Respuesta :
Answer:
There is a 0.058% probability that this student will pass the examination.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either it is correct, or it is wrong. This means that we can solve this problem using the binomial probability distribution.
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 combinatios 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.
In this problem
There are 12 questions, so [tex]n = 12[/tex].
The student guesses each question. There are five possible answers, only one which is correct, so [tex]p = 0.2[/tex].
What is the probability that a student who guesses at random on each question will pass the examination?
This is [tex]P(X \geq 8)[/tex]
[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{12,8}.(0.2)^{8}.(0.8)^{4} = 0.000519[/tex]
[tex]P(X = 9) = C_{12,9}.(0.2)^{9}.(0.8)^{3} = 0.000058[/tex]
[tex]P(X = 10) = C_{12,10}.(0.2)^{10}.(0.8)^{2} = 0.000004[/tex]
[tex]P(X = 11) = C_{12,11}.(0.2)^{11}.(0.8)^{1} = 0.0000002[/tex]
[tex]P(X = 12) = C_{12,12}.(0.2)^{12}.(0.8)^{0} = 0.000000004[/tex]
So
[tex]P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.000519 + 0.000058 + 0.000004 + 0.0000002 + 0.000000004 = 0.00058[/tex]
There is a 0.058% probability that this student will pass the examination.
