Bag C starts with [math]7[/math] blue counters.

We want Bag C to end up with an even number of blue counters.

This means the counter transferred from Bag B to Bag C must be blue, because:

[math]7 + 1 = 8 (even)[/math]

So we need the probability that a blue counter is transferred from Bag B to Bag C.

Case 1: A blue counter is moved from Bag A to Bag B.

Probability of choosing a blue counter from Bag A:

[math]\frac{3}{10}[/math]

Bag B now contains [math]7[/math] blue counters and [math]5[/math] yellow counters, so the probability of choosing a blue counter from Bag B is:

[math]\frac{7}{12}[/math]

Combined probability:

[math]\frac{3}{10} \times \frac{7}{12} = \frac{21}{120}[/math]

Case 2: A yellow counter is moved from Bag A to Bag B.

Probability of choosing a yellow counter from Bag A:

[math]\frac{7}{10}[/math]

Bag B now contains [math]6[/math] blue counters and [math]6[/math] yellow counters, so the probability of choosing a blue counter from Bag B is:

[math]\frac{6}{12}[/math]

Combined probability:

[math]\frac{7}{10} \times \frac{6}{12} = \frac{42}{120}[/math]

Add the probabilities:

[math]\frac{21}{120} + \frac{42}{120} = \frac{63}{120} [/math]

Therefore, the probability that Bag C ends up with an even number of blue counters is:

[math]\frac{63}{120}[/math]