Probability of a tough committee 
A committee of three judges is randomly selected from among ten judges. Four of the ten judges are tough; the committee is tough if at least two of the judges on the committee are tough.
A committee decides whether to approve petitions it receives. A tough committee approves 50% of petitions and a committee that is not tough approves 80% of petitions.
(a) Find the probability a committee is tough.
(b) Find the probability a petition is approved.
(c) Suppose a petition can be submitted many times until it is approved. If a petition is approved with probability 3/4 each time, what is the mean number of times it has to be submitted until it is approved?

For selection of tough committee, I approached taking (4c2 / 10 * 1/10) + 4c3 / 10 = 0.46. 
For B, I considered the below P(tough & approved) or P(not tough and approved) 0.46*0.50 + 0.80*(1-0.46)= 0.662.
I dont know if this is the right approacha dnim stuck at the third question
 A: A) Well the total number of committees is$\ 10C3$ and the number of 'tough' committees is$\ (4C2)*(6C1) + (4C3)$ so therefore the probability of a tough committee is$\ (4C2*6+4C3)/(10C3) = 1/3 $
B) There is a 1/3 chance the comittee is tough, which correlates to a 50% chance of approval, and a 2/3 chance of a not tough comittee, which leads to a 80% chance of approval.
Hence, the probability of approval is $\ (1/3)*(1/2) + (2/3)*(4/5)=70%$
A: 

A committee of three judges is randomly selected from among ten judges. Four of the ten judges are tough; the committee is tough if at least two of the judges on the committee are tough. A committee decides whether to approve petitions it receives. A tough committee approves 50% of petitions and a committee that is not tough approves 80% of petitions.
(a) Find the probability a committee is tough.

For selection of tough committee, I approached taking (4c2 / 10 * 1/10) + 4c3 / 10 = 0.46. 

Think again. You want the probability for selecting either: two from four tough and one from six soft judges, or three from four tough judges, when selecting any three from all ten judges.
$$\dfrac{{}^4\mathrm C_2\cdot \phantom{{}^6\mathrm C_1}+{}^4\mathrm C_3 \cdot\phantom{{}^6\mathrm C_0}}{{}^{10}\mathrm C_3} = \dfrac{\phantom{76}}{120} = \phantom{0.6\dot3}$$

For B, I considered the below P(tough & approved) or P(not tough and approved) 0.46*0.50 + 0.80*(1-0.46)= 0.662.

Yes.   That is the correct approach, just not the correct numbers.
