Probability help

[Lead academic]

I have a question and two different answers:
I make 50 parts and when checked I have 5 of them defective, so 45 are OK.
what is the chance the next one (51st), will also be defective ?

 

[Deleted member 4993]

I have a question and two different answers:
I make 50 parts and when checked I have 5 of them defective, so 45 are OK.
what is the chance the next one (51st), will also be defective ?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[Lead academic]

Thank you Subhotosh
as I said, I have two possible answers and I am not sure if either is correct.
Option A:
events = 5+1=6
sample = 50+1
therefore probability of finding a defective part = 6/51 = 2/17 = 0.11765
Option B:
as the 5 defective parts out of 50 give 5/50 = 1/10 = 0.1
then for the next parts to be defective it would have an event of 5+0.1 = 5.1
a sample of 50+1
so the probability of the next part being defective would be 5.1/51 = 1/10 = 0.1

which would be correct Option A or Option B ?
Thanks
Tim

 

[Dr.Peterson]

I have a question and two different answers:
I make 50 parts and when checked I have 5 of them defective, so 45 are OK.
what is the chance the next one (51st), will also be defective ?

What are your two answers? And are they your answers, or someone else's?

There is an obvious answer; but that depends on the assumption that the first 50 are a representative sample of all your work.

There is also a small ambiguity in your question, namely what "also" refers to.

We need to know what issue concerns you.

 

[Lead academic]

Thank you for your reply, Dr Peterson.
The questions were in the post :-
Option A = 6/51 = 2/17 = 0.11765
Option B = 5.1/51 = 1/10 = 0.1
there is no "also" in the post, so I don't understand this.

However, i have thought more about this question and decided that it might be to hard for the student, so i have decided to provide a different question.
thanks for your help.

 

[Dr.Peterson]

Thank you for your reply, Dr Peterson.
The questions were in the post :-
Option A = 6/51 = 2/17 = 0.11765
Option B = 5.1/51 = 1/10 = 0.1
there is no "also" in the post, so I don't understand this.

However, i have thought more about this question and decided that it might be to hard for the student, so i have decided to provide a different question.
thanks for your help.

I was answering the original question, in post #1. Your answers are in post #3, which was not visible when I wrote, due to moderation. This is what I was referring to:

I have a question and two different answers:
I make 50 parts and when checked I have 5 of them defective, so 45 are OK.
what is the chance the next one (51st), will also be defective ?

Note that you also didn't mention a context; I take it you are the teacher writing the problem, not a student trying to understand a problem given to you, as I had assumed.

As I see it, the only data you have from which to find a probability are the original 50 parts; you don't yet know what is true of the 51st. So the only answer that makes sense is 5/50 = 10%. The denominator of 51 is inappropriate as part of the work.

But I agree that the question is worded in a confusing way, so it would be good to change it.