Probability of Outcome

KAE

New member
Joined
Oct 2, 2013
Messages
1
I have the following story problem:

A professor is teaching a large class. He is going to break the class up into groups of 3. In the class, 53% of students have never taken a calculus course. 30% have taken 1 semester of calculus. 17% have taken 2 or more semesters. Question: What is the probability at least one of your group (of 3) have had more than 1 semester of calculus.

.17/3 = .056%

But, I think I am not including something.
 
I have the following story problem:

A professor is teaching a large class. He is going to break the class up into groups of 3. In the class, 53% of students have never taken a calculus course. 30% have taken 1 semester of calculus. 17% have taken 2 or more semesters. Question: What is the probability at least one of your group (of 3) have had more than 1 semester of calculus.

.17/3 = .056%

But, I think I am not including something.
By any chance, are you studying the Binomial Distribution? Lets treat it as a binary choice:
Probability of more than 1 semester = p = 0.17
Probability of 0 or 1 semester = q = 0.83
[Note that it is a "large" class, so these probabilities don't change as a group of 3 is chosen.]
Number of trials = n = 3
You can use the binomial formula if you know it, but it is just as easy to do it by logic.

P(0) = the probability that all 3 are in the category of 0 or 1 semester = . . .

P(at least 1) = 1 - P(0) = . . . [quite a bit more probable than your first try!]
 
Top