**Toolbox:**

- In a toss of baised die
- P(getting 4)=\(\Large\frac{1}{10}P=\;\frac{1}{10}\;q=\;\frac{1}{10}\)
- Die is tossed 2 ,X defined as no of '4' seen
- X can take X=\((0\;1\;2)\)
- varianceX=\(\sum\)P\(_i\)X\(_i\)\(^2\)+(\(\sum\)P\(_i\)X\(_i\))\(^2\)

\(P(x=0)\)=\(p(no4\))

=\(P(not 4 and not 4)\)

=\(\large\frac{9}{10}\;\times\;\frac{9}{10}\;=\;\frac{81}{100}\)

or\(P(x=0)\)=\(^2\)c\(_1\)(\(\large\frac{9}{10}\))\(^0\)(\(\large\frac{9}{10}\))\(^1\)

\(p(x=0\))=P(one 4)

=\(^2\)c\(_1\)(\(\large\frac{1}{10}\))\(^1\)(\(\large\frac{9}{10}\))\(^1\)

=2x\(\large\frac{9}{100}\)=\(\large\frac{18}{100}\)

=\(P(x=2)\)=\(^2\)c\(_2\)(\(\large\frac{1}{10}\))\(^2\)(\(\large\frac{9}{10}\))\(^0\)

=\(\large\frac{1}{100}\)

\(\sum\)P\(_i\)X\(_i\)=0\(\times\large\frac{81}{100}\)+1\(\times\large\frac{18}{100}\)+2\(\times\large\frac{1}{100}\)

=\(\large\frac{20}{100}\)

(\(\sum\)P\(_i\)X\(_i\))\(^2\)=0\(\times\large\frac{81}{100}\)+1\(^2\)\(\times\large\frac{18}{100}\)+2\(^2\)\(\times\large\frac{1}{100}\)

\(\large\frac{18}{100}\)+\(\large\frac{4}{100}\)=\(\large\frac{22}{100}\)

variance X=\(\large\frac{22}{100}\)-(\(\large\frac{22}{100}\))\(^2\)

=\(0.18\)