An unbiased coin is tossed six times in a row and four different such trials are conducted. One trial implies six tosses of the coin. If H stands for head and T stands for tail, the following are the observations from the four trials:

(1) HTHTHT (2) TTHHHT (3) HTTHHT (4) HHHT__ __.

Which statement describing the last two coin tosses of the fourth trial has the highest probability of being correct?

The correct answer is: One H and one T will occur.

Let's analyze the problem step-by-step:

Since the coin is unbiased, each toss of the coin is an independent event. The outcomes of the first four tosses of the fourth trial (which are HHHT) do not affect the probabilities of the remaining two tosses. Therefore, we only need to look at the probabilities of the possible outcomes for the last two tosses.

For two independent coin tosses (where H stands for head and T stands for tail), the sample space of all possible outcomes is:

$S=\{HH,HT,TH,TT\}$

Since the coin is fair, each of these four outcomes is equally likely, having a probability of:
$P=\frac{1}{4}=0.25$

Now, let's calculate the probability of each statement in the options:

1. Two T will occur:
This event corresponds to the single outcome {TT}.
Probability = $\frac{1}{4}=0.25$

2. One H and one T will occur:
This event corresponds to the outcomes {HT, TH} (since order is not specified).
Probability = $\frac{2}{4}=0.50$

3. Two H will occur:
This event corresponds to the single outcome {HH}.
Probability = $\frac{1}{4}=0.25$

4. One H will be followed by one T:
This event specifies a strict sequence, corresponding only to the outcome {HT}.
Probability = $\frac{1}{4}=0.25$

Comparing these values, the statement "One H and one T will occur" has the highest probability of 0.50 (or 50%).