Counting Techniques

[MarkSA]

Hello,

Just started a statistics class thinking it would be a relatively easy one to go with my other difficult classes, but here I am really stumped about what's going on.

Q: A display allows a customer to hook together any selection of components, one of each type. These are the types:
Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood
CD player: Onkyo, Pioneer, Sony, Technics
Speakers: Boston, Infinity, Polk
Cassette: Onkyo, Sony, Teac, Technics

d) How many ways can a selection be made if at least one Sony component is to be included?
e) If someone chooses components in a random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?

I've done the first 3 parts, just by simple multiplication, but I don't think I can do that with part d) and e). I'm really confused about how I would approach this problem.

 

[galactus]

Hello,

Just started a statistics class thinking it would be a relatively easy one to go with my other difficult classes, but here I am really stumped about what's going on.

Q: A display allows a customer to hook together any selection of components, one of each type. These are the types:
Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood
CD player: Onkyo, Pioneer, Sony, Technics
Speakers: Boston, Infinity, Polk
Cassette: Onkyo, Sony, Teac, Technics

d) How many ways can a selection be made if at least one Sony component is to be included?

There are 240 different ensembles one can make from the 4 different components. Why?.

The best way to find at least 1, is to find the number with no Sony and subtract from the total(240).

There are 108 without Sony. Therefore, 240-108=132 ways without any Sony's. See?.

For exactly 1 Sony

For one way, choose a Sony Receiver and no Sony's from the others. That would be choose the Sony from the receivers, then you have 3 from the CD, 3 from the speakers, and 3 from the cassette(27). Then do the same with the other components by excluding the Sony's and add them up. See what I mean?.

 

[MarkSA]

Yes thank you. That makes sense. For part e), I end up with 132/240 probability for at least one Sony, or 0.55. And I end up with 27+36+36/240 probability for exactly one Sony, which is 0.4125.

It almost seems like each problem is a puzzle.. is there a set of standard procedures to follow when solving these types of questions?

 

[mmm4444bot]

… started a statistics class thinking it would be a relatively easy one to go with my other difficult classes …

Oh no!

< For me, statistics was the second-most difficult math subject that I ever studied in school. In fact, I had to take my STAT240 final twice because the first attempt literally forced me home early, sick. >

 

[MarkSA]

Re:

< For me, statistics was the second-most difficult math subject that I ever studied in school. In fact, I had to take my STAT240 final twice because the first attempt literally forced me home early, sick. >

Well, i'm not sure what I was expecting out of the class. It's supposed to be Engineering Statistics, with a lot of A's given out (relative to other engineering courses), which usually indicates an easy class to me. From what little I was familiar with statistics before, I was thinking it would be along the lines of manipulating data. I thought I would be learning how regression works, standard deviation, etc - the types of things that I might do with data analysis in Excel. I feel really lost dealing with probability though.