Probability of getting 18 or more bulls eye out of 20 shots An Olympic archer can hit the bulls eye an average of 9 times out of 10. The probability of the archer scoring 18 or more bulls eye from 20 shots is what? I've tried it but my answer is wrong. the average probability would be 0.9 as 9/10. But if we divide 18/20, we also get 0.9, what is that incorrect?

This is a classic binomial distribution question. 
First of all, let's work out the probability of hitting the target (success) for a single shot: $$q = \frac{9}{10} = 0.9$$ 
Then, the probability of failure is $$p = 1-q = 1-0.9 = 0.1$$
Let's define the random variable $X$ that represents the number of time the archer misses. $$X∼B(20, 0.1)$$ 
Saying that the archer has hit $10$ or more is equivalent to saying that he's missed $10$ or less and then the required probability is $$P(X \le 10)$$ which could be found in the tables for $n=20$ and $p = 0.1$. 
Similarly, for the archer to hit $18$ or more targets, then he needs to miss $2$ or less. 
Hence the required probability is: $$P(X \le 2)$$ 
which then could be found in the tables for $n=20$ and $p=0.1$
