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There are 8 people applying the a community. The community can have either 4, 5, 6 members. There are two special people, George and Kashish

Find the probability that either George or Kashish (but not both) are on the committee.

We add $P(4) + P(5) + P(6)$ for the num of members.

4: So There are a total of $\binom{8}{4}$ ways to create the group. 1 way to pick George, $\binom{6}{3}$ ways to pick someone not Kashish.

5 Similarly, $\binom{8}{5}$ ways to create group, 1 way to pick George, $\binom{6}{4}$ to choose someone not Kashish.

6 $\binom{8}{6}$ create group, $1$ way to pick George, $\binom{6}{5}$ ways to pick others not Kashish.

$P_{\text{george}}= \sum P_i = \frac{\binom{6}{3}}{\binom{8}{4}} + \frac{\binom{6}{4}}{\binom{8}{5}} + \frac{\binom{6}{5}}{\binom{8}{6}}$

Is this correct for George, for the total we would just $\times 2$ right?

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    "The community can have either 4, 5, or 6 members..." "Find the probability that either george or kashish (but not both) are on the committee" We aren't told anything about how to expect probabilities to work in this scenario. Is it equally likely for the committee to be 4 people as it is for it to be 5 people? Or is each possible committee equally likely? – 2017-01-21
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    You will find that $\frac{\binom{6}{3}}{\binom{8}{4}} + \frac{\binom{6}{4}}{\binom{8}{5}} + \frac{\binom{6}{5}}{\binom{8}{6}} = \frac{43}{56}$, so multiplying that by two would give you a "probability" greater than one, clearly an impossibility. – 2017-01-21

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As I pointed out in the comments above, it is not clear how the decision is made what committee we will randomly select. Your work is fine up to a point...

$\binom{6}{3}/\binom{8}{4}$ is indeed the probability that George is on a four-person committee without Kashish given that the committee is of size four. Your work does not account for the probability that the committee is in fact of size four to begin with. With variable probabilities for having chosen a committee of a particular size, I'll call them $p_4,p_5,p_6$ the probability that George is on the committee and not Kashish will be

$$p_4\cdot \frac{\binom{6}{3}}{\binom{8}{4}} + p_5\cdot\frac{\binom{6}{4}}{\binom{8}{5}} + p_6\cdot \frac{\binom{6}{5}}{\binom{8}{6}}$$

In the case that our randomly selected committee is equally likely to be each size and then each committee of that size is equally likely, each of the $p_i$ then would be equal to $\frac{1}{3}$.

Another interpretation of the problem however would be where every possible committee is equally likely regardless of size. In this case, there are $\binom{8}{4}+\binom{8}{5}+\binom{8}{6}$ different committees, making the denominator of each fraction you calculated be the entire sum $\binom{8}{4}+\binom{8}{5}+\binom{8}{6}$ instead of just one of the terms.

This yields a total probability that George is on the committee and Kashish is not as

$$\frac{\binom{6}{3}+\binom{6}{4}+\binom{6}{5}}{\binom{8}{4}+\binom{8}{5}+\binom{8}{6}}$$

You may then multiply the result by two for the final answer since by a symmetrical argument one may find that the probability only Kashish and not George is on the committee is the same.

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    Each outcome is equally likely though (eg. size of 4 prob is the same as size of 5 prob, size of 6 prob etc...) – 2017-01-21
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    @racer11 those two statements (*each outcome is equally likely*) and (*size of 4 prob is same as size of 5 prob...*) are in direct conflict with one another. If each outcome is equally likely then the probabilities that it is of size four is not the same as the probability that it is of size five and vice versa. – 2017-01-21
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    @racer11 Think of a much smaller example for now so we can clear up how probabilities work. There are two people who want to be on a committee and the committee can be of size $1$ or of size $2$. The committees are $\{a\},\{b\},\{a,b\}$. Is $Pr(\{a\})=\frac{1}{3}$ (*since there are three committees possible*)? or is $Pr(\{a\})=\frac{1}{4}$ (*since size 1 is picked half of the time and half of the time size1 is picked it is $\{a\}$*)? – 2017-01-21