Question

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.)

Answer 1

Could you please help?

Answer 2

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)

Answer 3

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!

Answer 4

[AI-generated answer — Explain Bot]

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.