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?

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.
