A fair coin is tossed 11 times in succession. What is the probability that only the first two tosses result in heads and all of the remaining nine tosses result in tails?

Introduction / Context:
This question involves repeated independent trials of a fair coin and a very specific pattern of outcomes. You toss a fair coin 11 times and want the probability that the first two tosses are heads while every one of the remaining nine tosses is tails. Problems like this test your ability to work with independent events and to translate a pattern into a product of probabilities.

Given Data / Assumptions:

• A fair coin is tossed 11 times.
• Each toss is independent of the others.
• Probability of head on a single toss = 1/2.
• Probability of tail on a single toss = 1/2.
• We require the exact pattern: H, H, T, T, T, T, T, T, T, T, T.

Concept / Approach:
Because all tosses are independent, the probability of a particular sequence of heads and tails is the product of the probabilities for each toss. For each head, we multiply by 1/2, and for each tail, we also multiply by 1/2. Since the pattern is completely specified (no variations allowed), we do not multiply by any combinatorial factor such as a binomial coefficient.

Step-by-Step Solution:
Step 1: Probability that the first toss is a head = 1/2. Step 2: Probability that the second toss is a head = 1/2. Step 3: For each of the remaining nine tosses, the probability of tail on a single toss = 1/2. Step 4: Probability that all nine remaining tosses are tails = (1/2)^9. Step 5: Because tosses are independent, multiply all probabilities. Overall probability = (1/2) * (1/2) * (1/2)^9. Step 6: Combine exponents: (1/2) * (1/2) * (1/2)^9 = (1/2)^(1 + 1 + 9) = (1/2)^11.

Verification / Alternative check:
The total number of possible sequences of heads and tails in 11 tosses is 2^11, since each toss has 2 outcomes and they are independent. For this specific event, only one of those 2^11 sequences satisfies the condition of heads only on the first two tosses and tails everywhere else. Thus the probability of this single sequence among all equally likely sequences is 1 / 2^11, which is exactly (1/2)^11. This is another way to see the same result.

Why Other Options Are Wrong:
Option (1/2)^10 would correspond to a sequence involving 10 specified outcomes instead of 11 and ignores one toss. Option (1/2)^9 and (1/2)^8 also incorrectly count fewer tosses in the exponent. Option 11*(1/2)^11 would be correct for the probability of exactly two heads occurring anywhere in the 11 tosses, but here the positions of the heads are fixed to the first two tosses, so no combinatorial factor of 11 choose 2 is needed. Hence, only (1/2)^11 matches the exact event description.

Common Pitfalls:
Students often confuse the probability of exactly k heads in n tosses with the probability of a particular arrangement of outcomes. For a specific pattern, you use simple multiplication of probabilities with no extra combinatorial factor. Another pitfall is to misread only the first two tosses will yield heads and think that later tosses could also be heads, which would completely change the event. Always make sure you understand whether the pattern is exact or just a count of successes with no regard to order.

Final Answer:
The probability that only the first two tosses yield heads and all remaining tosses are tails is (1/2)^11.