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Expert-verified Found in: Page 492 ### Discrete Mathematics and its Applications

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

The expected score of Linda on the final is$$170$$.

See the 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 $$170$$. ### Want to see more solutions like these? 