A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random sample of 5 items is chosen to be inspected. Find the probability that the sample contains exactly one defective item (a) if the sampling is done with replacement. (b) if the sampling is done without replacement.

a. 39.55%

b. 44.02%

Step-by-step explanation:

We have the following data:

n = 5

x = 1

p = 5/20 = 0.25

to. If the sampling is done with replacement.

We apply the binomial distribution formula, which is as follows:

P = nCx * (p ^ x) * ((1-p) ^ (n-x))

Where nCx, is a combination, and is equal to:

nCx = n! / x! * (n-x)!

replacing we have:

5C1 = 5! / 1! * 4! = 5

replacing in the main formula:

P = 5 * (0.25 ^ 1) * ((1- 0.25) ^ (5-1))

P = 0.3955

that is, without replacing the probability is 39.55%

b. if the sampling is done without replacement.

Here it is a little different from the previous one, but what you should do is calculate three cases,

the first was the one at point a, when n = 5 and x = 1

5C1 = 5! / 1! * 4! = 5

the second is when n = 20 and x = 5, this is all possible scenarios.

20C5 = 20! / 5! * 15! = 15504

and the third is when n = 15 (20-5) and x = 4 (5-1), which corresponds to the cases when none were damaged

15C4 = 15! / 4! * 11! = 1365

In the end, it would be:

P = (5C1 * 15C4) / 20C5

Replacing:

P = 5 * 1365/15504

P = 0.4402

Which means that without replacing the probability is 44.02%

damm bro the struggle

step-by-step explanation:

step- i d k and i d c : )

you: : 0

bye