[tex]X[/tex] 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 [tex]u=2[/tex].
For any probability density, the probabilities of all possible outcomes must sum to 1. This means
[tex]p+\dfrac{15}{36}+\dfrac1{36}=1\implies p=\dfrac59[/tex]
The mean/expectation of [tex]X[/tex] is
[tex]E[X]=\displaystyle\sum_xx\,P(X=x)=0\cdot p+2\cdot\frac{15}{36}+3\cdot\frac1{36}=\frac{11}{12}[/tex]
The variance of [tex]X[/tex] is
[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
where the second moment is
[tex]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}[/tex]
[tex]\implies V[X]=\dfrac{23}{12}-\left(\dfrac{11}{12}\right)^2=\dfrac{155}{144}[/tex]
Then the standard deviation is the square root of the variance:
[tex]\sqrt{V[X]}=\dfrac{\sqrt{155}}{12}\approx1.037[/tex]
