Statistics, variance, E(X) and probability help

I have some stats homework and here's what I got for the answer but I think it's wrong. If anyone could please write out the complete steps and explain it for me I would very much appreciate it.

4. The cook in a restaurant stashes away a tub containing 15 oysters because he knows that there are pearls in 9 of the oysters. A busboy who also knows about the pearls finds the tub, but can make off with only 4 of the oysters before someone sees him. If you let X be the number of oysters that contain a pearl out of those the busboy has, [18 points]

Write down the probability function of X? (4 points)

X 0 1 2 3 4

P(X=x) = P(x=0) + P(x=1)+P(x=2)+ P(x=3)+P(x=4)

What’s the probability that he has 0, 1, 2, 3, or 4 oysters with a pearl? [Hint: Evaluate P(X=0); P(X=1); P(X=2); P(X=3); P(X=4).] (14 points) [Show work and simplify your answers]

P(X = 0) = 6/15 * 5/14 * 4/13 * 3/12 = 0.01099

P(X = 1) = 9/15 * 6/14 * 5/13 * 4/12 = .03297

P(X = 2) = 9/15 * 8/14 * 6/13 * 5/12 = .06593

P(X = 3) = 9/15 * 8/14 * 7/13 * 6/12 = .09231

P(X = 4) = 9/15 * 8/14 * 7/13 * 6/12 = .09231

Find the E(X) and VAR(X). [4 points] [Work must be shown]

E(X) = (0*0.01099) + (1*0.03297) + (2*0.06593) + (3*0.09231) + (4*0.09231)

E(X) = 0.811

VAR(X) = (0 - 0.811)2 * 0.01099

+ (1 - 0.811)2 * 0.03297

+ (2 - 0.811)2 * 0.06593

+ (3 - 0.811)2 * 0.09231

+ (4 - 0.811)2 * 0.09231

VAR(X) = 1.483

Re: Statistics, variance, E(X) and probability help

This is a binomial distribution with an underlying $p=\dfrac{9}{15}=\dfrac 3 5$

$Pr[X=k]=\left(\begin{array}{r} 4 \\ k \end{array}\right) p^k(1-p)^{4-k}$

with $p$ as above

$Pr\{X\} = \dfrac 1 {625} \{ 16, 96, 216, 216, 81\} = \{ 0.0256, 0.1536, 0.3456, 0.3456, 0.1296\}$

you can figure out $E[X]$ and $var[X]$ given this.