Wegman claims that their checkout scanners scan correctly 99.8% of the items sold. How many would you expect to buy, on average, to find one that scans incorrectly?

Question

anonymous · Answer

99

anonymous · Answer

wait no

anonymous · Answer

999

anonymous · Answer

yeah

youngster · Answer

i don't think it is that.
You have to use the equation$$P(n=r)=nCr*q^{r-1}*p^{1}$$
It is a binomial model

anonymous · Answer

$$P(correct)=p=0.998$$$$P(incorrect)=q=0.002$$
We want to find an n so that $$\binom{n}{0}(0.998)^n(0.002)^0<\binom{n}{1}(0.998)^{n-1}(0.002)^1$$

youngster · Answer

wait, what does$$\left(\begin{matrix}n \ 0\end{matrix}\right)$$ mean?

anonymous · Answer

It means nC0
$$\binom{n}{0}=1,\binom{n}{1}=1$$
$$0.998^n<n0.998^{n-1}0.002$$
$$\frac{0.998}{0.002}<n$$
=499

anonymous · Answer

$$\binom{a}{b}=aCb$$

youngster · Answer

oh, that makes sense. thanks so much!             
im going to post another question in  a couple of minutes, can u help me with that?

youngster · Answer

wait, actually ur supposed to use expected value for binormial...
i think you get the same answer, but its less confusing.
Thanks anyways

kropot72 · Answer

Probability of an incorrect scan = 1.000 - 0.998 = 0.002
Let the expected quantity required to get one incorrect scan = n
E(X) = 1 = 0.002n
$$n=\frac{1}{0.002}=?$$

youngster · Answer

@kropot72
500.. so that's the answer?

kropot72 · Answer

@Youngster Yes, that's the answer.

youngster · Answer

thanks!!

kropot72 · Answer

You're welcome :)