Answer
$\dfrac{1}{21}$
Work Step by Step
Here, $A$ denotes that Suzan grabbed all the red marbles and $B$ denotes that Suzan did not grabbed all the fluorescent pink marbles.
Our aim is to calculate the conditional probability $P(A|B)$.
This can be found as: $P(A|B)=\dfrac{P(A \cap B)}{P(B)} ~~~~(1)$
We can see that $A \cap B$ shows of all outcomes that the sets of marbles of which $3$ of them are red and with no fluorescent pink marble. This means that $n(A \cap B)=C(3 , 3) \times C(6,1)=6$
So, $P(A \cap B)=\dfrac{n(A \cap B)}{n(S)}=\dfrac{6}{210}$
and $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{126}{210}$
Thus, the equation (1) becomes:
$P(A~|~B)=\dfrac{P(A \cap B)}{P(B)} =\dfrac{6/210}{126/210}=\dfrac{1}{21}$