Help needed with statistics, probability

[rock.freak667]

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

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) = ⁿC_x p_x(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)= ¹⁰⁰C_x0.98^x0.02^100-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?

 

[vela]

P(A \cup B) 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.

 

[rock.freak667]

P(A \cup B) 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

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.

 

[vela]

I typed the formula wrong

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.

 

[rock.freak667]

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

 

[vela]

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?

 

[rock.freak667]

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.

 

[vela]

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) = ⁿC_x p_x(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) = ¹⁰⁰C_x0.98^x0.02^100-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.

 

[vela]

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 A \cap B, and so on. (A and B) or (A and C) or (B and C) corresponds to the union of those intersections.

 

[rock.freak667]

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?

 

[vela]

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.

 

[rock.freak667]

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

Both engines A and B working corresponds to A \cap B, 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.

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

 

[vela]

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.

 

[rock.freak667]

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)?

 

[vela]

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

 

[rock.freak667]

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

 

[vela]

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

 

[rock.freak667]

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

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

there is no way to simplify that term?

 

[vela]

Not that I know of.

 

[rock.freak667]

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.