Question 1166165
you are correct, as far as i know.
there are 24 bulbs in each shipment.
the probability that any one bulb, randomly selected, will be defective is .05.
you want to find the probability that the whole shipment will be accepted.
the formula for binomial probability is:
p(x) = p^x * q^(n-x) * c(n,x)
c(n,x) is the number of ways you can get x elements out of a set of n elements.
c(n,x) is equal to n! / (x! * (n-x)!).
to be acceptable, the number of defects has to be 0 or 1.
you want the sum of p(0) + p(1).
that tells you the probability that 0 or 1 are defective.
that also tells you the probability  that  the shipment will be accepted.
p(0) = p^0 * q^(24-0) * c(24,0)
p(1) = p^1 * q^(24-1) * c(24,1)
p = .05
q = 1 - .04 = .95
the formulas becomes:
p(0) = .05^0 * .95^(24) * c(24,0) = .2919890243
p(1) = .05^1 * .95^(23) * c(24,1) = .3688282413


the probability of having 0 or 1 defective bulbs is the sum of p(0) and p(1) = .6608172556.


that's also the probability that the shipment will be accepted, since the shipment will be accepted if 0 or 1 bulbs are defective, i.e. less than 2.