Respuesta :
Answer:
0.0082 = 0.82% probability that he will pass
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the students guesses the correct answer, or he guesses the wrong answer. The probability of guessing the correct answer for a question is independent of other questions. So we use the binomial probability distribution 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.
In this problem we have that:
[tex]n = 14, p = 0.3[/tex].
If the student makes knowledgeable guesses, what is the probability that he will pass?
He needs to guess at least 9 answers correctly. So
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{14,9}.(0.3)^{9}.(0.7)^{5} = 0.0066[/tex]
[tex]P(X = 10) = C_{14,10}.(0.3)^{10}.(0.7)^{4} = 0.0014[/tex]
[tex]P(X = 11) = C_{14,11}.(0.3)^{11}.(0.7)^{3} = 0.0002[/tex]
[tex]P(X = 12) = C_{14,12}.(0.3)^{12}.(0.7)^{2} = 0.000024[/tex]
[tex]P(X = 13) = C_{14,13}.(0.3)^{13}.(0.7)^{1} = 0.000002[/tex]
[tex]P(X = 14) = C_{14,14}.(0.3)^{14}.(0.7)^{0} \cong 0 [/tex]
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) = 0.0066 + 0.0014 + 0.0002 + 0.000024 + 0.000002 = 0.0082[/tex]
0.0082 = 0.82% probability that he will pass
