With one method of a procedure called acceptance sampling

Question

a sample of items is randomly selected without replacement and the entire batch is rejected if there is at least one defective unit. The ABCD Electronics Company has just manufactured 4500 write-rewrite CDs, and 170 are defective. If 6 of these CDs are randomly selected and tested, what is the probability that the entire batch will be rejected?

Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol.

Answer 1

Hi Park,

1. Total number of CDs = 4500
2. Number of defective CDs = 170
3. Number of CDs to be randomly selected and tested = 6

To find the probability of selecting only non-defective CDs, you can use the formula for combinations (nCr):

Probability of selecting only non-defective CDs =

(Number of ways to choose 6 non-defective CDs) / (Number of ways to choose 6 CDs from all) =

(Comb(4500 - 170, 6) / Comb(4500, 6))

Now, to find the probability of selecting at least one defective CD (which leads to the batch being rejected), you can use the following equation:

Probability of batch being rejected = 1 - Probability of selecting only non-defective CDs

After performing the calculations, the probability of the entire batch being rejected is 20.6%.

"Comb" refers to the combination formula often denoted as nCr, which is a way to calculate the number of different selections of r items from a set of n items.

In the context of this problem:

1. The number of ways to choose 6 non-defective CDs from the total non-defective CDs is calculated as the combination of (4500−170)(4500−170) items taken 6 at a time.
2. The number of ways to choose any 6 CDs from the total is calculated as the combination of 4500 items taken 6 at a time.

After calculating these combinations, you subtract the probability of selecting only non-defective CDs from 1 to get the probability of the batch being rejected, which is 20.6%.

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Ben

Answer 2

Hi Park,

This is a binomial distribution problem where we have two outcomes, success and failure. Success, in this case, is incidentally defined as being defective. Note that we can define success and failure any way we want for these type of problems. Now, the binomial formula:

P(X=x)=C(n,x)p^xq^n-x

Now let's break this down:

n=sample size

x=number of successes--defectives here

p=probabiity of success (defective)

q=1-p, probability of failure

Now, we can approach this two potential ways. One, we can compute probabilities of 1,2,3,4,5,and 6 defectives, and add them up. That means six different binomial calculations. Haw about no?! The other, more expedient option, is to apply the complement rule, or as I call it, the "one minus trick." Basically, this rule states:

P(Success)=1-P(Failure)

Failure in this case equates to having 0 defectives out of 6, so if we can get that probability, we can then get the probability of having at least 1 via the one minus trick. So, proceeding with binomial formula:

P(X=x)=C(n,x)p^xq^n-x

Defining our variables:

n=6

x=0

p=170/4500=0.038

q=1-p=1-0.038=0.962

Thus:

P(X=0)=C(6,0)0.038⁰*0.962⁶

Now,

C(6,0)=6!/(0!*6!)

0!=1, by definition, so we are left with:

C(6,0)=6!/6!=1

Returning to original formula:

P(X=0)=1*0.038⁰*0.962⁶

Recall that any number raised to the 0 power is 1, so:

P(X=0)=0.962⁶

P(X=0)=0.793

Now, we want P(X>0) since one or more defectives will result in batch rejection, so let's do the one minus trick:

P(X>0)=1-P(X=0)

P(X>0)=1-0.793

P(X>0)=0.207

As a percentage, it is :100(0.207)=

P(X>0)=20.7%

I hope this helps.