A manufacturer of light bulbs knows that 3% of the production of their 60W bulbs will be defective. What is the probability that exactly 5 bulbs in a carton of 144 will ne defective?

Did not understand the posted answer before, need it simplified.

One way for your problem could be

DDDDDGGG...GG , with 139 G's, where D stands for "defective" and is .03 and G stands for "good" and G is .97)

the prob of that particular case would be
(.03)^5 (.97)^139

but...
there are many ways to arrange DDDDDGGG...GG
There are 144 elements to arrange, 5 are alike of one kind, and 139 are alike of another.
The number of such ways = 144!/(5!139!)

That is how drwls got his answer of
(0.03)^5*(0.97)^139*144!/(139!*5!)

(I had .16735)

To find the probability that exactly 5 bulbs in a carton of 144 will be defective, we need to use the binomial probability formula. The formula is:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

Where:
- P(X = k) is the probability of exactly k successes (in this case, exactly 5 defective bulbs)
- n is the total number of trials (in this case, the number of bulbs in the carton, 144)
- p is the probability of success in a single trial (in this case, the probability that a single bulb is defective, which is 3% or 0.03)
- (n C k) represents the binomial coefficient, which is the number of ways to choose k successes from n trials

Now, let's plug in the values and calculate the probability:

P(X = 5) = (144 C 5) * (0.03)^5 * (1 - 0.03)^(144 - 5)

To simplify further, we can use a calculator or software that can calculate binomial coefficients. The result will give you the probability that exactly 5 bulbs in a carton of 144 will be defective.