Help needed with statistics, probability

In summary: P(X>x).All good until the last line.If P(X=x)=0.10, that means 10% of the cartons have x defects. There's no need to divide by 100. The probability is already the proportion.Also, the problem isn't asking for P(X=x). That would be the probability that a carton has exactly x defective parts. The problem asks for the probability that a carton has at least x defective... so the answer is P(X>x).
  • #1
rock.freak667
Homework Helper
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URGENT help needed with statistics, probability

Homework Statement



1)
A plane functions iff at least 2 of its 3 engines function.
P(each engine functions)=p, the engines operate independently of each other. Find the

probability that the plane functions.

The Attempt at a Solution



A=engine 1, B=engine 2, C=engine 3.

P(plane functions)= P(AuB)+P(AuC)+P(BuC)

P(AnBnC)=p3

P(AnB)=P(AnC)=P(BnC)=p2

P(AuBuC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)-P(AnBnC)=3p-3p2-p3

http://img168.imageshack.us/img168/2991/venn.jpg [Broken]

But the regions they want are P(AuB)+P(AuC)+P(BuC) which is just 3(p-p2-2p3)



Homework Statement


2)

2% of products in a factory are defective. Products are sold in packages of 100. What

proportion of cartons contain at least 'x' defective products? x=1,2,3,...,100. (use

binomial distribution)


Homework Equations



Binomial distribution: P(X=x) = nCx px(1-p)n-x , where p is the probability of success.

The Attempt at a Solution


P(defective)=0.02
P(not defective)=0.98

I need to get X~Bin(100,0.98)

P(X=x)= 100Cx0.98x0.02100-x

So my answer is just P(X=x)/100 ? (replacing P(X=x) with the above)

Homework Statement



3)
A shipment of 8 items contain 3 that are defective. A person makes a random selection of 2

of these items, find the probability distribution for the number of defectives X. Find the

cumulative functions of X as well.


The Attempt at a Solution



So P(X=x) is defined as follows:

0 for x<1

3/8 for 1<x<2

5/56 for 2<x<3 ( 2 defective is 3/8 * 2/7)

6/336 for x>3 (3 defective is 3/8 * 2/7 *1/6)

And to get the cdf I just integrate the functions (in the regions) between 'x' and -infinity?
 
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  • #2


[itex]P(A \cup B)[/itex] is the probability that at least one of engines A and B is functioning, but you want the probability that both are functioning at the same time. Same problem with the other pairs.

Also, you should have

P(AuBuC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC) + P(AnBnC)

not a negative sign on the last term.
 
  • #3


vela said:
[itex]P(A \cup B)[/itex] is the probability that at least one of engines A and B is functioning, but you want the probability that both are functioning at the same time. Same problem with the other pairs.

Also, you should have

P(AuBuC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC) + P(AnBnC)

not a negative sign on the last term.

I typed the formula wrong :blushing:

Then I would get P(AnB)+P(AnC)+P(BnC)=3p2, if that is the case, I don't understand why they told me to use that formula.
 
  • #4


rock.freak667 said:
I typed the formula wrong :blushing:

Then I would get P(AnB)+P(AnC)+P(BnC)=3p2, if that is the case, I don't understand why they told me to use that formula.
Now you're triple-counting the intersection of A, B, and C, so you need to subtract the extra from your result. You can just add up the probabilities you have in your Venn diagram from the relevant regions to see what the answer should be.
 
  • #5


vela said:
Now you're triple-counting the intersection of A, B, and C, so you need to subtract the extra from your result. You can just add up the probabilities you have in your Venn diagram from the relevant regions to see what the answer should be.

It would just be the intersections alone, so it'd be 3(p2-p3)+p3
 
  • #6


Yes, that's right. I see what they wanted you to do now. The plane functions if (A and B) or (A and C) or (B and C). Do you see how that's the union of 3 sets, so that that formula would apply?
 
  • #7


vela said:
Yes, that's right. I see what they wanted you to do now. The plane functions if (A and B) or (A and C) or (B and C). Do you see how that's the union of 3 sets, so that that formula would apply?


Wouldn't that give a different answer to the post above?

P(AuBuC)=3p-3p2+p3, I'd have a 'p' term here.
 
  • #8


rock.freak667 said:

Homework Statement


2) 2% of products in a factory are defective. Products are sold in packages of 100. What proportion of cartons contain at least 'x' defective products? x=1,2,3,...,100. (use binomial distribution)

Homework Equations



Binomial distribution: P(X=x) = nCx px(1-p)n-x, where p is the probability of success.

