Question

A die with a 10% chance of rolling a 6 is rolled 10 times in a row. What is the probability that a 6 is rolled exactly 2 times?

Answer 1

\[ P(\text{roll a "6"})=0.1\\
P(\text{roll NOT "6"})=0.9 \]

Thus:
P(roll two "6") = P(roll a 6)*P(roll a 6)*[P(roll not 6) * P(roll not 6) * ... * P(roll not 6)]

where there are eight "P(roll not 6)" probabilities. This is because you need to roll two sixes, and the remaining eight rolls cannot be a 6.

So:
\(P(\text{roll two 6}) = (0.1)(0.1)(0.9)^8\)

Answer 2

Using the binomial distribution:
\[P(2\ sixes\ out\ of\ 10\ rolls)=10C2\times0.1^{2}\times0.9^{8}=you\ can\ calculate\]

Answer 3

Oh yes @kropot72 is correct. Indeed there is a combinatorial factor in front. I somehow interpreted the question as rolling a 6 the first two times.