The answers are:
We know that if we have a set of N elements, the number of different groups of K elements that we can make, such that:
N ≥ K
Is given by:
[tex]C(N, K) = \frac{N!}{(N - K)!*K!}[/tex]
a) If we have 20 semiconductors and we make groups of 3, then we can have:
[tex]C(20, 3) = \frac{20!}{(20 - 3)!*3!} = \frac{20*19*18}{3*2} = 1,140[/tex]
b) The probability that a randomly picked chip is nonconforming is given by the quotient between the number of nonconforming chips and the total number of chips.
So, if the first selected chip is the nonconforming one, the probability of selecting it is:
P = 10/20 = 1/2
The next two ones work properly, the probability is computed in the same way, but notice that now there are 10 proper chips and 19 chips in total.
Q = 10/19
And for the last one we have:
K = 9/18
The joint probability is:
P*Q*K = (1/2)*(9/19)*(8/18) = 0.132
But this is only for the case where the first one is the nonconforming, then we must take in account the possible permutations (there are 3 of these) then the probability is:
p = 3*0.132 = 0.395
c) The probability of getting at least one nonconforming chip is equal to the difference between 1 and the probability of not getting a nonconforming chip.
That probability is computed in the same way as above)
P = (10/20)*(9/19)*(8/18) = 0.105
Then we have:
1 - P = 1 - 0.105 = 0.895
So the probability of getting at least one nonconforming chip is 0.895.
If you want to learn more about probability, you can read:
https://brainly.com/question/251701
