This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.
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Step 1: Define Expectation of a Random Variable. a) The Expectation* (or Expected Value) of a random variable is the weighted average of all possible values that the random variable can take. The weights are the probabilities of each value occurring. It represents the long-run average value of the random variable if the experiment is repeated many times. For a discrete random variable X, it is given by E[X] = x P(X=x). Step 2: State the Properties of Expectation of a Random Variable. b)* The key properties of the expectation of a random variable are: 1. Expectation of a Constant: If c is a constant, then E[c] = c. 2. Linearity (Constant Multiple): If c is a constant and X is a random variable, then E[cX] = cE[X]. 3. Linearity (Sum/Difference): If X and Y are random variables, then E[X ± Y] = E[X] ± E[Y]. 4. Expectation of a Sum of Independent Random Variables: If X_1, X_2, , X_n are random variables, then E[_i=1^n X_i] = _i=1^n E[X_i]. This property holds even if the random variables are not independent. 5. Expectation of a Product of Independent Random Variables: If X and Y are independent random variables, then E[XY] = E[X]E[Y]. Step 3: Construct the Probability Distribution Table and find the Expected Value of X. c)* Let S be the sample space for tossing a fair coin twice. The possible outcomes are: S = \HH, HT, TH, TT\ Each outcome has a probability of (1)/(4) since the coin is fair. The random variable X is defined such that X(e) = 0 if e is T (tail) and X(e) = 1 if e is H (head). This means X counts the number of heads in the two tosses. Let's map the outcomes to the values of X: HH X = 1+1 = 2 HT X = 1+0 = 1 TH X = 0+1 = 1 TT X = 0+0 = 0 The possible values for X are \0, 1, 2\. Now, we find the probability for each value of X: P(X=0): Occurs for outcome TT. So, P(X=0) = (1)/(4). P(X=1): Occurs for outcomes HT and TH. So, P(X=1) = (1)/(4) + (1)/(4) = (2)/(4) = (1)/(2). P(X=2): Occurs for outcome HH. So, P(X=2) = (1)/(4). The Probability Distribution Table for X is: |c|c|c|c| x & 0 & 1 & 2 \\ P(X=x) & (1)/(4) & (1)/(2) & (1)/(4) \\ Now, we calculate the Expected Value of X, E[X]: E[X] = x P(X=x) E[X] = (0 × P(X=0)) + (1 × P(X=1)) + (2 × P(X=2)) E[X] = (0 × (1)/(4)) + (1 × (1)/(2)) + (2 × (1)/(4)) E[X] = 0 + (1)/(2) + (2)/(4) E[X] = 0 + (1)/(2) + (1)/(2) E[X] = 1 The Expected Value of X is 1. Last free one today — make it count tomorrow, or type /upgrade for unlimited.