A bag contains marbles of three types, x, y, and z. The total number of marbles is 27. The ratio of x to y marbles is 1:2, and the ratio of y to z marbles is 1:3. Find the number of z marbles and the

Let $N_x, N_y, N_z$ be the number of marbles of type $x, y, z$ respectively.
The total number of marbles is $N_x + N_y + N_z = 27$.
The given ratios are $N_x : N_y = 1 : 2$ and $N_y : N_z = 1 : 3$.
From $N_x : N_y = 1 : 2$, we have $N_y = 2N_x$, which means $N_x = \frac{1}{2}N_y$.
From $N_y : N_z = 1 : 3$, we have $N_z = 3N_y$.

Substitute $N_x = \frac{1}{2}N_y$ and $N_z = 3N_y$ into the total marbles equation:

Combine the terms:

Multiply both sides by $\frac{2}{9}$:

Now, calculate the number of $z$ marbles using $N_z = 3N_y$:

The number of $z$ marbles in the bag is $\boxed{\text{18}}$.

From part (a), the number of $y$ marbles is $N_y = 6$.
The total number of marbles in the bag is 27.

The probability of selecting a $y$ marble on the first draw is the number of $y$ marbles divided by the total number of marbles.

After one $y$ marble is selected and not replaced, there are $6 - 1 = 5$ $y$ marbles remaining.
The total number of marbles remaining in the bag is $27 - 1 = 26$.
The probability of selecting a second $y$ marble is:

The probability of selecting two $y$ marbles one after the other without replacement is the product of the probabilities from Step 2 and Step 3.

Simplify the fractions:

Further simplify by dividing the numerator and denominator by 2:

The probability is $\boxed{\frac{5}{117}}$.