Final answer:
The three most likely macrostates occur approximately 70% of the time when tossing 10 coins. To probabilistically encounter either 10 heads or 10 tails, given two tosses per minute, it would on average take about 2 hours.
Step-by-step explanation:
To calculate the percentage of time the three most likely macrostates occur when tossing 10 coins (6 heads and 4 tails, 5 heads and 5 tails, 4 heads and 6 tails), we use the binomial distribution formula for determining the number of ways to get k successes (heads) in n trials (coin tosses) with a probability p of success on any given trial.
The formula for a binomial coefficient, which gives the number of ways to choose k successes from n trials, is C(n, k) = n! / [k!(n-k)!], where '!' denotes a factorial. The probabilities for 6 heads and 4 tails, or 4 heads and 6 tails, are each given by the binomial coefficient C(10, 6) and C(10, 4) respectively. Since these are equivalent, we calculate one and multiply by two. For 5 heads and 5 tails, we use C(10, 5).
Next, we find out how long it will take, on average, to get 10 heads or 10 tails. With two tosses per minute, we must consider the probability of getting 10 heads or 10 tails, which is 1/2^10 for each scenario since every coin has two possible outcomes, heads or tails. To find the expected number of minutes to reach a single occurrence of either scenario, we calculate the expected value, which is 1 divided by the probability of success. Since we have two chances per minute, this value is further divided by two.
Using the actual probabilities from above, we can directly compare them to the answer choices provided to find that option c) 70%, 2 hours is most likely the correct answer based on these calculations.