Probability of winning a basketball championship? **It is felt that the probabilities are $.2, .4, .3, .10$ that the b-ball teams of four Universities 
$T , U ,V ,W$ will win their championship. 
If University $U$ is placed on probation and declared ineligible for the championship,  what is probability university T will win the championship?** 
Thoughts
So the task here is to find probability of $T$.
Now if $U$ is removed that means the chances of the other teams winning increase.
So would you divide $40$ by $3$ and then add that average to the other universities to solve it.
 A: Lets talk marbles (classic I know)
we have 2 blue, 4 red 3 white and finally 1 black and then we compute probabilities of picking either colour then we have 
$$
P(B) = 0.2\\
P(R) = 0.4\\
P(W) = 0.3\\
P(Blk) = 0.1
$$
now if we remove say all the white balls then we have what?
A: You are probably expected to scale up the other probabilities by multiplying by a common factor so that the sum is $1$.  In essence, you distribute the fraction of the time that $U$ wins in proportion to the chances that each other one wins.  This rests on many unstated assumptions, but it is the best I can do.  If the probability that $T$ wins is originally $0$, why should it jump to $\frac {.4}3$?  That is what happens in your proposed computation, but maybe $T$ is totally inept and can't win a game. On the other hand, maybe the probation happens after the games are played and if $U$ won they flip a three sided coin. Then you are correct.
A: probabilities always add up to 1
 you need to have 1-.40=.60
 T=.20/.60=.3333, V=.30/.60=.5, W=.10/.60=.16666
when you add T+V+W=1
A: There are two equivalent ways of looking it.
The first is to recalculate the probabilities according to the reduced probability space, as per Chinny84's answer.
You get the same result if you divide the $0.40$ extra into $6$ parts, and reassigned them as per the current ratio.
So, each part is worth $\frac1{15}$, University T gets $2$ parts, University V gets $3$ parts and University W gets $1$ part.
We can see this gives the same results: $\dfrac2{10}+\dfrac2{15}=\dfrac13$, $\dfrac3{10}+\dfrac3{15}=\dfrac12$ and $\dfrac1{10}+\dfrac1{15}=\dfrac16$.
This is because, say the eliminated component is worth $e$ points, so the remaining components add to $1-e$ points. Say the remaining components are $c_1,\dots,c_k$, each probability is now $\dfrac{c_i}{1-e}$ by Chinny84's reasoning.
If we split the $e$ points into $1-e$ parts, and reassign proportionately, we get: $$c_i+\dfrac{c_i\dot e}{1-e}=c_i(1+\dfrac{e}{1-e})=\dfrac{c_i}{1-e}$$
