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

Introduction:
This probability question involves repeated independent tosses of a fair coin. You are asked to find the probability of a very specific outcome sequence: heads on the first two tosses and tails on all remaining tosses. Unlike problems that ask for "exactly two heads anywhere", this problem fixes the positions of heads and tails, leading to a straightforward calculation.

Given Data / Assumptions:

• The coin is fair, so P(Head) = 1/2 and P(Tail) = 1/2 on each toss.
• The coin is tossed 11 times.
• We require the event: toss 1 = H, toss 2 = H, tosses 3 to 11 = T.
• All tosses are independent.

Concept / Approach:
When outcomes are fixed in a specific order and the tosses are independent, the probability of the combined event is the product of the probabilities of each individual event. Since each toss has probability 1/2 of matching the required outcome (either H or T), the probability of any particular 11-toss sequence is (1/2)^11.

Step-by-Step Solution:
We want the specific sequence: H, H, T, T, T, T, T, T, T, T, T.Probability that toss 1 is H = 1/2.Probability that toss 2 is H = 1/2.For each of tosses 3 to 11, we need T, each with probability 1/2.There are 2 heads required and 9 tails required.Total number of tosses = 11.Because the tosses are independent, multiply probabilities:P(required sequence) = (1/2) * (1/2) * (1/2) * ... (11 factors) = (1/2)^11.

Verification / Alternative check:
Note that every exact 11-toss sequence (like HHTTTTTTTTT, THTHTHTHTHT, etc.) has the same probability (1/2)^11, because for each of the 11 tosses, there are 2 equally likely outcomes. The sequence specified in the question is just one of those many sequences, so its probability must be (1/2)^11, consistent with this reasoning.

Why Other Options Are Wrong:
9(1/2) is not even a probability (it is greater than 1). (11C2)(1/2)^9 would correspond to exactly two heads anywhere among 11 tosses, without fixing their positions, which is a different event with many more sequences. 1/2 and (1/2)^9 also refer to wrong events or incorrect exponent counts and do not represent the probability of this specific 11-toss sequence.

Common Pitfalls:
A common mistake is to treat this as a problem of "exactly two heads in 11 tosses" and introduce combinations (11C2), which overcounts the allowed sequences. Here the heads must occur specifically on the first two tosses, so there is exactly one favourable sequence. Always read carefully to see whether positions are fixed or not.

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