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.)
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)
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!
To find the probability for both parts (a) and (b), we can use the binomial probability formula. The binomial probability formula calculates the probability of a specific number of successes in a given number of trials, assuming each trial has the same probability of success.
The formula for the probability of exactly x successes in n trials is:
P(x) = (nCx)(p^x)((1 - p)^(n - x))
Where:
P(x) is the probability of x successes,
n is the number of trials,
p is the probability of success in a single trial, and
nCx is the combination formula that represents the number of ways to choose x items from a set of n items.
Let's calculate both probabilities using this formula:
(a) Find the probability that this shipment is accepted if 5% of the total shipment is defective.
In this case, the probability of accepting a single pen (p) is equal to 1 - probability of a defective pen.
So, p = 1 - 0.05 = 0.95
We need to find the probability of having no more than 2 defective pens in a sample of 18 pens. Therefore, we need to find the sum of three probabilities: P(0), P(1), and P(2).
P(0) = (18C0)(0.95^0)(0.05^18) = 1 x 1 x (0.05^18) = (0.05^18) ≈ 0.000000000286
P(1) = (18C1)(0.95^1)(0.05^17) = 18 x 0.95 x (0.05^17) ≈ 0.00000002145
P(2) = (18C2)(0.95^2)(0.05^16) = (18 x 17 / 2) x (0.95^2) x (0.05^16) ≈ 0.0000002706
Now, we can sum up these probabilities to get the probability of accepting the shipment:
P(shipment accepted) = P(0) + P(1) + P(2) ≈ 0.000000000286 + 0.00000002145 + 0.0000002706 ≈ 0.0000002923
Therefore, the probability that this shipment is accepted if 5% of the total shipment is defective is approximately 0.000000292.
(b) Find the probability that this shipment is not accepted if 15% of the total shipment is defective.
In this case, the probability of accepting a single pen (p) is equal to 1 - probability of a defective pen.
So, p = 1 - 0.15 = 0.85
We need to find the probability of having more than 2 defective pens in a sample of 18 pens. Therefore, we need to find the sum of probabilities from 3 to 18 (since having 0, 1, or 2 defective pens would result in accepting the shipment).
P(shipment not accepted) = P(3) + P(4) + ... + P(18)
To calculate this, we can subtract the probability of accepting from 1:
P(shipment not accepted) = 1 - P(shipment accepted)
Using the formula from part (a), P(shipment accepted) is approximately 0.000000292. Therefore:
P(shipment not accepted) = 1 - 0.000000292 ≈ 0.999999708
So, the probability that this shipment is not accepted if 15% of the total shipment is defective is approximately 0.999999708.