Basic probability question

  • Thread starter semidevil
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I dont remember the exact story problem, but it goes something like this.

You are suppose to make a pin that is 6 cm long. Unfortunately, the machine will produce pins that 6 + y cm long. where 0 <= y <= 2, and the function f(y) = K(y + y^2).

if the pin is more then 7 cm long, then it is unusable. Find the proportion that is produced that will be unusable.

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ok, so if y = 0, or y = 1, then it is fine, and if y = 2, then it is unusable.

so can I just integrate from 0 to 1 of f(y), and then subtract from 1, to get the answer?

the book says the answer is 23/28....but I dont know how it got that.
 

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Tom Mattson
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semidevil said:
ok, so if y = 0, or y = 1, then it is fine, and if y = 2, then it is unusable.
Not quite. It should read: If 0<=y<=1, then it is fine, and if 1<y<=2, then it is unusable.

so can I just integrate from 0 to 1 of f(y), and then subtract from 1, to get the answer?
Yes, but first, you have to find K. To get it you have to normalize the probability distribution by integrating f(y) over the sample space and setting it equal to 1. This will allow you to solve for K. Then you can do what you have proposed above.
 

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