Kenneth G. answered • 12/06/13
Tutor
New to Wyzant
Experienced Tutor of Mathematics and Statistics
It's not completely clear from the question what the event is. Is it asking the probability that exactly 17% of the tests are graded 100, or that at least 17% of the tests are graded 100.
Also, we have to assume that no student ever earns a score of 100 because otherwise a test could be scored 100 because the student earned it, and the chance of getting a score of 100 would be greater than 10% and we might not have sufficient information to solve that problem.
If the requirement is an exact 17% - It is not possible for exactly 17% of the 19 tests to be 100 because 17% of 19 is 3.23. You could ask the probability that exactly 3 or exactly 4 tests are graded 100 however.
If we restate the problem more clearly, like this: "what is the probability that at least 17% of those tests will be given a 100?" then it is possible to work this problem. To restate this, it is asking "what is the probability that AT LEAST 4 tests are graded 100?"
This problem is equivalent to a coin problem. You have a coin that has a probability .1 of heads and .9 of tails. If you flip that coin 19 times, what is the probability that you get 4 or more heads? Same question as above.
Lets write the number of ways to choose m objects from a set of n objects (number of combinations of n objects taken m at a time) as C(n,m).
Then the probability P(0) of getting no scores of 100 can be written as C(19,0)(.9^19)(.1^0) where ^ indicates an exponent. Here C(19,0) is equal to 1, and so is .1^0, so this probability is .9^19.
Similarly, the probability P(1) of getting exactly one score of 100 is C(19,1)(.9^18)(.1^1) or 1.9*(.9^18); likewise, the probabilities of getting exactly two or three scores of 100 are C(19,2)(.9^17)(.1^2) and C(19,3)(.9^16)(.1^3), or 19*9*.01*(.9^17) and 19*3*17*.001*(.9^16).
These are the cases where the objective (to have at least 4 scores of 100) is false. If we subtract them from 1 we get the required probability.
P(at least 4) = 1 - P(0) - P(1) - P(2) - P(3) which is approximately
1 - .135085172 - .285179807 - .285179807 - .179557656
So P(At least 4 scores of 100) = .114997558 (approximately)
If the requirement is an exact 17% - It is not possible for exactly 17% of the 19 tests to be 100 because 17% of 19 is 3.23. You could ask the probability that exactly 3 or exactly 4 tests are graded 100 however.
If we restate the problem more clearly, like this: "what is the probability that at least 17% of those tests will be given a 100?" then it is possible to work this problem. To restate this, it is asking "what is the probability that AT LEAST 4 tests are graded 100?"
This problem is equivalent to a coin problem. You have a coin that has a probability .1 of heads and .9 of tails. If you flip that coin 19 times, what is the probability that you get 4 or more heads? Same question as above.
Lets write the number of ways to choose m objects from a set of n objects (number of combinations of n objects taken m at a time) as C(n,m).
Then the probability P(0) of getting no scores of 100 can be written as C(19,0)(.9^19)(.1^0) where ^ indicates an exponent. Here C(19,0) is equal to 1, and so is .1^0, so this probability is .9^19.
Similarly, the probability P(1) of getting exactly one score of 100 is C(19,1)(.9^18)(.1^1) or 1.9*(.9^18); likewise, the probabilities of getting exactly two or three scores of 100 are C(19,2)(.9^17)(.1^2) and C(19,3)(.9^16)(.1^3), or 19*9*.01*(.9^17) and 19*3*17*.001*(.9^16).
These are the cases where the objective (to have at least 4 scores of 100) is false. If we subtract them from 1 we get the required probability.
P(at least 4) = 1 - P(0) - P(1) - P(2) - P(3) which is approximately
1 - .135085172 - .285179807 - .285179807 - .179557656
So P(At least 4 scores of 100) = .114997558 (approximately)
or you could say 11.5 %.
