Respuesta :
Answer:
The probability that the student will pass is 72.6%
Step-by-step explanation:
There are 100 questions total.
The order that the questions are selected makes no diffence. For example, if Q3, Q28, Q57 are selected in this order or as Q28, Q57, Q3, it is the same exam. So it is a combination problem.
Combination Formula:
A combination of n elements from a set of m objects has the following formula:
[tex]C_{(m,n)} = \frac{m!}{n!(m-n)!}[/tex]
Probability:
The probability is the number of desired outcomes divided by number of total outcomes.
The number of total outcomes is the combination of 3 questions from a set of 100 questions. So, we have n = 3, m = 100.
[tex]T = C_{(100,3)} = \frac{100!}{3!(100-3)!} = 161700[/tex]
There are 161700 total outcomes.
The number of desired outcomes is the combination of 3 questions from a set of 90 questions(the questions that the student knows the answers). So, we have n = 3, m = 90.
[tex]D = C_{(90,3)} = \frac{90!}{3!(90-3)!} = 117480[/tex]
Probability P = D/T = 117480/161700 = 0.726 = 72.6%
The probability that the student will pass is 72.6%
The probability that the student will pass the examination is 72.6%.
How to calculate the probability?
From the information given, the number of outcomes is the combination of three questions out of the total one hundred questions.
This combination is depicted below:
= 100! / 3!(100 - 3)!
= 161700
Also, the number of outcomes when there are 90 questions will be:
= 90! / (90 - 3)!
= 117480.
Therefore, the probability will be:
= 117480/161700
= 72.6%
Learn more about probability on:
https://brainly.com/question/25870256
