Question

Three fair dice are rolled at once. Let X denote the number of dice that land with the same number of dots on top as at least one other die. The probability distribution for X is x 0 u 3 P ( x ) p 15 36 1 36 Find the missing value u of X. Find the missing probability p. Compute the mean of X. Compute the standard deviation of X.

Answer 1

$X$ can be any of 0, 2, or 3, which is to say the value of the dice are all different, 2 of them match, or all 3 are the same. So $u=2$.

For any probability density, the probabilities of all possible outcomes must sum to 1. This means

$p+\dfrac{15}{36}+\dfrac1{36}=1\implies p=\dfrac59$

The mean/expectation of $X$ is

$E[X]=\displaystyle\sum_xx\,P(X=x)=0\cdot p+2\cdot\frac{15}{36}+3\cdot\frac1{36}=\frac{11}{12}$

The variance of $X$ is

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where the second moment is

$E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=0^2\cdot p+2^2\cdot\frac{15}{36}+3^2\cdot\frac1{36}=\frac{23}{12}$

$\implies V[X]=\dfrac{23}{12}-\left(\dfrac{11}{12}\right)^2=\dfrac{155}{144}$

Then the standard deviation is the square root of the variance:

$\sqrt{V[X]}=\dfrac{\sqrt{155}}{12}\approx1.037$