SOLUTION: Not quite sure where to start here...would really appreciate the help! A company manufacturing laptops believes that 5% of their computers are faulty. They take a sample of 30 com

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Question 998968: Not quite sure where to start here...would really appreciate the help!
A company manufacturing laptops believes that 5% of their computers are faulty. They take a sample of 30 computers. Showing your calculations, find the probability;
a) Two of the laptops are faulty.
b) More than two of the laptops are faulty.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
p = probability the laptop is faulty.
q = probability the laptop is not faulty.
p = .05
q = 1 - .05 = .95

since this appears to be a binomial probsbility type problem, then the formula is:

p(x) = c(n,x) * p^x * q^(n-x)

c(n,x) = n! / (x! * (n-x)!)

n = 30
p = .05
q = .95

p(2) = c(30,2) * .05^2 * .95^28 = 435 * .0025 * .2378268853 = .2586367377.

the probability that exactly 2 of them are defective is 25.86%.

to find the probability that more then 2 of them are defective, you would need to find 1 minus the probability that at most 2 of them are defective.

that would be equal to 1 - (p(0) + 0(1) + p(2)).

the attached excel file shows the calculations required for all of the proabilities and can be used to confirm that the solution is correct.

the excel file shows p(2) = 0.258636738

this agrees with my calculations above.

from the excel file:

sum of (p(0) + p(1) + p(2) = 0.812178813
1 - sum of (p(0) + p(1) + p(2) = 0.187821187

the probability of exactly 2 defective is .2586.

the probability of more than 2 is 1 minus the probability of at most 2 which is equal to 1 - .8122 = .1878.

i'll do one of the calculations for you manually so you can see how they're done.

n = 30
x = 2
p = .05
q = .95
c(30,2) = 30! / (2! * 28!)

p(2) = c(30,2) * .05^2 * .95^28

c(30,2) = 30! / (2! * 28!) = (30 * 29 * 28!) / (2! * 28!)

the 28! on top and bottom cancels out and you are left with:

(30 * 29) / 2! = 30 * 29 / 2 = 15 * 29 = 435.

p(2) becomes 435 * .05^2 * .95^28

use your calculator to get p(2) = .2586367377 which rounds to .2586.

similar calculations are done for p(0) and p(1).

add them up and you get sum of (p(0) + p(1) + p(2)) which is the probability that at most 2 will be defective.

take 1 minus that and you get probability that more than 2 will be defective.