Respuesta :
Answer:
a. 0.008 = 0.8% probability that the person will answer all three questions correctly.
b. 0.096 = 9.6% probability that the person will answer exactly two questions correctly.
c. 0.384 = 38.4% probability that the person will answer exactly one question correctly.
d. 0.512 = 51.2% probability that the person will answer no questions correctly.
e. The expected value of the person’s score for the three questions is -0.2.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the person answers it correctly, or they do not. The probability of a person answering a question correctly is independent of any other question. This 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Each question has five possible answers.
Person has no idea which option is correct, so the probability of answering correctly is:
[tex]p = \frac{1}{5} = 0.2[/tex]
Three questions:
This means that [tex]n = 3[/tex]
a. What is the probability that the person will answer all three questions correctly?
This is [tex]P(X = 3)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.2)^{3}.(0.8)^{0} = 0.008[/tex]
0.008 = 0.8% probability that the person will answer all three questions correctly.
b. What is the probability that the person will answer exactly two questions correctly?
This is [tex]P(X = 2)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.2)^{2}.(0.8)^{1} = 0.096[/tex]
0.096 = 9.6% probability that the person will answer exactly two questions correctly.
c. What is the probability that the person will answer exactly one question correctly?
This is [tex]P(X = 1)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{3,1}.(0.2)^{1}.(0.8)^{2} = 0.384[/tex]
0.384 = 38.4% probability that the person will answer exactly one question correctly.
d. What is the probability that the person will answer no questions correctly?
This is [tex]P(X = 0)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512[/tex]
0.512 = 51.2% probability that the person will answer no questions correctly.
e. Suppose that the person gets one point of credit for each correct answer and that 1/3 point is deducted for each incorrect answer. What is the expected value of the person’s score for the three questions?
The expected number of correct answer is:
[tex]E(X) = np = 3*0.2 = 0.6[/tex]
And the expected number of wrong answers is 3 - 0.6 = 2.4. So, the expected score is:
[tex]S(x) = 0.6 - \frac{2.4}{3} = 0.6 - 0.8 = -0.2[/tex]
The expected value of the person’s score for the three questions is -0.2.
