Roll 5 fair six sided dice. Let X be the number of sixes.
1) Find $\ P(X=2)$
2) Find $\ P(X <= 1)$
3) Find $\ E[X]$
4) Find $\ Var(X)$
I've been able to find the first two answers on Wolfram Alpha but I don't understand how to get them.
Here's my incorrect approach to #1: $\ (\frac{1}{6})^2(\frac{5}{6})^3 $
Any help would be appreciated
Best Answer
A random variable $X$ follows a binomial distribution when it describes a probability of obtaining $k$ successes out of $n$ trials, each of which succeeds with probability $p$:
$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Note that this is a distribution because, by the binomial theorem, it is normalized:
$$\sum_{k=0}^n P(X=k) = 1$$
For a) plug $n=5$, $k=2$, $p=1/6$ into the above formula.
For b) you need to sum the cases $k=0$ and $k=1$.
For c), recall that expected value is
$$E[X] = \sum_{k=0}^n k \, P(X=k)$$
There is a formula you can derive for $E[X]$, or you can simply do the sum.
For d) recall that variance is
$$Var[X] = E[X^2] - E[X]^2 = \sum_{k=0}^n k^2 \, P(X=k) - \left [ \sum_{k=0}^n k \, P(X=k) \right ]^2$$
Again, there is a formula, or you can just evaluate the sum.