A bag contains 15 red marbles, some blue marbles and some yellow marbles. A marble is chosen at random from the bag. The probability of choosing a blue marble is $\frac{1}{6}$ and the probability of choosing a yellow marble is $\frac{7}{12}$.

1. Find the probability of choosing a red marble.

Since the marble must be red, blue or yellow,
$P(\text{red})=1-P(\text{blue})-P(\text{yellow})$
$=1-\frac{1}{6}-\frac{7}{12}$
$=\frac{12}{12}-\frac{2}{12}-\frac{7}{12}=\frac{3}{12}=\frac{1}{4}.$

2. Use the red probability to determine the total number of marbles.

There are 15 red marbles, and these are $\frac{1}{4}$ of the total number of marbles.

Let the total number of marbles be $T$.
$\frac{15}{T}=\frac{1}{4}$
$T=15\times 4=60.$

1. Find the number of blue marbles in the bag.

From part (a), total number of marbles $=60$.

Number of blue marbles
$=\frac{1}{6}\times 60=10.$

2. Find the probability that both marbles chosen are blue.

Since the marbles are chosen without replacement,
$P(\text{both blue})=\frac{10}{60}\times\frac{9}{59}.$

$P(\text{both blue})=\frac{90}{3540}=\frac{3}{118}.$