Question

# A manufacturer of window frames knows from long experience that 5% of the production will have...

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
At least one will need adjustment
More than two will need adjustment

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

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