BINOMIAL Probability Distribution question?

Question

gawd i need help, lol:

the BINOMIAL Probability Distribution question reads:

"XYZ company randomly selects and tests 24 light bulbs, then accepts the whole batch if there is ONLY ONE OR NONE that doesn't work. If a particular shipment of thousands of bulbs actually has a 4% rate of defects, what is the probability that this whole shipment will be accepted?"


this is how i do:
let n=fixed number of trials
let x=specific number of successes in 'n' trials
let p=probability of success in one of the 'n' trials
let q=probability of failure (or q=1-p)

n = 24, x = 24, p = 1-0.04, or 0.96 (or 96%)
P (x is at least 23) = P (23) + P (24)= 0.391+0.375=0.766, or 77%.
[NOTE: i used the binomial probability formula above, which is given as n!/[(n-x)!x!] * p^x * q^(n-x)]

is this right?
please help, thanks

Answer 1

It looks like the right approach. You are assuming that the batch size is sufficiently large relative to the sample size that the probabilities are essentially the same as for sampling with replacement.

Zero or one defect is the same as 23 or 24 successes in 24 trials.

P(x ≥ 23) = P(24) + P(23) = p^24 + (24C1)(p^23)(q)

P(x ≥ 23) = .96^24 + 24(.96^23)(.04)

P(x ≥ 23) ≈ 0.3754 + .3754 = .7508
_________

It's a coincidence that P(23) = P(24).