What is the probability that the most successful out of these twenty will be right at least 9 times out of the next 10 predictions they make? Suppose that we have twenty economists, each of whom makes their predictions based 
on the toss of a fair coin. What is the probability that the most successful out of these twenty will be right 
at least 9 times out of the next 10 predictions they make? (Here, the “most successful” economist is the one 
who makes the highest number of correct predictions out of all twenty economists.) 
I don't understand where to start on this problem, please help?
How do I determine who the "most successful" economist is?
 A: Ingredients: The probability that at least one of the economists will be right at least $9$ times is $1$ minus  the probability that they all get $8$ or fewer right.
Let $p$ be the probability that a given economist gets $8$ or fewer right. Then the probability that they all get $8$ or fewer is $p^{20}$.
To find $p$, it is easiest to find $1-p$, the probability a given economist gets $9$ or more right.
We leave the cooking to you.
A: For each economist, the probability to be right at least $9$ times out of $10$ predictions is 
$$10\times 0.5^9 \times 0.5^1 + 0.5^{10} = 11/1024$$
So the probability that all $20$ economists fail to be right at least $9$ times out of the next $10$ predictions is $(1-11/1024)^{20} = 0.8057304$
So the answer is $1-0.8057304 = 0.1942696$
A: $$
Neconomists=20
$$
$$
Pdecision=0.5
$$
The probability for a single economist to attain an average accuracy of $0.9$ given a set of $10$ elements. 
$$
P1economist=(Pdecision)^9=(0.5)^9 \approx 0.001953125
$$
$$
P20economist=P1economist*20 \approx [0.0390625]
$$
Hopefully I have the right idea, but I am not quite confident in this conclusion.
Sincerely,
--
Jordan D. Ulmer
