Respuesta :
Answer:
42.1875% probability that the student gets all three questions wrong
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the student gets it wrong, or he does not. The probability of the student getting a question wrong is independent of other questions. So we use the binomial probability distribution to solve this question.
Binomial probabily 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.
On a multiple-choice test, each question has 4 possible answers.
One of these options is correct and the other 3 are wrong. We want to find the probability of getting questions wrong. So [tex]p = \frac{3}{4} = 0.75[/tex]
Three question:
This means that [tex]n = 3[/tex]
What is the probability that the student gets all three questions wrong?
This is P(X = 3).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.421875[/tex]
42.1875% probability that the student gets all three questions wrong
