Question

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct. The student has no clue as to the correct answer to any of the question! and merely guesses. [You can use binompdf or binomcdf for a), b), and c).] {5.4)

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)
b) What is the probability of the student getting 3 or less answers correct? [5 points)
c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers correct.] [4 point

Answer 1

Answer:

a) 0.11% probability that the student will get exactly 6 answers correct

b) 94.37% probability of the student getting 3 or less answers correct

c) 5.63% probability of the student getting 4 or more answers correct

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct.

This means that $n = 8, p = \frac{1}{5} = 0.2$

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)

This is P(X = 6).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{8,6}.(0.2)^{6}.(0.8)^{2} = 0.0011$

0.11% probability that the student will get exactly 6 answers correct

b) What is the probability of the student getting 3 or less answers correct? [5 points)

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{8,0}.(0.2)^{0}.(0.8)^{8} = 0.1678$

$P(X = 1) = C_{8,1}.(0.2)^{1}.(0.8)^{7} = 0.3355$

$P(X = 2) = C_{8,2}.(0.2)^{2}.(0.8)^{6} = 0.2936$

$P(X = 3) = C_{8,3}.(0.2)^{3}.(0.8)^{5} = 0.1468$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1678 + 0.3355 + 0.2936 + 0.1468 = 0.9437$

94.37% probability of the student getting 3 or less answers correct

c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

So

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9437 = 0.0563$

5.63% probability of the student getting 4 or more answers correct