Respuesta :
Answer:
There is a [tex]\frac{2097}{10000}[/tex] probability that she can solve all 7 problems.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we 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}.\pi^{x}.(1-\pi)^{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 [tex]\pi[/tex] is the probability of X happening.
For this problem, we have that:
In our sample there are 7 questions, so [tex]n = 7[/tex].
There are 10 problems, and she can solve 8 of them. So [tex]\pi = \frac{8}{10} = 0.5[/tex]
What is the probability that the student can solve all 7 problems on the exam?
This is [tex]P(X = 7)[/tex].
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 7) = C_{7,7}.(0.80)^{7}.(0.2)^{0} = 0.2097 = \frac{2097}{10000}[/tex]
There is a [tex]\frac{2097}{10000}[/tex] probability that she can solve all 7 problems.
Probability of an event shows the chances of occurrence of that event.
The probability that the student in the given context can solve all 7 problems on the exam is 0.2097 approximately.
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}[/tex]
Where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is the multiplication rule of probability for independent events?
Suppose there are n mutually independent events.
The probability of their simultaneously occurrence is given as
[tex]P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1) \times P(A_2) \times ... \times P(A_n)[/tex]
(This is true only if all those events are mutually independent).
For the given case, let we have
E = A student successfully solving a problem
Then
[tex]P(E) = \dfrac{8}{10} = 0.8[/tex]
Since each of the question has its own success and failure chances and all are independent of each other, thus, we have:
Probability of solving all 7 problems =
[tex]P(E_1 \cap E_2 \cap .. \cap E_7) = (0.8) \times (0.8) \times ... \times (0.8) \text{\:\:(7 times)} = (0.8)^7 \approx 0.2097[/tex]
(intersection shows simultaneous occurrence and [tex]E_i[/tex] is the event of solving ith problem)
Thus,
The probability that the student in the given context can solve all 7 problems on the exam is 0.2097 approximately.
Learn more about multiplication rule of probability here:
https://brainly.com/question/14399918
