Probability of acceptance

[Siles]

Hello, i need help please, English not very good. I write as in shown in exercise.

question : "The probability of person getting asked for revision of article after submission to news paper is 0.6 and if successfully revised, the probability of getting published in news paper is 0.8, what is the probability of this person getting 3 revise request out of 5 submissions and further get zero publication."

What are steps of calculation please.

Thanks very much for the help,

regards,

Andrè Silès

 

[Siles]

my try is below (i forgot to send according of conditions of website, sorry)

P(Revision) = 0.6,
P(Published/Revision) = 0.8,
P(Revision and Published) = P(Published/Revision)*P(Revision) = 0.6*0.8 = 0.48.
Revision and not published = 1-0.48 = 0.52
I think above steps correct - please make correct if wrong.

about the 3 request of 5 submission i dont know how solution looks.

Thanks very much for help.

Regards,

Andrè Silès

 

[tri]

question : "The probability of person getting asked for revision of article after submission to news paper is 0.6 and if successfully revised, the probability of getting published in news paper is 0.8, what is the probability of this person getting 3 revise request out of 5 submissions and further get zero publication."

It is not known if exactly 3 out of 5 is asked, or, 3 or more. Even 5 revise requests out of 5 submissions have 3 revise requests.

 

[Dr.Peterson]

question : "The probability of person getting asked for revision of article after submission to news paper is 0.6 and if successfully revised, the probability of getting published in news paper is 0.8, what is the probability of this person getting 3 revise request out of 5 submissions and further get zero publication."

What are steps of calculation please.

P(Revision) = 0.6,
P(Published/Revision) = 0.8,
P(Revision and Published) = P(Published/Revision)*P(Revision) = 0.6*0.8 = 0.48.
Revision and not published = 1-0.48 = 0.52
I think above steps correct - please make correct if wrong.

about the 3 request of 5 submission i dont know how solution looks.

Thanks for doing as we ask!

You've started correctly; P(Revision and Published) is 0.48. But subtracting from 1 does not give P(Revision and not Published).

But as has already been pointed out, the problem as given is ambiguous. Not only does it not say whether they mean exactly 3, or at least 3; but the wording of the final question sounds partly like "getting 3 revise requests and no publication", and partly like "no publication, given 3 revise requests". The latter is a more sensible thing to ask, and in that case "exactly 3" would be appropriate. In the former case, "at least 3" would be more sensible in real life.

You say "I write as in shown in exercise"; do you mean that you copied it exactly, word for word, or did you translate or paraphrase the problem? (In the latter case, it could be helpful to show us the actual original.)

But we can start with a simpler problem: How would you find the probability of getting 3 revise requests out of 5 submissions? What techniques have you learned that might apply?

 

[Steven G]

The complement of A and B = (A and B)' = A' U B'
That is why P(Revision and Published) ≠ 1 - P(Revision and not Published)

 

[Siles]

Thanks Mr. Steven. P, I understand, I mistake, in this situation, what will 1-P be = to? Thanks

 

[Siles]

The complement of A and B = (A and B)' = A' U B'
That is why P(Revision and Published) ≠ 1 - P(Revision and not Published)

Thanks for doing as we ask!

You've started correctly; P(Revision and Published) is 0.48. But subtracting from 1 does not give P(Revision and not Published).

But as has already been pointed out, the problem as given is ambiguous. Not only does it not say whether they mean exactly 3, or at least 3; but the wording of the final question sounds partly like "getting 3 revise requests and no publication", and partly like "no publication, given 3 revise requests". The latter is a more sensible thing to ask, and in that case "exactly 3" would be appropriate. In the former case, "at least 3" would be more sensible in real life.

You say "I write as in shown in exercise"; do you mean that you copied it exactly, word for word, or did you translate or paraphrase the problem? (In the latter case, it could be helpful to show us the actual original.)

But we can start with a simpler problem: How would you find the probability of getting 3 revise requests out of 5 submissions? What techniques have you learned that might apply?

Thanks Dr. Peterson, and Mrs. Tri, I have copied exactly but i missed one word, i made in bold - is ATLEAST. Please see correct below.

The probability of person getting asked for revision of article after submission to newspaper is 0.6 and if successfully revised, the probability of getting published in newspaper is 0.8, what is the probability of this person getting ATLEAST 3 revise request out of 5 submissions and further get zero publication."

I am thinking this is binomial probability. But i do not know how to proceed.

Regards,

Andrè

 

[Siles]

Thanks Mr. Steven. P, I understand, I mistake, in this situation, what will 1-P be = to? Thanks

I try more clear, if I do exactly as I did, what does that mean? What does 0.52 actually in this mean? Anything useful or with meaning?

 

[Dr.Peterson]

I am thinking this is binomial probability. But i do not know how to proceed.

Correct. I was hoping to learn that you had heard of that.

Start by finding the probability that one submission leads to a revise request and is not published. This is a P(A and B).

