Calculating Probability of Exactly One Faulty Product in 5 Days

[Xenix]

I am given this probability table:

x 0 1 2 3 4
P(X=x) 0.8 0.1 0.05 0.03 0.02

X is the amount of faulty products produced in a day.

I am being asked to find the probability of a exactly one product being foulty in a 5 day periode.

I am a bit confused. I know from the table that the probability of exactly one faulty product in 1 day is 0.1.
But for 5 days, is it that ease just to multiply 0.1 by 5? Or is it 0,1^5?

 

[Dick]

I am given this probability table:

x 0 1 2 3 4
P(X=x) 0.8 0.1 0.05 0.03 0.02

X is the amount of faulty products produced in a day.

I am being asked to find the probability of a exactly one product being foulty in a 5 day periode.

I am a bit confused. I know from the table that the probability of exactly one faulty product in 1 day is 0.1.
But for 5 days, is it that ease just to multiply 0.1 by 5? Or is it 0,1^5?

No, it's not that simple. One of the days has to have a single faulty product. The other four days have to have no faulty products.

 

[Xenix]

Ok, thank you for your answer, so then 0.1*(4*0.8)?

 

[Dick]

Ok, thank you for your answer, so then 0.1*(4*0.8)?

No, the probability of something with probability 0.8 happening on 4 days isn't (4*0.8). That's bigger than 1! This an example of a binomial distribution problem. Don't you have a lesson on that?

 

[Xenix]

Thanks again for the answer. It turns out there is a scheduling conflict at the uni and we are not supposed to cover that topic until next week, however the assignment is due Monday.
I managed to solve this though.
Thanks for your help! :)

 

[Dick]

Thanks again for the answer. It turns out there is a scheduling conflict at the uni and we are not supposed to cover that topic until next week, however the assignment is due Monday.
I managed to solve this though.
Thanks for your help! :)

Good work. I was wondering why you seemed to be missing a lot of the basics to tackle this problem.