Answer
$\dfrac{5}{21}$
Work Step by Step
Here, $A$ denotes that Suzan grabbed none the red marbles and $B$ denotes that Suzan 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 $1$ of them fluorescent pink marble and with no red marbles. This means that $n(A \cap B)=C(1,1) \times C(6,3)=20$
So, $P(A \cap B)=\dfrac{n(A \cap B)}{n(S)}=\dfrac{20}{210}$
and $n(B)=C(1,1) \times C(9,3)=84$
So, $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{84}{210}$
Thus, the equation (1) becomes:
$P(A~|~B)=\dfrac{P(A \cap B)}{P(B)} =\dfrac{20/210}{84/210}=\dfrac{5}{21}$
