1. ## Tried this problem and not getting the answers?? Help please

Hi,
Question is 4 out of 5 teens carry a wireless device.
If 3 teens are randomly selected, find the following probabilities:

A) - None of the teens have a wireless device. ANS is .2

B) - at least one has a wireless device ANS is .8

C) - all 3 have wireless devices. ANS is .512

A) I put 1/5 X 1/4 X 1/3 and I get .02 What did I do wrong?
B) I am not sure what to do here
C) I am not sure what to do here.
I am having trouble with stats and the teacher is not that good in explaining, so any help would be great, thanks.
Jo

2. ## Re: Tried this problem and not getting the answers?? Help please

Hi Jo,

I just wanted to clarify a few points in your question, because at the moment only the answer given to part C seems reasonable to me. This leads me to think that you may have rephrased the question, rather than posting in its original form. Are you told the size of the group of teenagers that you are selecting from? The question says "4 out of 5", but presumably that is saying "80% of teenagers", rather than "in a group of 5 teenagers, 4 have a wireless device", as in this latter case, part A would make no sense. I may have some more questions for you depending on your response.

3. ## Re: Tried this problem and not getting the answers?? Help please

Hi britmath

It's how it is worded in my book. The beginning just said a poll was taken, that's all. So then the answers at the back of my book are wrong then I guess?
No size is given, just that four out of 5 teens carry a wireless device. THAT"S IT!.

I tried them again, and I can't even get the last one. Something is not clicking in my head on this, and I don't know why.
Any help would be great, thanks.
Jo

4. ## Re: Tried this problem and not getting the answers?? Help please

Well, in that case here are the things you want to think about. You're told that a student carries a wireless device with probability 4/5 (call this A), which means that the probability that a student, chosen at random, does not have a wireless device is 1/5 (call this B). If you choose 3 students at random, the combinations of A and B will look like this:

AAA (i.e. the first student has a wireless device, and the second student, and the third student).
AAB
ABA
BAA
ABB
BAB
BBA
BBB

So, there are a total of 8 (2*2*2) different combinations that can occur. Consider the third part of the question, where you want to know the probability of all 3 students having a wireless device, you can see from the list of combinations that only one such combination exists: AAA. You know the probability of A, so hopefully you can now work this one out (the answer given by the book for part C is correct).

For the first two parts of the question I believe the answers are wrong, but you approach these parts in a similar fashion to part C: for the case where none of the students has a wireless device, you are interested in instances where only Bs are present, and again, only one such instance exists.

With the case of at least one student having a wireless device, you want to identify combinations that contain *at least* one A; note that more than one such combination exists (ABB, BAB, etc), so you'll need to calculate the probability of each combination to get the final result (there is another approach to this part of the question that is faster, see if you can figure it out).

I hope that gets you started!