need assistance

#1
Help, and I have no statistical background.

I am taking a course in safety and now have one class left, but ran into a big problem. This class has statistics and I don't have a clue. The text is of no help and doesn't explain anything but theory. I am over my head and need some directions with a number of problems.

1. A system has 5 sensors. The probability of any sensor failing is equal and 3 sensors are in a failed state. As an inspector if 1 sensor is selected randomly what is the probability that it wil be in a failed state?

******** If 2 senors are selected randomly what is the probablity of at least one in a failed state?

******** If 3 sensors are selected what is the probability that the system will be found in a failed state.

2. Consider 4 books on a shelf with 4 spaces. how many unordered ways may they be arranged.

3. Consider the 4 books, but only 3 spaces. how many unordered ways may these books be arranged.

4. A bin contains 7 parts. The probability that any part in the bin is defective is 3.5E-3. If 1 part is elected at random from the in, what is the probability that it is defective.

******* If 3 parts are selcted at random, what is the probability that exactly 2 are defetive?

5. A component has a failure rate per hour of 4.5E-3. The failures are governed by the exponential distribution. What is the probability of this component failing exactly once in 100 hours?

******** what is the probability of this component failing at least once in 100 hours?

6. A fire detection sensor is tested for its failure arte and found to have a mean failure rate of 3.6e-2 per hour of exposure with a standard deviation of .4E-2 per hour. The distribution of the failure rate is normal. If a sensor is selected at random from a large lot for installation in a sysyem, what is the probability that it will have a failure rate greater than 4.5e2 per hour.

I do not even know where to begin with these sample problems. Any assistance is appreciated.
 

JohnM

TS Contributor
#2
1.
part 1 --> p(1) = 3/5 = 0.6
part 2 --> p(at least 1) = 1-p(none) = 1- [ (2/5)*(1/4) ]
part 3 --> assuming that the system will be viewed as in a failed state if all 3 sampled sensors are defective --> p(3) = (3/5)*(2/4)*(1/3)

2. and 3.
use the combinations formula: nCr = n! / (r! (n-r)!)
n = number of books
r = number of spaces

4.
uses the binomial probability distribution
p(r) = nCr * p^r * q^(n-r)

part 1: you can assume that the 1 sampled part was taken directly from the process that made the parts
p = 3.5E -3 = .0035
q = 1-p = .9965
n = 1
r = 1

part 2: same assumption in part 1 is valid here as well
p = 3.5E -3 = .0035
q = 1-p = .9965
n = 3
r = 2

5.
working on it

6.
uses the normal distribution
mean = .0036
std dev = .004
x = .0045

z = (x - mean) / std dev
compute z, then use the normal distribution table to find the proportion of the area under the curve that is above z
 
#3
This is all I have and can figure out so far, I need some help, I have absolutely no stat background, I do not know what most of the symbols mean, or how to actually work out these problems. I have asked my instructor for help and he said "buy the cliff notes". I even had a grad student/educator try to help me and she said she didn't have a clue. Any assistance is appreciated.

1. A system has 5 sensors. The probability of any sensor failing is equal and 3 sensors are in a failed state. As an inspector if 1 sensor is selected randomly what is the probability that it wil be in a failed state?

p(1)=3/5=0.6
p=0.6



******** If 2 senors are selected randomly what is the probablity of at least one in a failed state?

p(at least)= 1-p(none)
= 1-[(2/5)*(1/4)]
p=0.9




******** If 3 sensors are selected what is the probability that the system will be found in a failed state.

p(3)=(3/5)*(2/40*(1/3)
1-p=(3/5)*(2/4)*(1/3)
p=.93



2. Consider 4 books on a shelf with 4 spaces. how many unordered ways may they be arranged.

This number is 1, because order is of no consequences.



3. Consider the 4 books, but only 3 spaces. how many unordered ways may these books be arranged.

4*3*2/3*2*1=24/6
=4



4. A bin contains 7 parts. The probability that any part in the bin is defective is 3.5E-3. If 1 part is elected at random from the in, what is the probability that it is defective.

p(r)=nCr*p^r*q^(n-r)
1*(.0035*.9965)
p=.0035



******* If 3 parts are selcted at random, what is the probability that exactly 2 are defetive?

p(r)=nCr*p^r*q^(n-r)
p(r)=6*(.0035*.0035)*.9965(1)
p(r)=6*(.000012)*.9965
P=.00072



5. A component has a failure rate per hour of 4.5E-3. The failures are governed by the exponential distribution. What is the probability of this component failing exactly once in 100 hours?

.0045*100


******** what is the probability of this component failing at least once in 100 hours?

?


6. A fire detection sensor is tested for its failure arte and found to have a mean failure rate of 3.6e-2 per hour of exposure with a standard deviation of .4E-2 per hour. The distribution of the failure rate is normal. If a sensor is selected at random from a large lot for installation in a sysyem, what is the probability that it will have a failure rate greater than 4.5e2 per hour.


m=.0036
Std. dev=.004
x=.0045
z=(.0045-.0036/.004)
z=.225
 

JohnM

TS Contributor
#4
Check your math on the third part of question number 1, and the second part of question number 4.

You computed z correctly on the last question - now you need to use the normal distribution table to find the probability of z >= .225.

In defense of your instructor, this course sounds like it has a stats prerequisite, and he probably feels like it's not his job to get you up to speed...
 
#5
still trying

On the third part of question 1-this is the re-work.
3/5*2/4*1/3
.60*.50*.33
P=.099

if I then put in the 1-p
that gives me p=.90

The second part of question 4
p(r)=6*(.0035*.0035)*.9965(1)
p(r)=6*(.000012)*.9965
p=.0000717

I looked at the nominal distribution table to find z
I am not quite sure what to look for, I looked down at 2.2, then went to .05 across the table and found .48776
 

JohnM

TS Contributor
#6
On the third part of question 1, you don't need to subtract the multiplication from 1.0 - it's simply:

P(3 defective parts) = P(part 1 defective AND part 2 defective AND part 3 defective)

= (3/5) * (2/4) * (1/3) = 6/60 = 0.1

In the second part of question 4, you should get 3 for 3C2:

3!/(2!(3-2)!) = 6/2 = 3

then your answer would be 3*(.0035^2)*(.9965) = .0000366

For question 7, to find the probability of z >= 0.225, it's 0.411