# Probability Distributions

1. Feb 23, 2008

### Somefantastik

On a multiple guess exam, there are 3 possible answers for each of the 5 questions. What is the probability that the student will get four or more correct answers just by guessing?

Is this hypergeometric or binomial?

2. Feb 23, 2008

### HallsofIvy

Staff Emeritus
Since on each question a student can either choose the correct answer or not, it is binomial. Do you see what is the probability a student will choose the correct answer on a specific question just by guessing?

3. Feb 23, 2008

### Somefantastik

I thought it was either this

$$\frac{(^{5}_{4}) (^{10}_{1})}{(^{15}_{5})}$$ choose 4 from 5 correct answers, choose 1 from 10 wrong answers divided by choosing 5 answers from 15

Or this (which I now know is probably the right answer)

$$(^{5}_{4})p^{4}q + (^{5}_{5})p^{5}$$ which is the probability of 4/5 right answers + probability of 5/5 right answers, where prob correct answer = 1/3

4. Feb 24, 2008

### HallsofIvy

Staff Emeritus
Yes, the second is the one you want (you are NOT choosing 4 or 5 correct answers from all 15- that would imply that you could choose 2 correct answer from one problem and none from another!). Now what number is $$(^{5}_{4})p^{4}q + (^{5}_{5})p^{5}$$?

5. Feb 24, 2008

### Somefantastik

p = 1/3 and q = 2/3 => 0.14 unless I calculated wrong which is entirely possible.