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 what is known as a binomial distribution, because we are looking at “sampling” with two possible outcomes, the probability of which we say is constant between trials. There is a prescribed formula for binomial distributions:
$$X\sim\mathrm{B}(n,p) \implies \operatorname{P}(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$
We will let $X$ be how many bull’s eyes the archer hits, $n=20$ be the number of times she shoots, and $p=0.9$ bet the probability that she successfully hits a bull’s eye.
Hence we have $$\operatorname{P}(X=x)=\binom{20}{x}(0.9)^x(0.1)^{20-x}$$
To determine the probability that she gets eighteen or more bull’s eyes, we need to add the probabilities of all the valid cases—$X=18,X=19,X=20$—which we can calculate as 
$$\sum_{x=18}^{20}\binom{20}{x}(0.9)^x(0.1)^{20-x}$$
or, if you have a Texas Instrument on hand, you can use the command
$$\mathtt{1-binomcdf(20,0.9,17)}$$
The answer comes out to $$0.676\,926\,805\,189$$
