Troubling Probability question about simple topics There are 6 examiners to pick from. Two of them are easy and will pass one with a probability of $p=.8$. The other four are hard and will pass with probability $q=.6$. All are independent. I can elect to have a panel of three examiners chosen at random or I can choose to have a 5 member panel selected by me. 
If I choose the 3-member panel, then I pass if two of three vote to pass me. 
With 5-member panel, pass if 3 of 5 pass me.
Which size panel should I pick? It is clear that I should pick the panel with the highest probability that will pass me. 
If choose random selection, 3 panel: $6\choose 3$ possible panels. $3 \choose 2$ that will pass me. I get stuck on how to incorporate the respective probabilities because I know to account for what we dont know, we use probabilities. But it is random which one will be picked. Since that is random, how determine prob associated with it.
If choose 5 member: Let us assume that I am rational and will pick the panel comprised of the two easy and 3 hard (I really dont think picking more than 3 hard judges will result in a pass). So I need the 3 to pass me. There is $5 \choose 3$ that will pass me. Same as the above part, how associate probabilities with this.
What is probability associated with this panel? This will be gotten from above, if I am not mistaken?
 A: We have 6 examiners to choose from, 2 p-type and 4 q-type.
Let $\mathrm T$ be the event of passing from a 5-member panel comprised of 2 p-type and 3 q-types.  A five member panel passes you if 3 or more do:
The number of p-types and q-types who pass you will each be binomially distributed. Let $X$ be the number of p-type members who pass you. Then we have by Law of Total Probability:
$$\begin{align}
\mathsf P(\mathrm T) & = \mathrm P(X=0)\cdot\mathsf P(\mathrm T\mid X=0)+\mathrm P(X=1)\cdot\mathsf P(\mathrm T\mid X=1)+\mathrm P(X=2)\cdot\mathsf P(\mathrm T\mid X=2)
\\ & =(1-p)^2 \cdot q^3 + {2\choose 1}p(1-p)\cdot({3\choose 2}q^2(1-q)+ q^3) + p^2\cdot(1-(1-q)^3)
\end{align}$$
Simplify.

Next we form 3-member panels at random.
Let $N$ be the number of p-types in the 3-member panel.  If choosen at random this gives: $$\mathsf P(N=n) = \dfrac{{2\choose n}{4\choose 3-n}}{6\choose 3}, \qquad\text{for }n\in\{0,1,2\}$$
A three member panel passes you if 2 or 3 elect to pass you.
Letting $\rm S$ be the event of passing, then the probability that you will pass given $n$ p-types in such a panel is:
$$\begin{align}\mathsf P(\mathrm S \mid N=0) & = {3\choose 2}q^2(1-q)^1 + q^3
\\
\mathsf P(\mathrm S \mid N=1) & = (1-p)q^2 + p ({2\choose 1}q(1-q)^1 + q^2)
\\
\mathsf P(\mathrm S\mid N=2) & = {2\choose 1}(1-p)p\cdot q + p^2
\end{align}$$
Hence using the Law of Total Probability, the probability that you will pass with a random 3-member panel is: $$\mathsf P(\mathrm S) = \mathsf P(\mathrm S\mid N = 0)\mathsf P(N= 0) + \mathsf P(\mathrm S\mid N = 1)\mathsf P(N= 1) + \mathsf P(\mathrm S\mid N = 2)\mathsf P(N = 2)$$
Substitute and evaluate.

Compare and contrast.
