The expression $\frac{\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)+\left(\genfrac{}{}{0ex}{}{2}{2}\right)}{\left(\genfrac{}{}{0ex}{}{9}{2}\right)}$ represents the probability that both drawn tiles are the same color, because:
- $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ counts ways to draw 2 red tiles,
- $\left(\genfrac{}{}{0ex}{}{3}{2}\right)$ counts ways to draw 2 blue tiles,
- $\left(\genfrac{}{}{0ex}{}{2}{2}\right)$ counts ways to draw 2 green tiles,
- and $\left(\genfrac{}{}{0ex}{}{9}{2}\right)$ counts all possible ways to draw 2 tiles from 9.

The event "two tiles are different colors" is the complement of "two tiles are the same color." Therefore,

Comparing this with

shows that $k=1$.

So the correct answer is $\boxed{1}$.

Let $E$ be the event that a sample came from Region A or is contaminated. The question asks for

First, count the samples in the numerator: Region A and not contaminated = $280$.

Next, count the samples in the given condition $E$.
- Region A total: $400$
- Contaminated total: $120+180=300$
- Region A and contaminated: $120$

Because samples that are both in Region A and contaminated were counted twice, use inclusion-exclusion:

So the conditional probability is

Therefore, the correct answer is $\frac{14}{29}$.

Let the number of blue tiles originally be $b$. Since probability is proportional to the number of tiles, the number of red tiles is $3b$ and the number of green tiles is $6b$. The original total is therefore $b+3b+6b=10b$. After 8 green tiles are added, the number of green tiles becomes $6b+8$, and the new total becomes $10b+8$. We are told that the new probability of green is $\frac{2}{3}$, so

Cross-multiply:

So the bag originally had 4 blue tiles, 12 red tiles, and 24 green tiles. Therefore the statement that must be true is that the bag contained 4 blue tiles before the 8 green tiles were added.

Use conditional probability. First find the probability that a randomly selected part is defective:

Given $P(A)=0.60$, $P(B)=0.40$, $P(D\mid A)=0.04$, and $P(D\mid B)=0.07$,

Now find the probability that a part came from Supplier B given that it is defective:

So the probability is approximately $0.54$.

To get two marbles of different colors, there are two possible orders: red then blue, or blue then red.

The probability of drawing a red marble first and then a blue marble is

The probability of drawing a blue marble first and then a red marble is

These two probabilities are equal, so the total probability is

Therefore, the equivalent expression is $2\left(\frac{5}{12}\cdot \frac{7}{11}\right)$.