# Find the probability distribution of the maximum of two scores obtained when a die is thrown twice .Determine also the mean of the distribution.

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=$$\frac{1}{36}$$/$$\frac{3}{36}$$/$$\frac{5}{36}$$/$$\frac{7}{36}$$/$$\frac{9}{36}$$/$$\frac{11}{36}$$

edited Jun 4, 2013