Probability Question????

Question

At a certain university, 1 in 7 students is an engineering major. Suppose we randomly select students until we find one who is an engineering major. What is the probability that we will find an engineering major within the first fourn students we select? (Assume a geometric distribution.)

Get get 10 points as Best Answer if its correct
2 points for trying

Explain how you get your numbers...please

Answer 1

why are you thinking about math now its summer!

Answer 2

Assuming that we are selecting students with replacement (that's where the "assume a geometric distribution" comes in), we can do the following:

P(NOT selecting an engineering major) = 6/7

So the probability of NOT selecting an engineering major in four tries is (6/7) * (6/7) * (6/7) * (6/7) ( or (6/7)^4 ) since the trials are independent.

We want the probability of selecting at least one engineering major, which is the complement of the event above.

Since P(A) = 1 - P(not A), we have

P(at least one engineer selected in four trials) = 1 - (6/7)^4, or approximately 46.02%.


If we were not selecting with replacement, then the probability would be slightly different depending on the size of the student body (but close enough for all practical purposes).

As an aside, the answer above which claims that you could find the answer using 1/7 + 1/6 + 1/5 +1/4 pains me as a teacher. A quick way to show the error in this reasoning is to note that if we were trying to select an engineer in 5 tries, this proposed method would yield a probability of 1/7 + 1/6 + 1/5 + 1/4 + 1/3, which is over 100%! Please don't fall into such traps!

Answer 3

The probability that you DON'T find an engineering major on any selection is
6 / 7.

The probability that you don't find an engineering major on any of four tries is
(6 / 7)^4
= 6^4 / 7^4
= 1296 / 2401

The probability that you find at least one engineering major in the first four selections, then, is
1 - (1296 / 2401)
= (2401 / 2401) - (1296 / 2401)
= 1105 / 2401, or about 46.0225%.

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If you want to be more specific about this problem, you can determine the probability of exactly how many engineering majors are picked in the first four selections. If n is the number of engineering majors selected in four tries,
P(n) = C(4,n) · (1 / 7)^n · (6 / 7)^(4 - n)

P(0) = C(4, 0) · (1 / 7)^0 · (6 / 7)^(4)
= 1 · 1 · (6 / 7)^4 = 1296 / 2401, or about 53.9775%.

P(1) = C(4, 1) · (1 / 7)^1 · (6 / 7)^(3)
= 4 · (1 / 7) · (6 / 7)³ = 864 / 2401, or about 35.9850%.

P(2) = C(4, 2) · (1 / 7)^2 · (6 / 7)^(2)
= 6 · (1 / 7)² · (6 / 7)² = 216 / 2401, or about 8.9963%.

P(3) = C(4, 3) · (1 / 7)^3 · (6 / 7)^(1)
= 4 · (1 / 7)³ · (6 / 7) = 24 / 2401, or about 0.9996%.

P(4) = C(4, 4) · (1 / 7)^4 · (6 / 7)^(0)
= 1 · (1 / 7)^4 · 1 = 1 / 2401, or about 0.0416%.

The problem of finding at least one engineering major can then be solved either of two ways:
1 - P(0)
= 1 - (1296 / 2401)
= 1105 / 2401.

or

P(1) + P(2) + P(3) + P(4)
= (864 / 2401) + (216 / 2401) + (24 / 2401) + (1 / 2401)
= 1105 / 2401.

Answer 4

The Proablility is the sum of each constituent event

Probability that 1st student is a major = 1/7
Probability that 2nd student is a major = 1/6
Probability that 3rd student is a major = 1/5
Probability that 4th student is a major = 1/4

Obviously, once the first one is selected; six remain to be chosen from. And then five and then four.

Now the probability that one of the first four is a major is the sum of each of the probabilities calculated above.

1/7 + 1/6 + 1/5 +1/4 = 0.75952381

Probability is 75.95%

Answer 5

The usual way to do one of these is to calculate the "complementary" probability of NOT choosing an engineering student in the first four. Since P(choosing one)+P(NOT choosing one)=1, this will tell you your answer.

The probability of NOT choosing one on any given try is 6/7. So the probability of NOT choosing four will be (6/7)^4--note that here you are ignoring the fact that with each choice you have eliminated one student from the population and so slightly altered the 6/7 nonengineer to total ratio. Or you could say that the total population is so large that this correction doesn't really matter.

Anyway, the best you can do without knowing how many total students is the complement to this, or 1-(6/7)^4.

Answer 6

1 - (6/7) ^ 4, or 46.0%

The probability you WON'T select an engineering major is 6/7.
Each time you select someone, it's a separate "pick", so it's 6/7 raised to the 4th power.
1 minus (6/7) to the 4th power is ~46%. (You need to do the inverse since you are looking for the probabilty you WILL find an engineering major).

there you've got it!

Answer 7

4 out of 7

Answer 8

How many students in the university?

Do you mean the first FOURTEEN students?

Hmmm.....
# of ways an event can occur
over the total # of possible outcomes = probability

Let's see.... say 100,000 students in the university.
14,286 engineering majors.

We choose fourteen students..... I'd say we have a 50% chance of finding an engineering major.

Answer 9

Come on!! It cannot be 4/7. Following this reasoning if you pick 7 students the probability should be 100%, which is obviously not true.

Answer 10

On these types of questions you can make a tree. at each level you can choose an engineer or not choose an engineer, if you choose an engineer stop, if you dont go on. go until you select 4 people and find the sum of the probablilities of successful tries. you should get 4 successes and get about .46

Answer 11

4/7