Question

A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require adjustment. What is the probability that in a sample of 20 window frames
None will need adjustment
At least one will need adjustment
More than two will need adjustment

Answer 1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.05

N = number of trials = 20

R = desired success

A)

None = P(0) = 20c0*(0.05^0)*(1-0.05)^20 = 0.35848592241

B)

At least 1 = 1 - P(0) (as sum of all the probabilities is =1

P(at least 1) = 0.6415140776

C)

P(more than 2) = 1 - (p(0) + P(1) + P(2)) = 0.07548367379