Question 1084349: 1. There are 18 players on the roster of the Ludlow wildcats baseball team. Of the 18 players 8 are currently enrolled in college. The coach decides to name tri-captains and to select them at random. The names of the players are placed in an old baseball hat and there selected at random. What is the probability none of those selected are college students? Follow up Question.What is the probability of selecting one college stident and two non-college students?
2. Company buy electric motors from two suppliers. Sixty percent are purchased from two suppliers. Sixty are purchased from the Hall electric and the rest from Harmon products. Five percent of the motors purchased from the Hall Electric need additional work, whereas 8% from Harmon products need additional work. An electric motor was selected at random and found to be defective. What is the probability it was purchased from Harmon product?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! out of 18 players, 8 are currently enrolled in college.
coach selects 3 player atrandom.
p(college student) = 8/18
p(not college student = 10/18
probability all 3 players selected are not college students is 10/18 * 9/17 * 8/16 = 720 / 4896.
this can be simplified to 5/34.
that's the probability that all 3 will not be college students.
in decimal form, that would be .1470588235
in percent form, that would be 14.70588235%
you can also use the combination formula of c(n,x) to get the same percent.
in combination form, the probability is c(10,3) / c(18,3)
that's the number of ways you can get a set of 3 students that are not college students divided by the number of ways you can get 3 students at all.
c(10,3) is the combination formula for getting a set of 3 out of 10 where order doesn't matter.
c(18,3) is the combination formula for getting a set of 3 out of 18 where order doesn't matter.
c(n,x) = n! / (x! * (n-x)!)
the combination type formula works when there is no replacement.
it doesn't work when there is replacement.
in this particular case, you would get c(10,3) / c(18,3) = .1470588235
if you multiply that by 34, you get 5
.1470588235/1 * 34/34 = 5/34.
that's the probability that none of the students selected is a college student.
the probability that one of the students selected is a college student would be:
9/18 * 10/17 * 9/16 * c(3,1) = .1654411765 * 3 = .4963235294
you multiplying by c(3,1) because there are that many ways you can get 3 players, one of which is a college student, and 2 of which are not college students.
the 3 ways are:
cnn
ncn
nnc
cnn would be selecting a college student first and then 2 non-college students.
ncn would be selecting a non-college student first, then a college student, then a non-college student.
nnc would be selecting 2 non-college students first and then a college student.
all 3 are possible.
the probability of each sequence is the same.
therefore multiply one of the possible ways by 3 to get the total possibility.
consider:
9/18 * 10/17 * 9/16 = (10 * 9^2) / (18 * 17 * 16)
10/18 * 9/17 * 9/16 = (10 * 9^2) / (18 * 17 * 16)
the probability is the same even though the selection sequence is different.
in combination terms, it would be c(9,1) * c(10,2) / c(18,3) = .4963235294
that's the number of ways you can get 1 college student out of 9 times the number of ways you can get 2 non-college students out of 10, all divided by the number of ways you can get 3 players out of 18.
i'm not sure i understand your second part.
you state 2 suppliers.
then you state 60% are from 2 suppliers.
does that mean the other 40% are from somebody other than the two suppliers?
i'm not sure i can answer that question anyway, but i'm willing to give it a shot after you clarify the problem statement since it doesn't make sense to me.
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