Question 1184337
If there are on average 4 defects per 18 square feet, then we expect to have 22/3 defects in 33 square feet of metal sheet.  So {{{lambda = 22/3}}}


===>  {{{p(x) = e^(-lambda)*(lambda^x/x!) = e^(-22/3)*((22/3)^x/x!)}}}


===> {{{P(X >= 5) = sum(p(x), x=5, infinity) = 1-sum(p(x), x=0, 4)


= 1 - e^(-22/3)*(1 + 22/3 + (22/3)^2/2!+ (22/3)^3/3! + (22/3)^4/4!)}}}


Now you should be able to carry out the necessary calculation.