A bag contains some green counters, some orange counters and some yellow counters.
The probability of picking a green counter at random is 0.2.
The probability of picking an orange counter at random is 0.1 more than the probability of choosing a yellow counter at random.

1. Represent the yellow and orange probabilities and use the total probability of all colours.

Let the probability of picking a yellow counter be $y$.

Then the probability of picking an orange counter is $y+0.1$.

Since the bag contains only green, orange and yellow counters,
$0.2+(y+0.1)+y=1$
$2y+0.3=1$
$2y=0.7$
$y=0.35$

1. Relate the number of yellow counters to the total number of counters using the yellow probability.

Let the total number of counters be $T$.

Since the probability of picking a yellow counter is $0.35$ and there are $14$ yellow counters,
$0.35T=14$

2. Solve for the total number of counters.

$T=\frac{14}{0.35}=40$

1. Find the original number of orange counters.

From part (a),
$P(\text{yellow})=0.35$
Since the probability of orange is $0.1$ more than the probability of yellow,
$P(\text{orange})=0.35+0.1=0.45$
From part (b), the total number of counters is $40$.
$\text{Number of orange counters}=0.45\times 40=18$

2. Update the total number of counters after removing 3 orange counters and find the new probability of yellow.

After removing $3$ orange counters,
$\text{new total number of counters}=40-3=37$
The number of yellow counters is still $14$.
$P(\text{yellow})=\frac{14}{37}$