Probability: best chance of picking a desired marble out of 10 following are the two questions I've made myself, but I need help in solving them.
1) Suppose there are 10 marbles in a box. One out of them is the desired marble, or you can say one is black others are white,etc etc. Case 1 : You pick 3 marbles altogether out of the box. What is the probability that you'll choose the black marble ? Case 2: You pick 3 marbles out, but one at a time. Now what is the probability of getting the black marble in those 3 attempts ?
2) Now suppose there are 10 marbles arranged linearly, each with position say 1,2....10. again, one is black & rest are white.You are to pick 3 marbles in order to have best chances of getting a black marble. Mathematically, is it better to pick 3 successive marbles,ie position 3,4,5 or would it be better to pick up randomly, like position 3,6,9 or it makes no difference whether you do it either way ??
Note: Although both questions can be put into one single question, it'd be better to solve it, & for me to understand it, if put up as two seperate questions.   
 A: The answer to both cases in your first question is $0.3$. There is no difference whether you pick the marbles all three at once or one by one. 
To convince you on this consider calculating the probability of the second case as:

We either draw the black marble on the first try, or the second try, or the third try.

Since these are disjoint events (only one can happen) we can just add their probabilities together. 


*

*What is the probability of drawing the black marble on the first try? It's simply $\frac1{10}$

*What is the probability of drawing the black marble on the second try? We have to not pick the black marble on the first try, and pick it on the second. That's $\frac9{10}\cdot \frac19 = \frac1{10}$

*What is the probability of drawing the black marble on the third try? We have to not pick the black marble on the first two tries, and pick it on the third. That's $\frac9{10}\cdot \frac8{9}\cdot \frac18 = \frac1{10}$


Total is $\frac3{10} = 0.3$
Perhaps the mistake you did was to calculate the probabilities like this:
$\frac1{10} + \frac9{10}\cdot \frac1{10} +\frac9{10}\cdot \frac9{10}\cdot \frac1{10} = 0.271$. I hope you can see now why this is a mistake. It's because we are not accounting for the removed marbles when we compute the probabilities for the second and third tries, and we treat these events as if there are still 10 marbles in the box. 
As far as the second question goes the answer is that it does not matter. Unless the marbles do not have a uniform probability to be the black one, order does not matter, and so which ones you pick does not matter too.
A: An intuitive explanation:
Picking "3 marbles altogether out of the box" is the same than randomly shuffling an ordered list of the ten marbles and keeping the three initial ones.
Now, if you shuffle such a list (with no bias of course), you expect all permutations of the list to share the very same probability. Then, the black marble can be at any position in the list with the same probability.
Thus probability for the black marble to be at the first position is 1/10, the same for all other positions. Which leads to the expected answer: the black marble is among the three initial ones with probability 3/10.
