Let the original total number of chips be $T$. Since the probability of selecting a red chip is $\frac{3}{5}$, the number of red chips was $\frac{3}{5}T$, and the number of blue chips was $\frac{2}{5}T$.

After 4 red chips are removed, the number of red chips becomes $\frac{3}{5}T-4$, and the total number of chips becomes $T-4$. The new probability of selecting a red chip is $\frac{1}{2}$, so

Cross-multiply:

Simplify:

Subtract $T$ from both sides:

Add 8 to both sides:

Multiply by 5:

So, the bag originally contained $20$ chips.

There are $4+3+5=12$ marbles in the bag. A marble that is not green must be either red or blue, so there are $4+3=7$ marbles that are not green. Thus, the probability of choosing a marble that is not green on one draw is $\frac{7}{12}$. Because the marble is returned each time, the 3 draws are independent and the probability stays the same on each draw. Therefore, the probability that all 3 chosen marbles are not green is

So the correct answer is ${\left(\frac{7}{12}\right)}^3$.

The question asks for the probability that a bulb is fluorescent given that it is defective. This is a conditional probability, so the sample space is only the defective bulbs.

From the table, there are 30 defective bulbs total, and 12 of those are fluorescent. Therefore,

So the correct answer is $\frac{2}{5}$.

Let the original total number of beads be $T$. Since the probability of selecting a blue bead was $\frac{3}{5}$, the original number of blue beads was $\frac{3}{5}T$, and the original number of green beads was $\frac{2}{5}T$.

After 4 blue and 6 green beads are added, the new total is $T+10$, and the new number of blue beads is $\frac{3}{5}T+4$. We are told the new probability of selecting a blue bead is $\frac{7}{11}$, so

Now solve:

Multiply both sides by 5:

So the original bag contained 15 beads.

Because the dart is equally likely to land anywhere on the board, the probability is the ratio of the area of the bull's-eye to the area of the entire dartboard. The bull's-eye has area $\pi (2{)}^2=4\pi$. The whole board has area $\pi (10{)}^2=100\pi$. So the probability is $\frac{4\pi }{100\pi }=\frac{4}{100}=\frac{1}{25}$. Therefore, the correct answer is $\frac{1}{25}$.