# More issues with probability distribution

#### jakedowdy

##### New Member
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

##### TS Contributor
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?
$P(X=x)=\frac{\binom{2}{x}\binom{4}{2-x}}{\binom{6}{2}}=\frac{6}{15},\frac{8}{15},\frac{1}{15}$

for x=0,1,2 respectively

$\begin{array}{c|ccc}x&0&1&2\\\hline P(X=x)&\frac{6}{15}&\frac{8}{15}&\frac{1}{15}\end{array}$

#### Riverdale27

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