A fair die is rolled 7 times, find the probability of getting " 66 dots" exactly 5 times.

 A fair die is rolled 7 times, find the probability of getting "

6$6$ dots" exactly 5 times.

Solution 

This is an example where although the outcomes are more than 2, we interested in only 2: "6" or "no 6".
The die is rolled 7 times, hence the number of trials is n=7$n=7$.
In a single trial, the outcome of a "6" has probability p=1/6$p=1/6$ and an outcome of "no 6" has a probability 1−p=1−1/6=5/6$1-p=1-1/6=5/6$
The probability of having 5 "6" in 7 trials is given by the formula for binomial probabilities above with n=7$n=7$, k=5$k=5$ and p=1/6$p=1/6$

P(5heads in 7 trials)=(75)(1/6)5(1−5/6)7−5=(75)(1/6)5(5/6)2$$\begin{gathered} {\displaystyle P(5\text{heads in 7 trials})={\displaystyle (\genfrac{}{}{0ex}{}{7}{5})(1/6{)}^5(1-5/6{)}^{7-5}}} \\ ={\displaystyle (\genfrac{}{}{0ex}{}{7}{5})(1/6{)}^5(5/6{)}^2} \end{gathered}$$

Use formula for combinations to calculate

(75)=7!5!(7−5)!=21${\displaystyle (\genfrac{}{}{0ex}{}{7}{5})={\displaystyle \frac{7!}{5!(7-5)!}}=21}$
Substitute
P(5"6" in 7 trials)=21(1/6)5(5/6)2=            $0.00187$

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