## Youngster 3 years ago 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?

1. koalamon

99

2. koalamon

wait no

3. koalamon

999

4. koalamon

yeah

5. Youngster

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

6. henpen

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

7. Youngster

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

8. henpen

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

9. henpen

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

10. Youngster

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

11. Youngster

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

12. kropot72

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 $n=\frac{1}{0.002}=?$

13. Youngster

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

14. kropot72

15. Youngster

thanks!!

16. kropot72

You're welcome :)