Stat Tree Diagram Help

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
HEllo there,

I needed some help with drawing this tree diagram for my statistics class. I'd appreciate it if anyone could help me out:

Q) A biomechanical device for medical emergencies can operate 0, 1 or 2 times a night. Construct a tree diagram to show taht there ar 10 different ways that it can be operated for a total of 6 times over 4 nights.

If I just draw one branch which branches out to 0 1 2, and then branches come out of each digit, 0 1 and 2 and each having their branch of 0 1 2 , then it just becomes way too large.. any idea how do I do this? i'm clueless..

Thanks

 

CottonRabbit

Golden Member
Apr 28, 2005
1,026
0
0
Reread the problem. It wants you to operate 6 times in 4 nights. Every night you can operate a maximum of 2 times. You don't want every possible combination, you just want the ones that total to 6
 

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
hmm I see.. But what do Ihave to show?

N1 > 0 1 2
N2> 0 1 2
N3 > 0 1 2
N4 > 0 1 2

Is that sufficient? That shows that it can be either 0 1 2 . and that it can be arranged in 10 diff ways.. how do i specify the ways using the diagram? or do i set it up like this?

N >

0 > 2 > > 2
1 > 2 > 2> 1
2 > 2> 2 > 0
 

CottonRabbit

Golden Member
Apr 28, 2005
1,026
0
0
The sum of 2 > 2 > 2 > 0 (4 nights) = 6. That's one possible way out of 10. You want a tree diagram with only the branches that sum to 6 showing.
 

esun

Platinum Member
Nov 12, 2001
2,214
0
0
I think you want something like:

Nights:
N0___N1___N2___N3
0 --> 2 --> 2 --> 2
1 --> 1 --> 2 --> 2
__--> 2 --> 1 --> 2
__--> 2 --> 2 --> 1
2 --> 2 --> 2 --> 0
__--> 0 --> 2 --> 2
__--> 2 --> 0 --> 2
__--> 1 --> 2 --> 2
__--> 2 --> 1 --> 2
__--> 2 --> 2 --> 1
 

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
Cool.. Thanks much for the help. Could you please verify the following answers too? Somehow, I'm kinda finding statistics hard compared to calc even though I've heard its supposed to be very easy...

1) P(A) = 0.35, P(B) = 0.73 and P(A intersect B ) = 0.14, find:

a) P ( A Union B ) = P(A) + P(B) - P ( A intersect B ) = 0.94
b) P ( complement of A intersect B ) = P(B) - ( A intersect B ) = 0.73 - 0.14 = 0.59 .. Is this correct?
c) P ( A intersect complement of B ) = P(A) - ( A intersect B ) = 0.35 - 0.14 = 0.21 .. is this correct?
d) P ( complement of A union complement of B ) = 1 - P ( A intersect B ) = 1 - 0.14 = 0.86.. A friend told me that this was wrong and that I must do 1 - ( A union B ) = 1 - 0.94 = 0.06

Thanks much..
 

MagnusTheBrewer

IN MEMORIAM
Jun 19, 2004
24,122
1,594
126
Stats is all about queue theory and has the same relationship to math as probability does i.e. none to speak of. Next to accounting, it was the most depressing classes I ever took.
 

esun

Platinum Member
Nov 12, 2001
2,214
0
0
Originally posted by: jai6638
Cool.. Thanks much for the help. Could you please verify the following answers too? Somehow, I'm kinda finding statistics hard compared to calc even though I've heard its supposed to be very easy...

1) P(A) = 0.35, P(B) = 0.73 and P(A intersect B ) = 0.14, find:

a) P ( A Union B ) = P(A) + P(B) - P ( A intersect B ) = 0.94
b) P ( complement of A intersect B ) = P(B) - ( A intersect B ) = 0.73 - 0.14 = 0.59 .. Is this correct?
c) P ( A intersect complement of B ) = P(A) - ( A intersect B ) = 0.35 - 0.14 = 0.21 .. is this correct?
d) P ( complement of A union complement of B ) = 1 - P ( A intersect B ) = 1 - 0.14 = 0.86.. A friend told me that this was wrong and that I must do 1 - ( A union B ) = 1 - 0.94 = 0.06

Thanks much..

Easiest way to verify is to draw a sample space, mark events A, B, and their intersection, and go from there.

But anyway, I'll check the formulas but not the numbers:

a) Correct
b) Correct
c) Correct
d) Correct (you are, not your friend)

EDIT: Oh, and saying probability has no relationship to math is so far off.
 

jai6638

Golden Member
Apr 9, 2004
1,790
0
0
Great.. thanks much... Last question,

Among the 12 solar collectors on display at a trade show, 9 are flat-plate collectors and the others are concentrating collectors. If a person visiting the show randomly selects 4 of the solar collectors to check out, what is the probability that 3 o fthem will be flat-plate collectors?

WOuld I use binomial distro for this or the hypergeometric distribution? I used the latter and got the answer to be 50.1%

THanks