# A die is tossed twice.A 'success' is getting an even number on a toss.Find the variance of the number of successes.

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• $$In\;a\;single\;toss\;of\;die\;getting\;even\;no\;=\;\Large\frac{3}{6}=\frac{1}{2}$$
• $$\large\;p(S)=\frac{1}{2}\;p(F)=\frac{1}{2}$$
• $$in\;a\;die\;tossed\;twice\;we\;define\;X=\;no\;of\;sucesses$$
• $$X=(0\;1\;2\;3\;)$$
• $$Mean\;=\sum\;P_i\;X_i$$
• $$\sigma\;=\sqrt{variance}=\sqrt{\sum\;P_iX^2_i\;-\;(\sum\;P_iX_i)^2}$$
$$P(X=0)=p(no\;success)$$=$$p(S\;S)$$
$$\Large\frac{3}{2}\times\frac{1}{2}$$
$$P(X=1)=p(1S\;,1F)or\;p(1F\;,1S)$$
$$\Large\;2(\frac{1}{2}\times\frac{1}{2})$$
$$P(X=2)=p(both\;failurs)$$=$$p(F\;,F)$$
$$\Large\frac{1}{2}\times\frac{1}{2}$$
$$E(X)=\sum\;P_iX_i$$
=$$\Large\;0\times\frac{1}{4}\;+\;1\times\;\frac{1}{2}\;+\;2\;\times\;\frac{1}{4}$$
=$$\Large\;1$$
$$\sum\;P_iX_i^2=\Large\;0\;\times\;\frac{1}{4}\;+\;1^2\;\times\;\frac{1}{2}\;+\;2^2\times\frac{1}{4}$$
=$$\Large\frac{1}{2}+1=\frac{3}{2}$$
$$var\;=\Large\frac{3}{2}-\;1^2$$
=$$\Large\frac{1}{2}$$

edited Jun 4, 2013