Given $7$ marbles in a box, $3$ are white and numbered from $1$ to $3$ and other $4$ are grey and numbered from $4$ to $7$. Calculate Probabilities Question: You have $7$ marbles in a box, $3$ are white and are numbered from $1$ to $3$ and the other $4$ are grey and are numbered from $4$ to $7$. You extract them one by one till you empty the box. Calculate the following probabilities:

*

*Probability that the first $3$ marbles extracted are white

*Probability that the first three marbles extracted are white and are also in the sequence $[1, 2, 3]$
My Approach
For the first part, I simulated $3$ extractions and I said: first extraction, I have $3$ white marbles out of $7$ total, second extraction I have $2$ out of $6$ and third extraction $1$ out of $5$, so:
$P=\tfrac 3 7\cdot \tfrac 2 6\cdot\tfrac 1 5$
Another idea was using the binomial coefficient: total ways I can extract $3$ marbles out of $7$ is $7\choose 3$, while the way I can extract $3$ white from $3$ total white is $3\choose 3$, so
$$P=\frac{{3\choose 3}}{\binom 7 3}$$
The second part took me some troubles. I thought to think this probability as
$\begin{align}P(\text{First 3 are white & in sequence [1,2,3]}) & ={{P(\text{First White and Number 1})}\\\cdot{P(\text{Second White and Number 2})}\\\cdot{P(\text{Third White and Number 3})}}\\[1ex] & =\tfrac 3 7\cdot\tfrac 1 3\cdot\tfrac 2 6\cdot\tfrac 1 3\cdot\tfrac 1 5\cdot\tfrac 1 3\end{align}$.
Are this calculations correct?
 A: From the first part, there are $3!$ sequences of having chosen all white balls, which are as follows: $$\begin{array}{ccc} 1 & 2 & 3\\ 1 & 3 & 2\\ 2 & 1 & 3\\ 2 & 3 & 1\\ 3 & 1 & 2\\ 3 & 2 & 1\end{array}$$
Out of this, only one sequence is in the order $1 2 3$, so, the probability is $$\frac{1}{6} \times P(\text{First Part}) = \frac{1}{6} \times \frac{6}{7\times 6 \times 5} = \frac{1}{210}$$ Hope it helps.
A: Your first calculation is entirely correct.
The second has the right idea but you evaluated the three probabilities incorrectly, and you should be using conditionals as the events are not independent.
Also note that number 1,2,3 are all white, so "white and number 1" is redundant.
$\begin{align}P(\text{drawn in sequence }[1,2,3]) & ={{P(1\text{ drawn first})}\\\cdot{P(2\text{ drawn second}\mid 1\text{ drawn first})}\\\cdot{P(3\text{ drawn third}\mid 1\text{ and }2\text{ first and second})}}\\[2ex] & =\bbox[white]{\color{white}{ \tfrac 1 7\cdot \tfrac 16\cdot\tfrac 15}}\end{align}$
A: Rohan's calucation is correct.  I'll try to explain how to arrive at the result using your own approach.  
Firstly, you could apply the reasoning you used for your 'simulations' in your first part:
In the first selection, there are 7 marbles, only one of which is White #1.
In the second selection, there are 6 marbles left, only one of which is White #2.
selection, there are 5 marbles left, only one of which is White #3.  
Therefore the probability is
$P=\frac{1}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}=\frac{1}{210}$
In your attempt at calculating this, you just made a couple of errors in not updating the number of white marbles remaining after each selection was made.  Your fourth factor should be $\frac{1}{2}$, since there are two white marbles remaining, and you are choosing one of them specifically.  Following this reasoning, your last factor should be simply $\frac{1}{1}$ or $1$, since there is only one white marble remaining at that point.
