More issues with probability distribution

[jakedowdy]

Now I have to construct a probability distribution table from the following information:

There are six pencils for sale and two are defective. If you buy two pencils chosen at random, let X be the number of defective pencils purchased. Construct the probability distribution for X.

I tried to compute p(x) by looking at the purchasing of the two pencils are separate events WOR. So probability of not getting a defect on the first purchase would be 2/3 and second purchase would be 4/5. So p(0)= 2/3 * 4/5 = .533

However, I know this is wrong, because it results in the following probability distribution:

x 0 1 2
p(x) .533 .267 .066

Since all outcomes for p(x) should sum to 1, I know I screwed up somewhere. What am I doing wrong?

[Martingale]

Now I have to construct a probability distribution table from the following information:

There are six pencils for sale and two are defective. If you buy two pencils chosen at random, let X be the number of defective pencils purchased. Construct the probability distribution for X.

I tried to compute p(x) by looking at the purchasing of the two pencils are separate events WOR. So probability of not getting a defect on the first purchase would be 2/3 and second purchase would be 4/5. So p(0)= 2/3 * 4/5 = .533

However, I know this is wrong, because it results in the following probability distribution:

x 0 1 2
p(x) .533 .267 .066

Since all outcomes for p(x) should sum to 1, I know I screwed up somewhere. What am I doing wrong?

http://latex.codecogs.com/gif.latex?P(X=x)=\frac{\binom{2}{x}\binom{4}{2-x}}{\binom{6}{2}}=\frac{6}{15},\frac{8}{15},\frac{1}{15}

for x=0,1,2 respectively

http://latex.codecogs.com/gif.latex?\begin{array}{c|ccc}x&0&1&2\\\hline%20P%28X=x%29&\frac{6}{15}&\frac{8}{15}&\frac{1}{15}\end{array}

[Riverdale27]

It's the hypergeometric distribution by the way... That's like the binomial, but without replacement...