The Attempt at a Solution


P(defective)=0.02
P(not defective)=0.98

I need to get X~Bin(100,0.98).

P(X=x) = 100Cx0.98x0.02100-x

So my answer is just P(X=x)/100 ? (replacing P(X=x) with the above)
All good until the last line.

If P(X=x)=0.10, that means 10% of the cartons have x defects. There's no need to divide by 100. The probability is already the proportion.

Also, the problem isn't asking for P(X=x). That would be the probability that a carton has exactly x defective parts. The problem asks for the probability that a carton has at least x defective parts.
 
  • #9


rock.freak667 said:
Wouldn't that give a different answer to the post above?

P(AuBuC)=3p-3p2+p3, I'd have a 'p' term here.
No, because the three sets aren't A, B, and C.

Both engines A and B working corresponds to [itex]A \cap B[/itex], and so on. (A and B) or (A and C) or (B and C) corresponds to the union of those intersections.
 
  • #10


vela said:
All good until the last line.

If P(X=x)=0.10, that means 10% of the cartons have x defects. There's no need to divide by 100. The probability is already the proportion.

Also, the problem isn't asking for P(X=x). That would be the probability that a carton has exactly x defective parts. The problem asks for the probability that a carton has at least x defective parts.

So then if I find P(X=x-1), I'd get exactly 'x-1' defective, and then if I put 1-P(X=x-1), I should get 'x or more defective', which would be at least x, right?
 
  • #11


No, that would give you the probability of 0, 1, ..., x-2 defects or x, x+1, x+2, ... 100 defects. You only want the second part, not the 0 through x-2 part.
 
  • #12


vela said:
No, because the three sets aren't A, B, and C.

Both engines A and B working corresponds to [itex]A \cap B[/itex], and so on. (A and B) or (A and C) or (B and C) corresponds to the union of those intersections.

Oh now I get it.

vela said:
No, that would give you the probability of 0, 1, ..., x-2 defects or x, x+1, x+2, ... 100 defects. You only want the second part, not the 0 through x-2 part.

This part has me confused :confused:
 
  • #13


1-P(X=x) is the probability of not having exactly x defects. That means you could have more or less than x defects, which isn't the same as having at least x defects.
 
  • #14


The problem wants me to get P(X≥x), I have P(X=x). I need to get 1-P(X<x). How would I get P(X<x)?
 
  • #15


You would sum the probabilities P(0)+P(1)+P(2)+...+P(x-1).
 
  • #16


vela said:
You would sum the probabilities P(0)+P(1)+P(2)+...+P(x-1).

I know it would be a binomial expansion, but of what? The normal bin distribution is (p+q)n where q=1-p
 
  • #17


It'll be part of a binomial expansion. As far as I know, there's no closed form expression for it.
 
  • #18


vela said:
It'll be part of a binomial expansion. As far as I know, there's no closed form expression for it.



[tex]P(X \geq x) = 1 - \sum_{x=0} ^{x-1} ^{100}C_x 0.98^x 0.02^{100-x}[/tex]

there is no way to simplify that term?
 
  • #19


Not that I know of.
 
  • #20


vela said:
Not that I know of.

Then I have to leave my answer like that then.

any help for the 3rd one?


EDIT: I got out the 3rd one.
 
Last edited:

1. What is the difference between statistics and probability?

Statistics is the study of collecting, analyzing, and interpreting data to make informed decisions. Probability, on the other hand, is the measure of the likelihood of an event occurring based on mathematical principles.

2. How can I determine which statistical test to use for my data?

The choice of statistical test depends on the type of data you have collected and the research question you are trying to answer. Consult with a statistician or refer to a statistical guidebook to determine the appropriate test for your data.

3. What is the purpose of conducting a hypothesis test?

Hypothesis testing is used to determine if there is a significant difference between two or more groups or variables. It allows researchers to make conclusions about the population based on a sample of data.

4. How do I interpret p-values in statistical analysis?

P-values indicate the probability of obtaining the observed results of a study by chance. A smaller p-value suggests that the results are unlikely to occur by chance and are therefore considered statistically significant. The commonly accepted threshold for significance is a p-value of 0.05 or less.

5. Can I use statistics to prove causation?

No, statistics can only show correlation between variables. To prove causation, a controlled experiment must be conducted to establish a cause-and-effect relationship between variables.

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