binomial distribution

[ssmmss]

70% of disk drives made from Company X function properly. If we have a collection of 5 disk drives, what is the probability that:

1. At least 1 disk drive doesn’t function?
2. At least 1 disk drive is functioning?

This is what I came up with for the first question:
C(5,1) *.3^1 *.7^4 + C(5,2) *.3^2 *.7^3 + C(5,3) *.3^3 *.7^2 + C(5,4) *.3^4 *.7^1 + C(5,5) *.3^5 *.7^0

This is what I came up with for the 2nd question:
C(5,1) *.7^1 *.3^4 + C(5,2) *.7^2 *.3^3 + C(5,3) *.7^3 *.3^2 + C(5,4) *.7^4 *.3^1 + C(5,5) *.7^5 *.3^0

 

[pka]

70% of disk drives made from Company X function properly. If we have a collection of 5 disk drives, what is the probability that:

1. At least 1 disk drive doesn’t function?
2. At least 1 disk drive is functioning?

This is what I came up with for the first question:
C(5,1) *.3^1 *.7^4 + C(5,2) *.3^2 *.7^3 + C(5,3) *.3^3 *.7^2 + C(5,4) *.3^4 *.7^1 + C(5,5) *.3^5 *.7^0

This is what I came up with for the 2nd question:
C(5,1) *.7^1 *.3^4 + C(5,2) *.7^2 *.3^3 + C(5,3) *.7^3 *.3^2 + C(5,4) *.7^4 *.3^1 + C(5,5) *.7^5 *.3^0

Why not use complement probability?
1) \(\displaystyle 1-(0.7)^5\)

2) \(\displaystyle 1-(0.3)^5\)

 

[ssmmss]

Why not use complement probability?
1) \(\displaystyle 1-(0.7)^5\)

2) \(\displaystyle 1-(0.3)^5\)

The "at least" always gets me.

 

[ssmmss]

Why not use complement probability?
1) \(\displaystyle 1-(0.7)^5\)

2) \(\displaystyle 1-(0.3)^5\)

Is this different from my other post "Rolling a die and getting at least....."
which you responded to. I used the same approach. Is there any difference?

 

[pka]

Is this different from my other post "Rolling a die and getting at least....."
which you responded to. I used the same approach. Is there any difference?

Actually "At least one" is very very different from "at least four".

The event of At least on is the complement of none.

 

[ssmmss]

Actually "At least one" is very very different from "at least four".

The event of At least on is the complement of none.

Thanks pka. So this could re-written as:

The probability that all 5 disk drives are functioning is ?(?=0)=?(5,0) * 0.3^0 * (0.7)^5=0.16807
?(?≥1)= 1 - 0.16807

The probability that all 5 disk drives don’t function is ?(?=0)=?(5,0) * 0.7^0 * (0.3)^5=0,00243
?(?≥1)= 1 - .00243

Which is exactly what you suggested. But this is just much longer way.

 

[mmm4444bot]

What are the little boxes?

(Click thumbnail for larger image)

 

[ssmmss]

What are the little boxes?

(Click thumbnail for larger image)


View attachment 4168

Sorry I didn't realize it wasn't showing properly. Here it is:

The probability that all 5 disks are functioning is : P(x=0)=c(5,0) * 0.3^0 * (0.7)^5 = .16807
P(x>=1) = 1 - 0.16807

The probability that all disks don't function is P(x=0)=C(5, 0) * 0.7^0 * (0.3)^5 = 0.00243
P(X>=1) = 1 - 0.00243