**Toolbox:**

- Two biased setting dice
- 1\(^{st}\) dice P(6)=\(\frac{1}{2}\)
- others are equally likely
- P(1)=\(p(2)\)=\(p(3)\)=\(p(4)\)=\(p(5)\)=\(\large\frac{1}{2}\)/5=\(\large\frac{1}{10}\)
- 2\(^{nd}\) dice\(P(1)\)=\(\large\frac{2}{5}\) others eqally likely
- \(p(other\; than \;one)\)=\(\large\frac{3}{5}\)
- X denots number of 'Y' seen
- X=(\(0\;1\;2)\) when the two die are thrown togather

=\(P(X=0)\)=P(neither shows 1)

=P(1\(^{st}\)shows\(6\;5\;4\;3\;2\; and\; 2^{nd} shows\; other\; than\; one\))

=(\(\large\frac{1}{2}\)+\(\large\frac{4}{10}\))\(\times\large\frac{3}{5}\)=\(\large\frac{27}{50}\)=\(0.54\)

=\(P(X=1)\)=P(either shows 1)

=P(1\(^{st}\)shows1 and 2\(^{nd} \)shows other)+P(1\(^{st}\)shows other 2 shows'1')

=(\(\large\frac{1}{10}\;\times\;\frac{3}{5}\;+\frac{4}{10}\;+\;\frac{1}{2}\;\times\;\frac{2}{5}\))

=\(\large\frac{3}{50}\;+\;\frac{6}{10}\;=\;\frac{33}{50}\;=(0.42)

\(p(x=2)\)=P(both shows '1')

=\(\large\frac{1}{10}\)\(\times\large\frac{2}{5}\)=\(\large\frac{2}{50}\)

=\(0.04\)

Probability distrubation is

(\(X\;0\;1\;2\))

(\(p(x)\;0.54\;0.42\;0.04\))