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a dice is thrown five times. what is the probability that we get number 6 four times??
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This is a binomial probability problem where your successful trial is a 6 and a fail trial is anything other than 6. So define, \[p=\frac{1}{6}\]and\[q=\frac{5}{6}\]Now the probability of obtaining a 6, four times, is\[p(X=6)=^5C_4p^4q=\frac{5!}{4!(5-4)!}\left( \frac{1}{6} \right)^4\left( \frac{5}{6} \right)=5\frac{5}{6^5}=\frac{25}{1296}\]
thanx, now I finally get it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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