# Thread: How was this probability distribution calculated?

1. ## How was this probability distribution calculated?

I'm reading my textbook, and I don't follow how the following probability distribution was calculated.

Can you help me understand how we got the f(x) values in the probability distribution? What formulas were used? Is it some kind of counting calculation?

2. ## Re: How was this probability distribution calculated?

Originally Posted by Fischercr
I'm reading my textbook, and I don't follow how the following probability distribution was calculated.

Can you help me understand how we got the f(x) values in the probability distribution? What formulas were used? Is it some kind of counting calculation?
$f(k)=\dbinom{3}{k}\left( \dfrac{1}{4}\right)^k\left( \dfrac{3}{4}\right)^{3-k}$___$k=0,1,2,3$

$\dbinom{3}{k}=\left( \dfrac{3!}{(k!)(3-k)!}\right)$

3. ## Re: How was this probability distribution calculated?

Equivalently, we are told that "this process produces 25% defectives" so for any one item the probability it is defective (D) is 0.25= 1/4, the probability it is non-defective (N) is 075= 3/4.
The probability of NNN, all three are non-defective, since we are also told these are "independent", is (3/4)(3/4)(3/4)= 27/64. With X the number of defective items, P(X= 0)= 27/64.

The probability the first chosen is defective, the other two are non-defective, NND, in that order, is (3/4)(3/4)(1/4)= 9/64, But it is easy to see that the probability of "non-defective, defective, non-defective", NSN, in [b]that[/b ] order is (3/4)(1/4)(0.75)= 9/64 also and similarly "non-defective, non-defective, defective", NND, is also 9/64. P(X= 1)= 9/64+ 9/64+ 9/64= 3(9/64)= 27/64.

The probability the both the first two chosen are defective, the third non-defective, DDN, in that order, is (1/4)(1/4)(3/4)= 3/64. Again, it is easy to see that the probabililties of "defective, non-defective, defective", DND, and "non-defective, defective, defective", NDD are (1/4)(3/4)(1/4)= 3/64, and of (3/4)(1/4)(1/4)= 3/64 also. P(X= 2)= 3/64+ 3/64+ 3/64= 3(3/64)= 9/64.

Finally, the probability of "DDD", all three defective, is (1/4)(1/4)(1/4)= 1/64. P(X= 3)= 1/64.