Question 1196876: A student knows the answer to 17 questions out of a list of 20. A test consists of 3 questions, each selected at random from the aforementioned list. To pass the test, the student must get at least 2 correct answers. What chance does the student have of passing the test?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let A,B,C represent the 3 questions in which the student does not know the answer to.
Since the order doesn't matter, there's one way to select those 3 questions.
Formally you can use the nCr combination formula with n = 3 and r = 3 to get 3C3 = 1.
Let x = 1 since we'll be using it later.
Now let's consider the number of ways to select two of the three unknown questions.
The possibilities are:
{A,B}
{A,C}
{B,C}
Again, order doesn't matter.
You can use the nCr combination formula with n = 3 and r = 2 to find that 3C2 = 3
There are 3 ways to pick exactly 2 of the A,B,C
Then there are 17 other questions that the student knows
That makes 3*17 = 51 ways to select 3 questions where 2 are unknown and the third is known
Let y = 51 since we'll be using it later.
Let's add the results so far:
x+y = 1 + 51 = 52
There are 52 ways for the student to fail the test.
Either s/he gets all three unknown questions (x = 1) OR the student gets 2 unknown questions (y = 51)
Use the nCr combination formula to find that there are 17C3 = 680 ways to pick three questions from a pool of 17, where order doesn't matter.
This means there are 680-52 = 628 ways to pass the test
This is when the student gets at most 1 unknown question (i.e. 1 or fewer).
Either the student gets 1 unknown question and 2 known, or the student gets 3 known questions.
Divide the number of ways to pass over the number of ways to pick the three questions
628/680 = 0.9235294117647
Answer: The probability of the student passing is roughly 0.9235
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