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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

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. What is the probability that the first student to fall victim will be the third student who is sent a phishing e-mail that appears to originate from their high school main office? (A) \((0.4)^{3}\) (B) \((0.6)^{3}\) (C) (0.6)\((0.4)^{2}\) (D) \((0.6)^{2}(0.4)\) (E) \(\left(\begin{array}{l}5 \\ 2\end{array}\right)(0.6)^{2}(0.4)\)

Sixty-five percent of all divorce cases cite incompatibility as the underlying reason. If four couples file for a divorce, what is the probability that exactly two will state incompatibility as the reason? (A) 2(0.65)(0.35) (B) \(2(0.65)^{2}(0.35)^{2}\) (C) \(4(0.65)^{2}(0.35)^{2}\) (D) \(6(0.65)^{2}(0.35)^{2}\) (E) 0.65

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) \((0.9)^{3}+3(0.9)^{2}(0.1)\)

According to a CBS/New York Times poll taken in \(1992,15 \%\) of the public have responded to a telephone call-in poll. In a random group of five people, what is the probability that exactly two have responded to a call-in poll? (A) \(10(0.15)^{2}(0.85)^{3}\) (B) \(5(0.15)^{2}(0.85)^{3}\) (C) \((0.15)^{2}(0.85)^{3}\) (D) \((0.15)^{2}\) (E) \(5(0.15)^{2}\)

Suppose you toss a fair coin ten times and it comes up heads every time. Which of the following is a true statement? (A) By the law of large numbers, the next toss is more likely to be tails than another heads. (B) By the properties of conditional probability, the next toss is more likely to be heads given that ten tosses in a row have been heads. (C) Coins actually do have memories, and thus what comes up on the next toss is influenced by the past tosses. (D) The law of large numbers tells how many tosses will be necessary before the percentages of heads and tails are again in balance. (E) The probability that the next toss will again be heads is 0.5

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