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Chapter 7: Problem 25

A student studying for a vocabulary test knows the meanings of 12 words from a list of 20 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows?

Short Answer

Expert verified
The probability that the student knows at least 8 words from the test is calculated by summing the probabilities of three scenarios: knowing 8, 9, or 10 words from the test. To find these probabilities, we use the combination formula: \(C(n,r) = \frac{n!}{r! (n-r)!}\). Probability of knowing at least 8 words = \(\frac{C(12, 8) * C(8, 2)}{C(20, 10)}\) + \(\frac{C(12 ,9) * C(8, 1)}{C(20, 10)}\) + \(\frac{C(12, 10)}{C(20, 10)}\) Calculating the probabilities and summing them up will give us the required probability.

Step by step solution

01

Determine the possible scenarios

We are given that there are 20 words in total, out of which the student knows 12 words and the test contains 10 words. We have to find the probability that the student knows at least 8 words from the test. So, the possible scenarios are: the student knows 8, 9, or 10 words from the test.
02

Calculate the combinations for each scenario

In this step, we will calculate the combinations for each of the scenarios mentioned above using the formula nCr, where n is the total number of items, and r is the number of selected items. \(C(n,r) = \frac{n!}{r! (n-r)!}\) Scenario 1: Student knows 8 words from the test. In this case, we need to choose 8 known words and 2 unknown words from the list. Combinations for known words: \(C(12, 8) = \frac{12!}{8! (12-8)!}\) Combinations for unknown words: \(C(8, 2) = \frac{8!}{2! (8-2)!}\) Scenario 2: Student knows 9 words from the test. In this case, we need to choose 9 known words and 1 unknown word from the list. Combinations for known words: \(C(12, 9) = \frac{12!}{9! (12-9)!}\) Combinations for unknown words: \(C(8, 1) = \frac{8!}{1! (8-1)!}\) Scenario 3: Student knows all 10 words from the test. In this case, we need to choose all 10 known words from the list. Combinations for known words: \(C(12, 10) = \frac{12!}{10! (12-10)!}\)
03

Calculate the probability for each scenario

To find the probability for each scenario, we need to divide the product of the combinations of known and unknown words for each scenario by the total possible combinations of selecting 10 words from the list of 20 words. Total combinations of selecting 10 words from the list: \(C(20,10) = \frac{20!}{10! (20-10)!}\) Probability for scenario 1: \(\frac{C(12, 8) * C(8, 2)}{C(20, 10)}\) Probability for scenario 2: \(\frac{C(12 ,9) * C(8, 1)}{C(20, 10)}\) Probability for scenario 3: \(\frac{C(12, 10)}{C(20, 10)}\)
04

Determine the probability of the student knowing at least 8 words

Finally, to find the probability of the student knowing at least 8 words on the test, we will add the probabilities for each of the scenarios mentioned above. Probability of knowing at least 8 words = Probability of knowing 8 words + Probability of knowing 9 words + Probability of knowing 10 words Calculating the probabilities and summing them up will give us the required probability.

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Most popular questions from this chapter

A medical test has been designed to detect the presence of a certain disease. Among those who have the disease, the probability that the disease will be detected by the test is \(.95\). However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is .04. It is estimated that \(4 \%\) of the population who take this test have the disease. a. If the test administered to an individual is positive, what is the probability that the person actually has the disease? b. If an individual takes the test twice and both times the test is positive, what is the probability that the person actually has the disease? (Assume that the tests are independent.)

The admissions office of a private university released the following admission data for the preceding academic year: From a pool of 3900 male applicants, \(40 \%\) were accepted by the university and \(40 \%\) of these subsequently enrolled. Additionally, from a pool of 3600 female applicants, \(45 \%\) were accepted by the university and \(40 \%\) of these subsequently enrolled. What is the probability that a. A male applicant will be accepted by and subsequently will enroll in the university? b. A student who applies for admissions will he accepted by the university? c. A student who applies for admission will be accepted by the university and subsequently will enroll?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a subset of \(B\) and \(P(B)=0\), then \(P(A)=0\).

The accompanying data were obtained from the financial aid office of a certain university: \begin{tabular}{lccc} \hline & \multicolumn{3}{c} { Not } & \\ & Receiving Financial Aid & Receiving Financial Aid & Total \\ \hline Undergraduates & \(4.222\) & 3,898 & 8,120 \\ \hline Graduates & \(1.879\) & 731 & 2,610 \\ \hline Total & \(6.101\) & 4,629 & 10,730 \\ \hline \end{tabular} Let \(A\) be the event that a student selected at random from this university is an undergraduate student, and let \(B\) be the event that a student selected at random is receiving financial aid. a. Find each of the following probabilities: \(P(A), P(B)\), \(P(A \cap B), P(B \mid A)\), and \(P\left(B \mid A^{\circ}\right)\) b. Are the events \(A\) and \(B\) independent events?

Determine whether the given experiment has a sample space with equally likely outcomes. A loaded die is rolled, and the number appearing uppermost on the die is recorded.
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