What is the probability that there will be no defective computer chip found

A piece of electronic equipment contains 6 computer chips, 2 of which are defective. 3 chips are selected at random and inspected.

To illustrate the problem, let's say that the 4 good chips are called a1, a2, a3, and a4, and the two bad chips are called d1 and d2. The total amount of chips (elements) can be shown this way:

{a1, a2, a3, a4, d1, d2}

Three chips are selected at random, so some possibilities of these groups are:

{(a1, a2, a3), (a1, a2, a4), (a1, a3, d1), (a2, a4, d2), etc.......}

Notice that in this case (a1, a2, a3) and (a3, a2, a1) are the same group, so the order does NOT matter, and a possibility like (a1, a1, a2) does not exist, so we have a case of combinations without repetition. The formula for these cases can be expressed as:

C = m! / [n!(m - n)!]

Let's find out first how many possibilities there are of randomly choosing 3 good chips. There are 4 elements (good chips) and 3 elements (good chips) per group.

m = total number of elements, so m = 4
n = number of elements per group, so n = 3

C = 4! / [3!(4 - 3)!] = 4! / (3!*1!) = (4*3!) / 3! = 4

So there are 4 possibilities of randomly choosing three good chips. We can actually see them by inspection:

{(a1, a2, a3), (a1, a2, a4), (a1, a3, a4), (a2, a3, a4)}

Now let's find out the full number of possible groups of 3 chips that can be formed from the 6 chips. The process is the same, but:

m = total number of elements, so m = 6
n = number of elements per group, so n = 3

C = m! / [n!(m - n)!] = 6! / [3!(6 - 3)!] = 6! / (3!*3!) = (6*5*4*3!) / (3!*3!) = 120/6 = 20

Some of these groups will be:

{(a1, a3, a4), (a2, a4, d1), (a3, d1, d2), etc.......}

So, based on the results found, the probability that there will be no defective computer chip found is 4/20 = 1/5

If you can come up with the same result using more than one approach then that's the ideal situation to prove the solution, but this solution is correct.

I hope this helps...

A piece of electronic equipment contains 6 computer chips, 2 of which are defective. 3 chips are selected at random and inspected.
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six total chips, 2 defective, but more importantly: 4 good.

first draw: good chip, prob 4/6.
second draw: good chip, 3/5.
third draw: 2/4.

The probability that no defective chip will be found

(i.e. all good) is the product of all three probabilities.

4/6 x 3/5 x 2/4 = (4 x 3 x 2)/(6 x 5 x 4) = 24/120, or 1/5.

The probability is 1 in 3.