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Problem 94: A multiple-choice quiz consists of ten questions, each with five possible answers of which only one is correct. A student passes the quiz if seven or more correct answers are obtained. What is the probability that a student who guesses blindly at all of the questions will pass the quiz? What is the probability of passing the quiz if, on each question, a student can eliminate three incorrect answers and then guesses between the remaining two?

Problem 94: A multiple-choice quiz consists of ten questions, each with five possible answers of which only one is correct. A student passes the quiz if seven or more correct answers are obtained. What is the probability that a student who guesses blindly at all of the questions will pass the quiz? What is the probability of passing the quiz if, on each question, a student can eliminate three incorrect answers and then guesses between the remaining two?

Solution:


a. the probability that a student who guesses blindly at all of the questions will pass the quiz
Given that $x \ge 7, n=10$
$p(1 \; correct\; answer\;out\;of\;5\;possible\;answers) = \frac{1}{5} = 0.2$
$P(x\ge 7) = P(x=7) + P(x=8) + P(x=9) +p(x=10)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x= 7) = b(7;10,0.2) = \left(\begin{array}{c}10\\ 7\end{array}\right){0.2}^{7}(1-0.2)^{10-7} = 0.000786$
$P(x= 8) = b(8;10,0.2) = \left(\begin{array}{c}10\\ 8\end{array}\right){0.2}^{8}(1-0.2)^{10-8} = 7.37E^{-05}$
$P(x= 9) = b(9;10,0.2) = \left(\begin{array}{c}10\\ 9\end{array}\right){0.2}^{9}(1-0.2)^{10-9} = 4.1E^{-06}$
$P(x= 10) = b(10;10,0.2) = \left(\begin{array}{c}10\\ 10\end{array}\right){0.2}^{10}(1-0.2)^{10-10} = 1.02E^{-07}$
$P(x\ge 7) = P(x=7) + P(x=8) + P(x=9) +p(x=10) = 0.000864.$

b. the probability of passing the quiz if, on each question, a student can eliminate three incorrect answers and then guesses between the remaining two
Given that $x \ge 7, n=10$
$p(1 \; correct\; answer\;out\;of\;2\;remianing\;answers) = \frac{1}{2} = 0.5$
$P(x\ge 7) = P(x=7) + P(x=8) + P(x=9) +p(x=10)$
$b(x;n,p)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}(1-p)^{n-x}$
$P(x= 7) = b(7;10,0.5) = \left(\begin{array}{c}10\\ 7\end{array}\right){0.5}^{7}(1-0.5)^{10-7} = 0.117188$
$P(x= 8) = b(8;10,0.5) = \left(\begin{array}{c}10\\ 8\end{array}\right){0.5}^{8}(1-0.5)^{10-8} = 0.043945$
$P(x= 9) = b(9;10,0.5) = \left(\begin{array}{c}10\\ 9\end{array}\right){0.5}^{9}(1-0.5)^{10-9} = 0.009766$
$P(x= 10) = b(10;10,0.5) = \left(\begin{array}{c}10\\ 10\end{array}\right){0.5}^{10}(1-0.5)^{10-10} = 0.000977$
$P(x\ge 7) = P(x=7) + P(x=8) + P(x=9) +p(x=10) = 0.171875.$

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