I was solving the below problem, and I had a few questions:
An urn contains five red marbles and three blue marbles. Four marbles are chosen without replacement from the urn and their colors are noted. Describe the probability distribution for X, the number of red marbles selected. You may either develop an algebraic functional form or a table. Calculate the expected value and variance for X.
I started by making a table, where: Trial 1: Prob of being red: 5/8, probability of blue 3/8 Trial 2: Prob red, 4/7, prob blue 2/7 Trial 3: prob red 3/6, prob blue 1/6 Trial 4: prob red 2/5, prob blue 0.
Afterwards, what should I do?
For the next question, I said it's hyper geometric and the expected value is 4*5/8. For variance, I still have not figured it out.
Thanks!!
Producing the general formula for the variance of the hypergeometric is not easy. Of course you can look it up, by searching for hypergeometric distribution in Wikipedia. Or perhaps your book gives the information.
However, I think that this is not what you are expected to do. Since the numbers are small, you can explicitly write down the required probabilities, and compute the mean and variance from these probabilities.
The possible values of $X$ are $1,2,3,4$. (We cannot have $X=0$, since there are too few blues).
We will use a general formula. There are $\binom{8}{4}$ equally likely ways to choose $4$ balls. There are $\binom{5}{1}\binom{3}{3}$ ways to choose $1$ red and $3$ blue. Thus $$\Pr(X=1)=\frac{\binom{5}{1}\binom{3}{3}}{\binom{8}{4}}.$$ This turns out to be $\frac{1}{14}$.
Similarly, $$\Pr(X=2)=\frac{\binom{5}{2}\binom{3}{2}}{\binom{8}{4}}.$$ This turns out to be $\frac{6}{14}$. We leave it to you to calculate $\Pr(X=3)$ and $\Pr(X=4)$. These turn out to be $\frac{6}{14}$ and $\frac{1}{14}$.
Now we find the mean of $X$. You found that correctly, but a more basic way of doing it is $$E(X)=1\cdot \frac{1}{14}+2\cdot\frac{6}{14}+3\cdot\frac{6}{14}+4\cdot\frac{1}{14}.$$ The expression simplifies to $\frac{5}{2}$.
For the variance of $X$, you have probably seen the formula $\text{Var}(X)=E(X^2)-(E(X))^2$. We know the second term. And $$E(X^2)=1^2\cdot \frac{1}{14}+2^2\cdot\frac{6}{14}+3^2\cdot\frac{6}{14}+4^2\cdot\frac{1}{14}.$$
Remark: In view of your computations, we find the probability that $X=1$ in a different way. We can get $1$ red and therefore $3$ blue in $4$ different ways, RBBB, BRBB, BBRB, and BBBR. To find $\Pr(X=1)$, we find the probability of each of these and add up.
The probability of RBBB is $(5/8)(3/7)(2/6)(1/5)$. Calculate the probabilities of the other three. They will all turn out to simplify to the same thing, and when you add up you will get $\frac{1}{14}$.
In a similar way, you can find $\Pr(X=2)$, $\Pr(X=3)$, and $\Pr(X=4)$. A little painful, and with risks of error, but it will work.