Probability Problem

candor

Hi I was wondering if anyone can help me with this? Tried to figure it out and am stumped at the moment.

An industrial process manufactures components with a 5% defective rate. To control
the quality, each day products are inspected sequentially until the rst defective is obtained.
Let
i. What is the probability that the rst defective will occur in the 4th component
inspected?
ii. What is the probability that the rst defective will occur in one of the rst 4
components inspected?
iii. What is the probability that more than 4 components will be inspected before a
defective is found?
iv. If 4 components have already been inspected without a defect, what is the
probability that more the rst error will occur in the next 3?
v. Write down the R instructions for calculating (i) and (ii).

Fremen

It's better if you post up your attempt and where you get stuck - you'll learn more that way.

A couple of things to think about to get started:
If there's a 5% defective rate, what's the probability the first product we inspect is defective? what's the probability it's not defective?

The probability that the first defective product is found on the fourth inspection is the same as the probability that the first, second and third products are not defective and the fourth product is defective. If we assume inspections are independent, how would we work out this expression?

candor

Fremen wrote: »

It's better if you post up your attempt and where you get stuck - you'll learn more that way.

A couple of things to think about to get started:
If there's a 5% defective rate, what's the probability the first product we inspect is defective? what's the probability it's not defective?

The probability that the first defective product is found on the fourth inspection is the same as the probability that the first, second and third products are not defective and the fourth product is defective. If we assume inspections are independent, how would we work out this expression?

Sorry wasn't sure where to start bar knowing that 95% was not defective.

I guess the first product has a probability of 5% of being defective and a probability of 95% of being not defective.

I am thinking you could use bayes theorm to make the expression?

Fremen

It's helpful to use fractions rather than percentages, so Probability(defective) = 1/20.

If A and B are independent events, then Probability(A and = Probability(A)xProbability(B). How can we use this fact in the current situation?

candor

Fremen wrote: »

It's helpful to use fractions rather than percentages, so Probability(defective) = 1/20.

If A and B are independent events, then Probability(A and = Probability(A)xProbability(B). How can we use this fact in the current situation?

Probability(A and = Probability(A)xProbability(B)

Ok. So you take A to be defective rate and B to be correct rate.

Multiply them both together and you get 19/400 which is 0.0475.

Then do you take 19/400 and multiply by 19/20 (correct rate) and so on so fourth taking the new probability value til you get to the 4th iteration? (Hope I've explained that ok!)

Fremen

You're on the right track, but you need to express

Probability(first defective is found on fourth inspection)

as

Probability(something AND something else AND something else AND something else),

where the somethings are independent events. How would you do that?

Fremen

You wrote "rst" in the question - did you mean "first" or that there are r occurences?

candor

Fremen wrote: »

You're on the right track, but you need to express

Probability(first defective is found on fourth inspection)

as

Probability(something AND something else AND something else AND something else),

where the somethings are independent events. How would you do that?

Ok so:

The probability (first defective is found on fourth inspection) =

probability((1/20 x 19/20) + (1/20 x 19/20) + (1/20 x 19/20) + (1/20 x 19/20))

candor

Fremen wrote: »

You wrote "rst" in the question - did you mean "first" or that there are r occurences?

First yes sorry.

Fremen

Why 19/20 x 1/20, and why add them?

Try to express the event "first defective is found on fourth inspection" in terms of things that happen on the first, second, third and fourth inspections. Try to do it in words first, then convert into "events" with known probabilities.

Since the inspections are independent, you can then apply
Probability(A and = Probability(A)xProbability(B).

candor

Fremen wrote: »

Why 19/20 x 1/20, and why add them?

Try to express the event "first defective is found on fourth inspection" in terms of things that happen on the first, second, third and fourth inspections. Try to do it in words first, then convert into "events" with known probabilities.

Since the inspections are independent, you can then apply
Probability(A and = Probability(A)xProbability(B).

19/20 to represent 95% probability of a correct product.

So I'm guessing 19/20 x 19/20 x 19/20 x 1/20?

Thank you for bearing with me, I really appreciate it.

Fremen

Yeah, that's right.

Probability(fourth inspection is defective)

= Probability(first inspection is not defective
AND second inspection is not defective
AND third inspection is not defective
AND fourth inspection IS defective)

and then you use the multiplication rule. Nicely done!

Just like the AND-multiplication rule, there's an OR-addition rule. If A and B are events, P(A OR = P(A) + P(B) - P(A AND . If (A AND is impossible (for instance, flipping a coin and getting heads and tails), then
P(A OR = P(A) + P(B). You need this for part two. Any thoughts?

candor

Fremen wrote: »

Yeah, that's right.

Probability(fourth inspection is defective)

= Probability(first inspection is not defective
AND second inspection is not defective
AND third inspection is not defective
AND fourth inspection IS defective)

and then you use the multiplication rule. Nicely done!

Just like the AND-multiplication rule, there's an OR-addition rule. If A and B are events, P(A OR = P(A) + P(B) - P(A AND . If (A AND is impossible (for instance, flipping a coin and getting heads and tails), then
P(A OR = P(A) + P(B). You need this for part two. Any thoughts?

Something like this?

0.05 + 0.95*0.05 + 0.95*0.95*0.05 + 0.95*0.95*0.95*0.05

Fremen

Yeah, looks like you've got your head around it now

candor

Fremen wrote: »

Yeah, looks like you've got your head around it now

Thanks so much, I really appreciate the help. Happy New Year