Suppose you work for the quality control department at Hewlett Packard. You know that ink jet cartridges are sometimes defective. The current defective rate is 8 cartridges per package of 100 (i.e. 0.08) What is the probability of sampling 12 cartridges and finding:

Less than 3 are defective:
More than 2 are defective:
At least 4 are defective:
Exactly 1 is defective:
2 or more, but less than 5 is defective:

If you could help me on at least 1 I would greatly appreciate that, or even point me where I need to go, I have no clue what to do here.

this is a binomial probability ... defective or not defective

probability (d) = .08 , probability (n) = .92 ... (n + d)^12

expand the binomial and substitute the probabilities in the desired terms
... n^12 + 12 n^11 d + 66 n^10 d^2 + ... + 66 n^2 d^10 + 12 n d^11 + d^12

less than 3 ... .92^12 + (12 * .92^11 * .08) + (66 * .92^10 * .08^2)

p(more than 2) = 1 - p(less than 3)

I'm so sorry, but is there any way you can break down the whole (12 * .92^11 * .08) + (66 * .92^10 * .08^2) part for me. I'm just not understanding where all of these values came from. aside from 0.92 and 0.08

when you expand the binomial , the 12, 66, etc. are the coefficients of the terms

look up Pascal's Triangle for background

So if I'm understanding this right, for the probability of it being less than 3 my answer should be 0.93 right?

yes ... .9348

More than 2 would be 0.06?

Can you walk me through what I need to do for the last 3 please?

at least 4 ... 1 - (first 4 terms)

Exactly 1 ... second term

2 or more, but less than 5 ... (first 5 terms) - (first 2 terms)

Less than 3 are defective

---> none defective + one defective + 2 defective
= C(12,0)(.08)^0 (.92)^12 + C(12,1)(.08)^1 (.92)^11 + C(12,2)(.08)^2 (.92)^10
= .367666... + .38365... + .18348..
= appr .9348

More than 2 are defective
---> 3 defective + 4 def + ... + 12 def.
or 1 - (none def + 1 def + 2 def)

= 1 - (C(12,0)(.92^0)(.08^12) + C(12,1)(.92^1)(.08^11) + C(12,2)(.92^2)(.08^10) )
= ...

do you get the idea?

To calculate the probabilities in this scenario, we can use the binomial probability formula. The binomial probability formula is:

P(x) = C(n, x) * p^x * q^(n-x)

Where:
P(x) is the probability of getting exactly x successes
C(n, x) is the number of ways to choose x successes out of n trials, calculated using the combination formula n! / (x! * (n-x)!)
p is the probability of success in a single trial (defective rate in this case)
q is the probability of failure in a single trial (1 - p)
n is the total number of trials (number of cartridges in this case)

Now, let's calculate the individual probabilities:

1. Probability of less than 3 defective cartridges:
Since we need to find the probabilities for x = 0, 1, and 2 (less than 3), we can add these probabilities together.

P(0) = C(12, 0) * (0.08^0) * (0.92^12)
P(1) = C(12, 1) * (0.08^1) * (0.92^11)
P(2) = C(12, 2) * (0.08^2) * (0.92^10)

P(Less than 3) = P(0) + P(1) + P(2)

2. Probability of more than 2 defective cartridges:
Since we need to find the probabilities for x = 3, 4, 5, ..., 12 (more than 2), we can add these probabilities.

P(3) = C(12, 3) * (0.08^3) * (0.92^9)
P(4) = C(12, 4) * (0.08^4) * (0.92^8)
P(5) = C(12, 5) * (0.08^5) * (0.92^7)
...
P(12) = C(12, 12) * (0.08^12) * (0.92^0)

P(More than 2) = P(3) + P(4) + P(5) + ... + P(12)

3. Probability of at least 4 defective cartridges:
Since "at least 4" includes 4 or more defective cartridges, we need to find the probabilities for x = 4, 5, 6, ..., 12.

P(4) = C(12, 4) * (0.08^4) * (0.92^8)
P(5) = C(12, 5) * (0.08^5) * (0.92^7)
P(6) = C(12, 6) * (0.08^6) * (0.92^6)
...
P(12) = C(12, 12) * (0.08^12) * (0.92^0)

P(At least 4) = P(4) + P(5) + P(6) + ... + P(12)

4. Probability of exactly 1 defective cartridge:
We need to find the probability for x = 1.

P(1) = C(12, 1) * (0.08^1) * (0.92^11)

5. Probability of 2 or more, but less than 5 defective cartridges:
We can calculate the probability for x = 2, 3, and 4 and add them together.

P(2) = C(12, 2) * (0.08^2) * (0.92^10)
P(3) = C(12, 3) * (0.08^3) * (0.92^9)
P(4) = C(12, 4) * (0.08^4) * (0.92^8)

P(2 or more, but less than 5) = P(2) + P(3) + P(4)

To calculate these probabilities, you can use a statistical software or a calculator with binomial probability functions. Alternatively, you can use spreadsheets like Excel, which have built-in functions to calculate binomial probabilities.