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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jan 2008 19:38:14 IST
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using binomial distribution,
prob. of 3 occuring exactly once = 1- 4C0 (5/6)^4 = (671/1296)
sample space = 6^4 = 1296
no. of favourable outcomes = (prob.) * (sample space)
= 1296 *( 671/ 1296)
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