Combination and Statistics

Shinzaku

New member
Joined
Feb 27, 2012
Messages
1
So I need some insight on my homework:

1) Treating a patient for high blood pressure, a physician has a choice of five different drugs, two of which are experimental. She can also select one of four drugs, two involve indoor activities and the other two outdoor. Lastly, she can choose one of three diets, one being completely salt-free.

a) How many treatments of one drug, one exercise, and one diet are possible?
b) How many treatments in part A involve the use of the experimental drug?
c) All treatments in part A are considered equally desireable. If a treatment is selected at random, what is the probability that it will involve an experimental and an outdoor exercise program?
d) If a specific one of the experiemental drugs is dangerous when used with the salt-free diet, what is the probability that such a treatment will be prescribed by chance?


I'm using C(n, x) notation, where C is a combination of n using x samples. So C = n!/(x!(n - x)!)

A) I got 60. Using combinations, I took C(5,1)*C(4,1)*C(3,1). Is this right?

B) I'm assuming I just take the combination, C(5,2) to get the combination of the experimental drugs. But that doesn't seem right - do I need to use the total I found in some way?

C) Since a treatment has to consist of one of each, I use that in calculating it. So C(5,2)*C(4,2)*C(3,1)/C(60,1). This gives me a whole number and not a fraction/probability... so I'm using the numbers terribly wrong.

D) I have no clue where to start.


Any ideas?
Am I wrong with assuming these are combinations? I know permuations required that they are ordered, but I feel like order doesn't matter here so they are combinations for probabilities...

I'm pretty sure once I get this one I can get the rest of my homework.
 
1) Treating a patient for high blood pressure, a physician has a choice of five different drugs, two of which are experimental. She can also select one of four drugs, two involve indoor activities and the other two outdoor. Lastly, she can choose one of three diets, one being completely salt-free.

a) How many treatments of one drug, one exercise, and one diet are possible?
b) How many treatments in part A involve the use of the experimental drug?
c) All treatments in part A are considered equally desireable. If a treatment is selected at random, what is the probability that it will involve an experimental and an outdoor exercise program?
d) If a specific one of the experiemental drugs is dangerous when used with the salt-free diet, what is the probability that such a treatment will be prescribed by chance?


I'm using C(n, x) notation, where C is a combination of n using x samples. So C = n!/(x!(n - x)!)

A) I got 60. Using combinations, I took C(5,1)*C(4,1)*C(3,1). Is this right?

B) I'm assuming I just take the combination, C(5,2) to get the combination of the experimental drugs. But that doesn't seem right - do I need to use the total I found in some way?

Problem A) looks fine.

Problem B) is not C(5,2). C(5,2) would simple be the number of ways you could choose 2 things out of 5. Instead, you MUST choose only 1 experimental drug, and there are only 2 of those. So your fist choice is C(2,1). Make sense? But you must still choose the activity and diet, so

C(2,1)*C(4,1)*C(3,1)

This was not asked, but if you wanted to know the probability of randomly choosing a treatment that had an experimental drug, it would be the number of treatments with the exp. drugs divided by the total number of treatment options:

C(2,1)*C(4,1)*C(3,1) / C(5,1)*C(4,1)*C(3,1)

As you can see, that will simplify to just 2/5, which reflects the fact that there are 2 experimental drugs and 5 total drugs.

Hope that helps.
 
Top