Let S be a set and P be a property of the elements of the set,...

Let S be a set and P be a property of the elements of the set, such that each element either has property P or not. For example, maybe S is the set of your classmates, and P is "likes Japanese food." Then if s ∈ S is a classmate, he/she either likes Japanese food (so s has property P) or does not (so s does not have property P). Suppose Pr(s has property P) = p for a uniformly chosen s ∈ S. Suppose Ofer the Oracle has magic powers that allow him to check property P for any s, but it doesn’t always work since each time he relies on some (independently) generated random numbers to help him. If he says “Yes,” then s has the property. If s has the property, then he says “Yes” with probability at least q.

Suppose we let Ofer check property P on an element s a total of N times, and he responded “No” each time. Find a lower bound (i.e. a smaller number, but a useful one) for Pr(s does not have property P). Suppose p = 99/100 and q = 1/2, how many times should you let Ofer check probability P before you are 99% confident that s does not have property P?

Expert Solution

Let S be a set and P be a property of the elements of the set,
such that each element either has property P or not.
For example, maybe S is the set of your classmates, and P is likes Japanese food.
Then if s ∈ S is a classmate, he/she either likes Japanese food
(so s has property P) or does not (so s does not have property P).
Suppose Pr(s has property P) = p for a uniformly chosen s ∈ S.
Suppose Offer the Oracle has magic powers that allow him to check property P for any s,
but it doesn’t always work since each time he relies on some (independently) generated random numbers to help him.
If he says “Yes,” then s has the property.
If s has the property, then he says “Yes” with probability at least q.
Suppose we let Offer check property P on an element s a total of N times, and he responded “No” each time. Find a lower bound (i.e. a smaller number, but a useful one) for Pr(s does not have property P).
Suppose p = 99/100 and q = 1/2,
let Offer check probability P before you are 99% confident that s does not have property P
sample size is number of times n

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.5
ME = 0.99
n = ( 2.576 / 0.99 )^2 * 0.5*0.5
= 1.693 ~ 2

milcah answered 1 year ago

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