A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 18 randomly selected pens yields no more than two defective pens.
(a) Find the probability that this shipment is accepted if 5% of the total shipment is defective. (Use 3 decimal places.)
(b) Find the probability that this shipment is not accepted if 15% of the total shipment is defective. (Use 3 decimal places.)
3 answers
Could you please help?
this is a binary situation ... a pen is defective or not
let d = defective and n = not defective
(n + d)^18 = n^18 + 18 n^17 d + 153 n^16 d^2
the 1st three terms of the expansion covers the "not more than two defective"
(a) p(n) = .95 , p(d) = .05
... p(accept) = .95^18 + (18 * .95^17 * .05) + (153 * .95^16 * .05^2)
(b) similar to (a) , but ... p(not accept) = 1 - p(accept)
... p(n) = .85 , p(d) = .15
... follow same calculation format as (a)
let d = defective and n = not defective
(n + d)^18 = n^18 + 18 n^17 d + 153 n^16 d^2
the 1st three terms of the expansion covers the "not more than two defective"
(a) p(n) = .95 , p(d) = .05
... p(accept) = .95^18 + (18 * .95^17 * .05) + (153 * .95^16 * .05^2)
(b) similar to (a) , but ... p(not accept) = 1 - p(accept)
... p(n) = .85 , p(d) = .15
... follow same calculation format as (a)
Scott is correct. No idea why he got thumbs down. I worked it like he shows, and I got the correct answers according to Cengage a homework program! Thanks Scott!!! You saved me since the tables don't include 17, 18, or 19!
1 answer