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Q7E

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Discrete Mathematics and its Applications
Found in: Page 492
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Question: The final exam of a discrete mathematics course consists of \(50\) true/false questions, each worth two points, and\(25\)multiple-choice questions, each worth four points. The probability that Linda answers a true/false question correctly is \(0.9\), and the probability that she answers a multiple-choice question correctly is \(0.8\). What is her expected score on the final?

Answer:

The expected score of Linda on the final is\(170\).

See the step by step solution

Step by Step Solution

Step 1: Given information

The final exam of a discrete mathematics course consists of \(50\) true/false questions, each worth two points, and \(25\)multiple-choice questions, each worth four points.

The probability that Linda answers a true/false question correctly is\(0.9\), and the probability that she answers a multiple-choice question correctly is\(0.8\).

Step 2: Theorem for expected number of successes

The expected number of successes when n mutually independent Bernoulli trials are performed, where \(p\) is the probability of success on each trial, is\(np\).

Formula used:

\(E(X + Y) = E(X) + E(Y)\)

Step 3: Calculate the expected value

Let X be the random variable giving the score on the true-false questions and let Y be the random variable giving the score on the multiple-choice questions.

The expected number of true-false questions Linda gets right is the expectation of the number of successes when\(50\)Bernoulli trials are performed with\(p = 0.9\) \(p = 0.9\)

By given theorem, the expectation for the number of successes is np is \(50*0.9 = {\rm{ }}45\)

Since each problem counts \(2\)points, the expectation of\(X\)is \(45*2 = 90\)

Similarly, the expected number of MCQs she gets right is the expectation of the number of successes when \(25\) Bernoulli trials performed \(\left( {p is 0.8} \right), i.e., 25*0.8 = 20\)

Since each problem counts four point, the expectation of Y is \(20*4 = 80\)

Therefore, the expected score on the exam is

\(\begin{array}{l}E(X + Y) = E(X) + E(Y)\\ = 90 + 80\\ + 170\end{array}\)

The expected score of Linda is \(170\).

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