Respuesta :
Answer:
0.946 = 94.6% probability that he will answer at least 2 questions correctly.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly 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.
12 questions.
This means that [tex]n = 12[/tex]
Each question has three answers, of which only one is correct.
This means that [tex]p = \frac{1}{3} = 0.3333[/tex]
Assuming that Richard guesses on all 12 questions, find the probability that he will answer at least 2 questions correctly.
Either he answers less than 2 questions correctly, or he answers at least 2 questions correctly. The sum of the probabilities of these outcomes is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. So
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = x) = C_{12,0}.(0.3333)^{0}.(0.6667)^{12} = 0.008[/tex]
[tex]P(X = x) = C_{12,1}.(0.3333)^{1}.(0.6667)^{11} = 0.046[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.008 + 0.046 = 0.054[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.054 = 0.946[/tex]
0.946 = 94.6% probability that he will answer at least 2 questions correctly.
