1. chaeck for solution

Hi,
I have two probability problems and I need them to be checked please:

problem-1
A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
My solution:
lambda=0.03*1=0.03
f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)

2- if 50 panesl is randomly sampled from the production process, what is the probability it has no flaws.
No solution here. I am not sure but it is the same as above but with lambda=0.03*50??

3- What is the expected number of panels that need to be sampled before flaws are found?
I am not sure of what I have done for this question:
lambda=0.03
(0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66

problem 2

the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:

1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)

2-P(X=4)
P(X=4)=b(4;6,0.6)

3-mean
mean=n*p=3.6

4-variance
Var^2=n*p*(1-p)

5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)

B.

Hi,
I have two probability problems and I need them to be checked please:

problem-1
A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
My solution:
lambda=0.03*1=0.03
f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)
Yes, but simplify (or rather calculate the approcimate value of this):

p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704

2- if 50 panesl is randomly sampled from the production process, what is the probability it has no flaws.
No solution here. I am not sure but it is the same as above but with lambda=0.03*50??
Probability that there are no flaws in a batch of 50 is:

P(0) = [p(0)]^50 ~= 0.223

3- What is the expected number of panels that need to be sampled before flaws are found?
I am not sure of what I have done for this question:
lambda=0.03
(0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66
I'm not quite sure what this is asking for, I might have to think about what it is asking,
but I expect what they want as an answer is the batch size for which the expected
number of flawed pannels = 1.

Let the batch size be N, probability that n are flawed is:

b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

RonL

3. Originally Posted by CaptainBlack
Yes, but simplify (or rather calculate the approcimate value of this):

p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704

Probability that there are no flaws in a batch of 50 is:

P(0) = [p(0)]^50 ~= 0.223

I'm not quite sure what this is asking for, I might have to think about what it is asking,
but I expect what they want as an answer is the batch size for which the expected
number of flawed pannels = 1.

Let the batch size be N, probability that n are flawed is:

b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

RonL
there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

- What is the expected number of panels that need to be sampled before two flaws are found.
I guess the steps are the same but for E(x)=2.

CaptainBlack, for the question 2, why did you do p(0)^50?

Hi,
I have two probability problems and I need them to be checked please:

problem-1
A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
My solution:
lambda=0.03*1=0.03
f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)
Originally Posted by CaptainBlack
Yes, but simplify (or rather calculate the approcimate value of this):

p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704
3- What is the expected number of panels that need to be sampled before flaws are found?
I am not sure of what I have done for this question:
lambda=0.03
(0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66
Originally Posted by CaptainBlack
I'm not quite sure what this is asking for, I might have to think about what it is asking,
but I expect what they want as an answer is the batch size for which the expected
number of flawed pannels = 1.

Let the batch size be N, probability that n are flawed is:

b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

RonL
If probability of at least one flaw in a panel is p = .02955, then the number X of panels sampled before a flawed one is found has a geometric distribution Pr(X=k) = p(1-p)^(k-1) for k = 1,2,3,... The expected number of panels sampled before a flawed panel is found is the mean of this distribution, which is 1/p = 33.8. Same answer but different reasoning.

CaptainBlack, for the question 2, why did you do p(0)^50?
Because this is the probability that there are no flaws on the first panel, times the probability that there are no flaws on the second panel times ..
times there are no flaws on the 50th panel = p(0)^50.

RonL

6. Originally Posted by JakeD
If probability of at least one flaw in a panel is p = .02955, then the number X of panels sampled before a flawed one is found has a geometric distribution Pr(X=k) = p(1-p)^(k-1) for k = 1,2,3,... The expected number of panels sampled before a flawed panel is found is the mean of this distribution, which is 1/p = 33.8. Same answer but different reasoning.

RonL

7. Thank you

In the first problem there was a typo. I dont know if you pay attention when I said:[quote]there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

- What is the expected number of panels that need to be sampled before two flaws are found.

I guess the steps are the same but for E(x)=2.

In this case is it still a geometric distribution?

Now for problem 2:

the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)

2-P(X=4)

P(X=4)=b(4;6,0.6)

3-mean
mean=n*p=3.6

4-variance
Var^2=n*p*(1-p)

5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)

Thank you

In the first problem there was a typo. I dont know if you pay attention when I said:
there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.
- What is the expected number of panels that need to be sampled before two flaws are found.

I guess the steps are the same but for E(x)=2.

In this case is it still a geometric distribution?
Yes, with p now the probability of 2 or more flaws.

In the first problem there was a typo. I dont know if you pay attention when I said:
there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

- What is the expected number of panels that need to be sampled before two flaws are found.

I guess the steps are the same but for E(x)=2.

In this case is it still a geometric distribution?

Now for problem 2:

the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)

2-P(X=4)

P(X=4)=b(4;6,0.6)

3-mean
mean=n*p=3.6

4-variance
Var^2=n*p*(1-p)

5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)