# Thread: Need help with geometric distribution question

1. ## Need help with geometric distribution question

I don't have much confidence in my answers and would really appreciate some input. I apologize about the length of this, I'm pretty scared I got things wrong.
The question is:

A survey was carried out checking the percentage of aged 6-18 students suffering from shortsightedness. The results were:
35% of the students are ages 6-10, and 10% of them use glasses.
40% of the students are ages 10-15, and 20% of them wear glasses.
The rest of the students are ages 15-18 and 30% of them wear glasses.

Students are checked randomly until one is found to wear glasses.
What's the probability that in total 6 students will be checked?

I described the question's details in a tree and used the complete probability law to find out the probability of those who wear glasses in all the ages. So 0.35*0.1+0.4*0.2+0.25*0.3=0.19

So what I did was defining a random variable X, as 'the number of checks until a student is found to wear glasses', X~G(0.19)
then:
P(X=6)=0.19*〖0.81〗^5=0.066

Now the question I really struggle with is:
20 students are chosen randomly and it appears that all of them wear glasses. Whats the probability that exactly 8 out of the 20 are ages 10-15?
This is obviously conditional probability so at first I calculated the probability that a student aged 10-15 will be wearing glasses: 0.2*04=0.08
but this is where I get stuck. I don't know how to express 'age 10-15|wearing glasses'
I know the rest basically means using the probability I struggle finding in a binomial distribution.

Anyone can help me please? again really sorry for the length.

Thanks!

2. ## Re: Need help with geometric distribution question

P(A|B) = P(A and B)/P(B). Let A be the event that age is between 10-15 and B be the event that they're wearing glasses. You have P(A and B) and P(B).

3. ## Re: Need help with geometric distribution question

So basically:
0.08/0.19=0.421
Is my approach correct in my answers though?

Thank you!

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