Then, since "at least 3" means "3, 4, or 5", use the binomial formula for x=3, 4, and 5, and add them.

 

[Siles]

Thank you Dr. Peterson,

So Accordingly, I have P (A and B) = P(A)(Revision)*P(B)(Not Published) = 0.6*(1-0.8) = 0.6*0.2 =0.12 (?).

Binomial Probability is C(n,x)* P(A)^x * P(B)^(n-x)

x= 3, 4 and 5
n = 5
P(A)(Revision)= 0.6
P(B)(Not Published) = 0.12

x=3
C(5,3)*(0.6)^3* (0.12)^(5-3)
10*0.216*0.0144=0.032
-----------------
x=4
C(5,4)*(0.6)^4* (0.12)^(5-4)
5*0.13*0.12=0.078
-----------------
x=5
C(5,5)*(0.6)^5* (0.12)^(5-5)
1*0.77*1=0.77
-----------------
Can they be assume as individual (mutually exclusiv) events? so add is just algebraic add? 0.032+0.078+0.77 = 0.88 =88% ?
(It seems much just looking at numbers, I am making thinking mistake, no? Are they not mutually exclusiv (correct term?)

Regard,

Andrè

 

[Dr.Peterson]

Can they be assume as individual (mutually exclusiv) events? so add is just algebraic add? 0.032+0.078+0.77 = 0.88 =88% ?
(It seems much just looking at numbers, I am making thinking mistake, no? Are they not mutually exclusiv (correct term?)

The events "exactly 3", "exactly 4", and "exactly 5" are mutually exclusive; if you think about it, that should be obvious. (But it's wise in this field not to assume that what seems obvious is really true!)

So, yes, you can add them.

However, the numbers you want to add are incorrect.

So Accordingly, I have P (A and B) = P(A)(Revision)*P(B)(Not Published) = 0.6*(1-0.8) = 0.6*0.2 =0.12 (?).

This is correct, though your notation is not exactly right. Don't write things like P(A)(Revision), but rather P(A), where A = revision.

I would have said, if A = revision requested, and B = published, then you are given P(A) = 0.6 and P(B | A) = 0.8. (Technically this assumes that getting the revision request implies that you will make the revision!)

Then you want P(A and B') = P(A)*P(B' | A) = P(A)*(1 - P(B | A) = 0.6*0.2 = 0.12.

Binomial Probability is C(n,x)* P(A)^x * P(B)^(n-x)

This is incorrect! It should be C(n,x)* p^x * q^(n-x), where p is the probability of the individual event (0.12), and q is its complement, 1-p. Did you have some reason to think what you wrote would be correct?

Thanks Mr. Steven. P, I understand, I mistake, in this situation, what will 1-P be = to? Thanks

I try more clear, if I do exactly as I did, what does that mean? What does 0.52 actually in this mean? Anything useful or with meaning?

He actually answered this question:

The complement of A and B = (A and B)' = A' U B'
That is why P(Revision and Published) ≠ 1 - P(Revision and not Published)

Taking A = revision requested, and B = published, your 1-0.48 = 0.52 is 1 - P(A and B) = P((A and B)') = P(A' U B'). That is the probability that either you don't get a revision request, or it is not published (or both).

What you wanted is P(A and B').

 

[Siles]

The events "exactly 3", "exactly 4", and "exactly 5" are mutually exclusive; if you think about it, that should be obvious. (But it's wise in this field not to assume that what seems obvious is really true!)

So, yes, you can add them.

However, the numbers you want to add are incorrect.

Thank you Dr. Peterson, I question myself many times for making sure it is mutually exclusive or not mutually exclusive. In the end i go to beginning and see dependency and know it is which one.

This is correct, though your notation is not exactly right. Don't write things like P(A)(Revision), but rather P(A), where A = revision.

I would have said, if A = revision requested, and B = published, then you are given P(A) = 0.6 and P(B | A) = 0.8. (Technically this assumes that getting the revision request implies that you will make the revision!)

Then you want P(A and B') = P(A)*P(B' | A) = P(A)*(1 - P(B | A) = 0.6*0.2 = 0.12.

This is incorrect! It should be C(n,x)* p^x * q^(n-x), where p is the probability of the individual event (0.12), and q is its complement, 1-p. Did you have some reason to think what you wrote would be correct?

I get confused with my notation and this is reason for using wrong value p. I learn new to make clear notation. thank you again.

Taking A = revision requested, and B = published, your 1-0.48 = 0.52 is 1 - P(A and B) = P((A and B)') = P(A' U B'). That is the probability that either you don't get a revision request, or it is not published (or both).

What you wanted is P(A and B').

After new explaining from you, i understand better.

For problem, answer is
0.0134+0.0009+0.00002 = 1.42%











My thanks to you Dr. Peterson, Steven G and Mrs. Tri, I learn and now I teach little brother