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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

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Consider if the order of the sixes doesn't matter, how many different configurations could you have of where to put the pair of 6 values for the first question. –  JB King Feb 5 '13 at 23:16

2 Answers 2

up vote 1 down vote accepted

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.

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$X$ is a Binomial random variable with parameters $n=5$ and $p=1/6$. You might find this Wiki page useful.

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