By mistake 3 faulty fuses are put into a box containing two good ones and mixed up. You take 3 fuses from the?

Question

continuing the question:
box with replacement.
a/ find the probability distribution for the random variable x where x is the number of faulty fuses.

Pls: can the answer be with a little explanation?

Answer 1

Alan M is correct but you also asked for an explanation.
You can tackle the question in two ways:
a. select 3 fuses without replacing them, that is taking out three fuses and not putting them back or
b. taking out a fuse, then putting it back and taking one again, and so on.
a. seems to be what the question is about, so fuse no. 1 has a probability of 3/5 or 0.60 of being faulty. That leaves 4 fuses and the probability of any one of them being faulty is now 3/4 or 0.75.
That leaves 3 fuses in the box, so the probability of the next one is 3/3 = 1.0. Now, multiply all three probabilities together:
0.6 x 0.75 x 1.0 = 0.45
That means that the probability of any faulty fuse being selected when selecting three from five without replacement = 0.45

Alan M's answer was correct if you replaced each fuse every time. I hope this helps.

Answer 2

Probability of a faulty fuse is 3/5, probability of a functional fuse 2/5. Personally I'd use a multimeter to filter out and bin the duff ones and have a probability of 2/2.

Answer 3

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