X defined as maximum of two scores when two die are thrown.

X can take values X=(\(1\;2\;3\;4\;5\;6)\)

When two die are thrown sample space S={(\(1\;1)(1\;2)\dots(6\;6)\)}

E(X)=\(\sum\)P\(_i\)X\(_i\)

Pprobability of getting any number on a single through of die=\(\frac{1}{6}\)

P(X=1)=\(\large\frac{1}{36}\)

P(maximum of two number is 2)

=P[(\(2\;1)or(1\;2)\)]

=\(\large\frac{1}{6}\;\times\;\frac{1}{6}\;+\;\frac{1}{6}\;\times\;\frac{1}{6}\)

=\(\large\frac{3}{36}\)

P(X=3)=P[(\(3\;1)(3\;2)(1\;3)(2\;3)(3\;3\)]

=5\(\times\large\frac{1}{36}\)

P(X=4)=P[(\(1\;4)(4\;1)(2\;4)(4\;2)(3\;4)(4\;3)(4\;4\)]

=7\(\times\large\frac{1}{36}\)

P(X=5)=P[(\(1\;5)(5\;1)(2\;5)(5\;2)(3\;5)(5\;3)(4\;5)(5\;4)(5\;5)\)]

=9\(\times\large\frac{1}{36}\)

P(X=6)=P[(\(1\;6)(6\;1)(2\;6)(6\;2)(3\;6)(6\;3)(4\;6)(6\;4)\dots(6\;6)\)]

p(X=6)=1\(\times\)\(\large\frac{1}{36}\)

px=\(\frac{1}{36}\)/\(\frac{3}{36}\)/\(\frac{5}{36}\)/\(\frac{7}{36}\)/\(\frac{9}{36}\)/\(\frac{11}{36}\)