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We can calculate that the probability that an employee with a score greater than 90 is 0.45. Suppose that 10 employees undergo the evaluation. What is the probability that exactly half of the employees score greater than 90?
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10 choose 5 * p^5 * (1-p)^(10-5)
What about if the probability was no more than 3 employees scoring greater than 90
no more than 3 is exactly 0 or exactly 1 or exactly 2 or exactly 3
\[\binom{10}{0}p^0(1-p)^{10-0}+\binom{10}{1}p^1(1-p)^{10-1}\\~~~~~~~~~~~~~~+\binom{10}{2}p^2(1-p)^{10-2}+\binom{10}{3}p^3(1-p)^{10-3}\]
Thanks that's what I thought
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