Probability help.

Question

11% of all widgets are defective an inspector randomly sample 14 from a box. What is the probability of finding at least 2 defective widgets?

Answer 1

This should get you started:

 

If the box has a large number of widgets we can assume the resulting "experiment" has a binomial distribution.

 

P(success)=.11; 

 

---- added steps -----

 

correction: I see the questions asks for at least 2 so that should be P(2 or 3 or 4 or ... or 14)

 

which can be more quickly done using the complementary event: 1 - P(0 or 1)

 

P(0 or 1)= C(14,0)p⁰(1-p)¹⁴ + C(14,1)p¹(1-p)¹³    where C(n,r) represents combinations and we add because these two are mutually exclusive

 

Plug is the value of p above to finish the calculation.

 

 

If you need more help comment back.