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Assume that a fund manager has no idea of hat he is doing and is essentially flipping a coin in order to make decisions. Let's say it's a 50/50 proposition that the manager will have a "good" year. What is the probability that the fund manager will have 6 or more good years out of 10?

Is it (1/2)^6?

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    That's the probability that six specified years are good. Read up on [binomial distributions](https://en.wikipedia.org/wiki/Binomial_distribution). – 2017-02-14
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    Can you provide me with the answer please so I can at least check my work? – 2017-02-14
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    Please provide your work. As I say, your answer is not at all correct. – 2017-02-14
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    So I could view this problem as, what is the probability of getting at least 6 heads on 10 flips, correct? That would be ((1/2)^6)*((1/2)^4)? – 2017-02-14
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    @Marley that is the probability of getting **exactly** the first six flips to be heads and the remaining flips to be tails in ten flips., not the probability of getting **at least** six heads and not the probability of getting exactly six heads and four tails **in any order** in ten flips – 2017-02-14
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    Your formula is wrong even for exactly six...it gives the probability that six specified years are good and the other four specified years are bad. to get the answer for exactly six out of ten you need to multiply by $\binom {10}6$. – 2017-02-14
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    I'm sorry, but what is the notation for (10 6) you have above. Is that (10!/6!). Thank you both for the help. – 2017-02-14
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    $\binom{n}{r}$ is the [binomial coefficient](https://en.wikipedia.org/wiki/Binomial_coefficient) "$n$ *choose* $r$" and counts the number of ways of selecting a subset of $r$ objects out of a set of $n$ distinct objects (*i.e. order doesn't matter*) and can be simplified algebraically as $\binom{n}{r}=\frac{n!}{r!(n-r)!}$. The information on what the binomial coefficient is should have been included in the link to the binomial distribution which we have provided several times already to you in the past hour which you should have had the opportunity to read already by now. – 2017-02-14
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    What is with all the attitude @Jmoravitz? Yes I know you have provided the wikipdia link to Binomial Distribution page. I am currently working a full time 10 hour/day job while trying to complete this homework. I don't have time to comb through wikipedia pages. So take it easy on me. – 2017-02-14
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    So I did (10!/4!*6!)(1/2)^6*(1/2)^4.....to get about .2 or 20%. Is this correct? – 2017-02-14
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    That is still the probability of exactly six, not at least six. Read the answer below. "At least six" corresponds to "exactly six or exactly seven or exactly eight or ..." – 2017-02-14
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    So I have to add to probabilities of exactly 7 and so on? Is the answer 37.70%? – 2017-02-14
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    Yes $~~~~~~~~~$ – 2017-02-14
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    Together we can achieve more @Jmoravitz. Now, let's go tackle the other one. I think i can apply some the same principles. – 2017-02-14

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Hint: The probability you are looking for is $$ \sum_{k=6}^{10} \binom{10}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{10-k}. $$

Left for the reader to simplify the above further.

As @lulu said, read on https://en.wikipedia.org/wiki/Binomial_distribution to arrive at the above answer.