Probability of Filling 100 Orders w/ Defective Components

[natethegreatX]

Homework Statement

A manufacturer has 100 customers and needs to make one component per customer. However 2% of the components manufactured come out defective. The components can be assumed to be independent.

If the manufacturer stocks 100 components, what is the probability that the 100 orders can be filled without re-ordering new components?

If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without re-ordering new components?

Homework Equations

Not sure, but possibly:

(n!)/(x!(n-x)!)*(p^x)(1-p)^(n-x)

p=probability of failure
n=number of tries
x=number of independent variable

The Attempt at a Solution

The only real problem I'm having is with this equation, the first question goes to 1, and the second goes to zero. I feel those arn't right at all. Wouldn't the first question end up 98%?
So I'm lost on if the binomial distribution function is even supposed to be used or if I just can't plug numers in a calc properly. Oh and this is my official first post on the forums.

 

[NeoDevin]

Please clarify that equation, what does it equal?

 

[dacruick]

if you don't mind telling me the "section" of probability that this question is under, I'm sure I could find an answer for you.

 

[dacruick]

okay what is the probability that a coin will be flipped 100 times, and there will be no heads. this is the same concept as the first question.

The second question is like, what is the probability that a coin is tossed 102 times, and there will be at least 100 heads. Or, in other words, what is the probability a coin is tossed and there are less than 3 tails.

 

[DeeNos]

I would approach this problem by using the binomial distribution function to calculate the probability of filling all 100 orders without re-ordering new components. The equation you provided is correct, but there are a few things that need to be clarified.

First, the probability of failure (p) should be 0.02, since 2% of the components are defective.

Second, the number of tries (n) should be 100, since there are 100 orders to be filled.

Third, the number of independent variables (x) should be 0, since we want to know the probability of filling all 100 orders without any failures.

Using these values, the equation would be:

P(x=0) = (100!)/(0!(100-0)!)*(0.02^0)(1-0.02)^(100-0) = (100!)*(0.98)^100 = 0.1326

Therefore, the probability of filling all 100 orders without any failures is approximately 13.26%.

For the second question, if the manufacturer stocks 102 components, the probability of filling all 100 orders without any failures would be:

P(x=0) = (102!)/(0!(102-0)!)*(0.02^0)(1-0.02)^(102-0) = (102!)*(0.98)^102 = 0

This means that there is a 0% chance of filling all 100 orders without any failures if the manufacturer only stocks 102 components.

In conclusion, the binomial distribution function can be used to calculate the probability of filling all 100 orders without re-ordering new components. However, it is important to ensure that the values for p, n, and x are correctly identified and plugged into the equation. In this case, the probability of success (filling an order without a defective component) is 98%, the number of orders is 100, and the number of successful orders (x) is 0.