What is the probability that a good design will be rejected

[Cyrus]

3) An engineer has designed a modified welding robot. The robot will be considered good enough to manufacture if it misses only 1% of its assigned welds. And it will be judged a poor performer if it misses 5% of its welds. (In-between possibilities are not considered.) The new design will be accepted if the number of missed welds, R, is 2 or less and rejected otherwise. A test is performed using 100 welds.
a. What is the probability that a good design will be rejected
b. What is the probability that a poor design will accepted

Here is what I know,

It is good if $$X=x=R \leq 2$$
n = 100
p=1% is rated Good
p=5% is rated bad

Im guessing I have to use a Binomial distribution here??

$$p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x}$$

So,

$$X\sim Bin(100,0.01)$$

The fact that we want 2 or fewer welds means

$$P(X \leq 2) = P(X=0-or-X=1-or-X=2)$$

PART A)

The probability that a good deisgn will be rejected will be equal to:

$$1 -p(x) = 1- P(X=x) = 1 -\frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x}$$

$$1 -\frac{ 100!}{x!(100-x)!} (.01)^x(1-.01)^{100-x}$$

for x=0,1,2:

$$P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2)$$

Because of the EXCESSIVE n value, its not going to work on my Ti-83 calculator and I don't feel like busting out matlab. So I am going to use the cumulative table provided by the prof online that I just found by mistake (nice of him to mention needing to use it :grumpy:)

$$p(x \leq 2)=0.92063$$

So the probability that a good design gets rejected is

$$1 - p(x \leq 2) = 0.07937$$

which is equivalent to, 7.937% good parts accidentially rejected.

PART B)

The same thing as Part a, however the table now refers to values for

$$X\sim Bin(100,0.05)$$

$$p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x}$$

$$\frac{ 100!}{x!(100-x)!} (.05)^x(1-.05)^{100-x}$$

for x=0,1,2:

$$P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2)$$

$$p(x \leq 2)=0.011826$$

so

1.1826% are accepted from the bad batch.