Probability of 0-6 Students in Class of 20 - Help!

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The discussion focuses on calculating the probability of having 0 to 6 students in a class of 20 who fall into a specific category, given that 8% of students belong to that category. The binomial distribution formula is used, where the probability of success (a student being in the category) is 0.08 and the probability of failure is 0.92. Participants clarify the correct usage of LaTeX formatting in the equations and discuss the concept of expected value in probability distributions. The conversation also touches on misconceptions about probabilities in larger groups, emphasizing that even with more students, the chance of having none or all in a category remains. Overall, the thread provides insights into binomial probabilities and common errors in mathematical notation.
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I haven't done a probability problem in a long time. Thank you.

Assume that 8 % of the students fall in some particular
category. We have 20 students in our class. What is the
probability that we have 0, 1, 2, 3, 4, 5, or 6 of those students in
our class?
 
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Binomial distribution. If the probability of a "success" in one "trial" is p, then the probability of a "failure" is 1-p. The probability of exactly i "successes" in n "trials" (and so n-i "failures") is _nC_i p^i q^{n-i} where _nC_i is the "binomial coefficient" \frac{n!}{i!(n-i)!}).

The probability that anyone student is in that category is 0.08 so the probability a student is NOT in that category is 0.92. The probability exactly i students out of 20 are in that category is \frac{20!}{i!(20-i)!}i^{0.08}(20-i)^{0.92}.

Calculate that for i= 0, 1, 2, 3, 4, 5, 6 and add.

(Sorry, I left out the "[ tex ]" originally)
 
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Ok

Thanks a lot I needed it
 
Wait

What did you mean by frac in the beginning of the equation and tex in tha last part.?
 
dranger35 said:
What did you mean by frac in the beginning of the equation and tex in tha last part.?
I think it's just a typo. The correction would be \frac{20!}{i!(20-i)!}i^0.08(20-i)^0.92
 
I don't know anything about probability but I would like to and I just had a thought.

If there were 100 students then the chance of 8 students being in that category would be 100% no? However, the chances of 7 students or 9 would being in that category would be lass than 100%. So their is a maximum and and it seems you could graph the probabilities verses the number of students and it would be parabola?
 
I think that's the idea of the bell curve.
 
dranger35 said:
What did you mean by frac in the beginning of the equation and tex in tha last part.?

I accidently left out the beginning "tex" tag. I edited to fix that.

honestrosewater: It's good to know I'm not the only one who messes up latex! (Or did you do that intentionally to make me feel better?)

You need { } in "i^{0.08}"

slug:"If there were 100 students then the chance of 8 students being in that category would be 100% no?"

No, if the probability of a student being in a certain category is any number less than 1.00, no matter how many students you have in a class, there is always some probability that NONE of the students are in that category and some probability that ALL are. You are, however, correct that the probability is highest at the "expected value" which, for a binomial distribution is np. In the orginal problem that is (20)(0.08)= 1.6 (round to 2) and if n= 100, 8.
But it's not a parabola (for one thing probability is never negative!)- it is, as philosophking said, a bell-shaped curve: basically given by e^{-x^2}.
 
HallsofIvy said:
honestrosewater: It's good to know I'm not the only one who messes up latex! (Or did you do that intentionally to make me feel better?)

You need { } in "i^{0.08}"
Actually, I just copy-pasted what you had written and added the beginning [ tex ]. Oddly enough I did wonder if that was correct- but you had just written _nC_i p^i q^{n-i} correctly so figured you knew to include the braces. Funny. I do mess up though- that's why I preview before posting. :biggrin:
 
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