Dustin W. answered • 06/18/19

Super Teacher

You would want to use the formula for Bernoulli trial: (n,k)*P^{k }*(1-P)^{(n-k) }where:

P=probability of success

n=# of tirals

k= # of a specific outcome

and (n,k), also can be written an nCk, is the binomial coefficient and is equal to:

nCk= n!/[k!*(n-k)!]

For this problem, it will be easier to find the inverse or the chance of accepting the shipment ( meaning either 0, 1, or 2 defecting pens) and subtracting that likelyhood from 1.

A) P=.9 chance of working pen

1-P=.1=chance of defective pen

The case with 0 failures is (17,0)*.9^{17}*.1^{0}=[17!/(0!*(17-0)!]*.9^{17}*1=(17!/17!)*.9^{17}=1*.9^{17}=.167

Repeating for 1 failure gives (17,1)*.9^{16}*.1^{1}=17*.9^{16}*.1^{1}=.315

Repeating for 2 failures gives (17,2)*.9^{15}*.1^{2}=136*.9^{15}*.1^{2}=.280

The probability of accepting a package of 17 pens .762

The probability of rejecting a package of 17 pens is 1-.762=.238

B) Repeat part A, but with P=.8 and (1-P)=.2

The result is .299 probability of accepting and a .701 chance of rejecting