Mathematics, 13.06.2021 07:30, evelynontiveros2016

1. A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge of the correct answers to any of the questions. She has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball
for each question and replaces the ball in the box.

The marking on the ball will determine her answer
to the question. There are five multiple choice questions on the exam. What is the probability that she will get:

a. Five questions correct?

b. At least four questions correct?

c. No questions correct?

d. No more than two questions correct?

Answers: 2

Show answers

Answers

answered: dayanawhite64

a) 0.001 = 0.1% probability that she will get five questions correct.

b) 0.0156 = 1.56% probability that she will get at least four questions correct.

c) 0.2373 = 23.73% probability that she will get no questions correct.

d) 0.8965 = 89.65% probability that she will get no more than two questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she gets it correct, or she does not. The probability of getting a question correct is independent of any other question, which means that the binomial probability distribution is used to solve this question.

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.

There are five multiple choice questions on the exam.

This means that n = 5

She has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball for each question and replaces the ball in the box.

This means that $p = \frac{1}{4} = 0.25$

a. Five questions correct?

This is P(X = 5) . So

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

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

0.001 = 0.1% probability that she will get five questions correct.

b. At least four questions correct?

This is:

$P(X \geq 4) = P(X = 4) + P(X = 5)$

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

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0146 + 0.001 = 0.0156$

0.0156 = 1.56% probability that she will get at least four questions correct.

c. No questions correct?

This is P(X = 0). So

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

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

0.2373 = 23.73% probability that she will get no questions correct.

d. No more than two questions correct?

This is:

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

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

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

$P(X = 1) = C_{5,0}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965$

0.8965 = 89.65% probability that she will get no more than two questions correct.

answered: Guest

step-by-step explanation:

a. since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation:

given that megan owns a clothing boutique.

the profits of 2010, 11, are recorded and shown as follows:

years since 2010 0 1 2 3 4 5

profits made in 1000's 39 84 111 120 111 84

we find that profit gradually increases form 2010 to 2013 upto 120, then starts decreasing.

the profits for 2012 and 2014 are the same while that of 2011 and 2015 are also the same.

this shows that there is symmetry about the line profit = 120000

hence answer is:

a. since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

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1. A student is taking a multiple-choice exam in which each question has four choices. Assume that t...