Let's solve this problem step by step.

Step 1: Total number of counters

There are 3 green, 1 yellow, and 6 blue counters. So, the total number of counters is:

$3 + 1 + 6 = 10$

Step 2: Probability of picking the same color

To pick two counters of the same color, there are three possible scenarios:

• Both counters are green
• Both counters are yellow
• Both counters are blue

Since the counter is replaced after each pick, the events are independent, so we can calculate the probabilities as follows:

Probability of both being green:

$P(\text{green, green}) = \left(\frac{3}{10}\right) \times \left(\frac{3}{10}\right) = \frac{9}{100}$

Probability of both being yellow:

$P(\text{yellow, yellow}) = \left(\frac{1}{10}\right) \times \left(\frac{1}{10}\right) = \frac{1}{100}$

Probability of both being blue:

$P(\text{blue, blue}) = \left(\frac{6}{10}\right) \times \left(\frac{6}{10}\right) = \frac{36}{100}$

Step 3: Total probability of picking the same color

Now, we sum the probabilities for each case:

$P(\text{same color}) = \frac{9}{100} + \frac{1}{100} + \frac{36}{100} = \frac{46}{100}$

Step 4: Simplify the fraction

$\frac{46}{100} = \frac{23}{50}$

So, the probability that the counters chosen are of the same color is:

$\boxed{\frac{23}{50}}$

Would you like any further details or have any questions?

Here are 5 related questions for further practice:

1. What is the probability of selecting two different-colored counters?
2. How would the probability change if the counters were not replaced after each pick?
3. What is the probability of picking two counters, both of which are either green or blue?
4. How would the probability change if there were 4 yellow counters instead of 1?
5. What if three counters are picked with replacement? What is the probability that all three are blue?

Tip: When dealing with probability, always double-check whether the events are independent (with replacement) or dependent (without replacement).