Question

Let \( X \) be the number of heads obtained in a sequence of 10 independent tosses of a fair coin. The fair coin is tossed again \( X \) number of times independently, and let \( Y \) be the number of heads obtained in these \( X \) number of tosses. Then \( E(X + 2Y) \) equals ............

Hint:

When calculating the expected value of a sum of random variables, use the linearity of expectation: \( E(X + Y) = E(X) + E(Y) \).

Answer 1

Correct Answer: 10

Step 1: Find \( E(X) \).
Since \( X \) is the number of heads in 10 independent tosses of a fair coin, \( X \) follows a binomial distribution with parameters \( n = 10 \) and \( p = 0.5 \). The expected value of \( X \) is: \[ E(X) = n \times p = 10 \times 0.5 = 5. \] Step 2: Find \( E(Y \mid X) \).
Given that \( X = x \), the random variable \( Y \) represents the number of heads in \( x \) independent tosses of a fair coin. Thus, \( Y \mid X = x \) follows a binomial distribution with parameters \( n = x \) and \( p = 0.5 \). The expected value of \( Y \mid X = x \) is: \[ E(Y \mid X = x) = x \times 0.5. \] Thus, the unconditional expected value of \( Y \) is: \[ E(Y) = E(E(Y \mid X)) = E\left(\frac{X}{2}\right) = \frac{1}{2} E(X) = \frac{1}{2} \times 5 = 2.5. \] Step 3: Find \( E(X + 2Y) \).
Now, we can compute \( E(X + 2Y) \): \[ E(X + 2Y) = E(X) + 2E(Y) = 5 + 2 \times 2.5 = 5 + 5 = 10. \] Final Answer: \[ \boxed{10}. \]