# A biased die is such that $P(4)=\Large \frac{1}{10}$ and other scores being equally likely.The die is tossed twice.If X is the number of fours seen',find the variance of the random variable X.

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=$$\large\frac{22}{100}$$-($$\large\frac{22}{100}$$)$$^2$$
=$$0.18$$

edited Jun 4, 2013