# Thread: probability distribution question - sampling without replacement.

1. ## probability distribution question - sampling without replacement.

A student is preparing for an upcoming exam. The professor for the course has given the class 30 questions to study from and plans to select 10 of the questions for use on the actual exam. suppose that the student knows how to solve 25 of the 30 questions.

(a) What is the probability that the student will get perfect on the exam?

(b) What is the probability that the student will get at least 8 questions correct on the exam?

I was thinking part a shud be 30C5/30C10, but it does not make sense, the probability shud be higher if the student know 25 of the questions, any help will be appreciated.

2. ## Re: probability distribution question

Try this.

p = 25/30 = 0.83

n = 10

And x the number of correct answers being binomial

a) find the probability that P(x=10)

b) find the probability that P(x>=8) = P(x=8)+P(x=9)+P(x=10)

3. ## Re: probability distribution question

Originally Posted by wopashui
A student is preparing for an upcoming exam. The professor for the course has given the class 30 questions to study from and plans to select 10 of the questions for use on the actual exam. suppose that the student knows how to solve 25 of the 30 questions.

(a) What is the probability that the student will get perfect on the exam?

(b) What is the probability that the student will get at least 8 questions correct on the exam?

I was thinking part a shud be 30C5/30C10, but it does not make sense, the probability shud be higher if the student know 25 of the questions, any help will be appreciated.
(a) (25C10)(5C0)/30C10 = ....

(b) (25C10)(5C0)/30C10 + (25C9)(5C1)/30C10 + (25C8)(5C2)/30C10 = ....

4. ## Re: probability distribution question

Originally Posted by pickslides
Try this.

p = 25/30 = 0.83

n = 10

And x the number of correct answers being binomial

a) find the probability that P(x=10)

b) find the probability that P(x>=8) = P(x=8)+P(x=9)+P(x=10)
Since it is sampling without replacement, I don't think the binomial distribution will be valid. The hypergeometric distribution will apply.