{2}\) (E) \(0.75+(0.75)(0.25)\)

Suppose that, for any given year, the probabilities that the stock market declines, that women's hemlines are lower, and that both events occur are, respectively, \(0.4,0.35,\) and \(0.3 .\) Are the two events independent? (A) Yes, because \((0.4)(0.35) \neq 0.3 .\) (B) No, because \((0.4)(0.35) \neq 0.3\). (C) Yes, because \(0.4>0.35>0.3\). (D) No, because \(0.5(0.3+0.4)=0.35\). (E) There is insufficient information to answer this question.

An inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole lot if at least two of the three items are in perfect condition. If in reality \(90 \%\) of the whole lot are perfect, what is the probability that the lot will be accepted? (A) \(3(0.1)^{2}(0.9)\) (B) \(3(0.9)^{2}(0.1)\) (C) \((0.1)^{3}+(0.1)^{2}(0.9)\) (D) \((0.1)^{3}+3(0.1)^{2}(0.9)\) (E) \((0.9)^{3}+3(0.9)^{2}(0.1)\)
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