Final answer:
To find the expected value, E(X), of the number of times the sequence HT appears when tossing a fair coin 10 times, we can use indicator random variables. Each indicator random variable represents whether the sequence HT appears starting at a specific position. The expected value of each indicator random variable is 0.25, and thus the overall expected value is 2.25.
Step-by-step explanation:
To find the expected value, E(X), of the number of times the sequence HT appears when you toss a fair coin 10 times, we can use indicator random variables. Let's define an indicator random variable, Xi, for each position i from 1 to 9, which takes the value 1 if the sequence HT appears starting at position i and 0 otherwise.
The expected value E(X) can then be calculated as the sum of the expected values of these indicator random variables:
E(X) = E(X1) + E(X2) + E(X3) + ... + E(X9).
Since each of these indicator random variables follows a Bernoulli distribution with a probability of success p = 0.25 (since there are 4 possible outcomes for each coin toss and we're interested in the sequence HT), the expected value of each indicator random variable is:
E(Xi) = p = 0.25.
Therefore, the expected value E(X) is:
E(X) = E(X1) + E(X2) + E(X3) + ... + E(X9) = 0.25 + 0.25 + 0.25 + ... + 0.25 = 9 * 0.25 = 2.25.