9 fair coins are independently tossed in a row. Let X be the random variable denoting the number of instances in which a Head is immediately followed by a Tail during these 9 tosses. The variance of X has the form ab, where a and b are coprime integers. What is the value of a+b?

Details and assumptions
Note: If we have 5 coins which are tossed as (HT)T(HT), there are 2 instances in which a Head is immediately followed by a Tail; they have been grouped.

To find the variance of the random variable X, which represents the number of instances in which a head is immediately followed by a tail, we need to consider the possible outcomes of each coin toss and calculate the probabilities.

Let's break down the problem step by step:

1. First, let's consider the possible outcomes of a single coin toss. There are 2 possibilities - either heads (H) or tails (T). Since we are only interested in the number of instances where a head is immediately followed by a tail, the outcomes we are interested in are: {HT} (Head followed by Tail) or {HH, TT} (Head followed by Head or Tail followed by Tail).

2. Now, let's consider the possible outcomes for 2 consecutive coin tosses. There are 4 possible outcomes: {HH, HT, TT, TH}. Again, we are only interested in the number of instances where a head is immediately followed by a tail, so the outcomes we are interested in are {HT} (1 instance) or {HH, TT} (0 instances).

3. To generalize this, we consider the probability of getting a head followed by a tail in any two consecutive tosses. This probability is 1/4 since only one of the 4 possible outcomes satisfies our condition.

4. Now, let's consider the possible outcomes for 3 consecutive coin tosses. There are 8 possible outcomes: {HHH, HHT, HTH, HTT, TTT, TTH, THT, THH}. Again, we are only interested in the number of instances where a head is immediately followed by a tail, so the outcomes we are interested in are {HTT, HTH} (1 instance) or {HHH, HHT, HTT, TTT, TTH, THT, THH} (0 instances).

5. To generalize this, we consider the probability of getting a head followed by a tail in any three consecutive tosses. This probability is 1/8 since only two of the 8 possible outcomes satisfy our condition.

6. Following this pattern, we can deduce that the probability of a head followed by a tail in any n consecutive tosses is 1/(2^n).

7. Now, to calculate the variance of X, we need to sum up the probabilities multiplied by the squared differences from the mean. The mean of X is found by multiplying the probability of a head followed by a tail in any two consecutive tosses (1/4) by the number of two consecutive tosses sequences in 9 coin tosses, which is (9 - 1) = 8.

8. Using the formula for variance:

Variance(X) = Σ [ (x - μ)^2 * P(x) ]

where x takes values from 0 to 8 (in this case) and P(x) is the probability of X taking on the value x.

Variance(X) = (0 - μ)^2 * P(0) + (1 - μ)^2 * P(1) + ... + (8 - μ)^2 * P(8)

9. Substituting the values, we have:

Variance(X) = (0 - 8)^2 * (1/(2^2)) + (1 - 8)^2 * (1/(2^3)) + ... + (8 - 8)^2 * (1/(2^9))

Simplifying, we get:

Variance(X) = 64 * (1/4) + 49 * (1/8) + ... + 0 * (1/512)

10. Calculating the sum of the above expression, we get:

Variance(X) = 272/8 = 34/1 = 34

11. The variance of X is 34, which means a = 34 and b = 1. Therefore, the value of a + b = 34 + 1 = 35.

So, the value of a + b is 35.