In a meeting 60\% of the members favour and 40\% oppose a certain proposal. A member is selected at random and we take $X = 0$ if he opposed and $X = 1$ if he is in favour, then $\text{Var}(X) =$
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Anytime a random variable is strictly defined as $1$ for an event happening and $0$ for it not happening, it is an Indicator (or Bernoulli) variable. You can bypass tables and immediately use $\text{Mean} = p$ and $\text{Variance} = p(1-p)$.
Step 1: Understanding the Question:
We are dealing with a discrete random variable $X$ that can only take two values: $1$ (success/favour) and $0$ (failure/oppose). We need to calculate its variance.
Step 2: Key Formula or Approach:
This scenario perfectly models a Bernoulli distribution, where a single trial has two possible outcomes.
For a Bernoulli random variable:
Probability of success ($X=1$) is $p$.
Probability of failure ($X=0$) is $q = 1 - p$.
The formula for the variance of a Bernoulli distribution is simply: $\text{Var}(X) = p \times q$.
Alternatively, use the general variance formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$.
Step 3: Detailed Explanation:
From the problem description:
Probability of favouring, $p = P(X=1) = 60\% = 0.6$.
Probability of opposing, $q = P(X=0) = 40\% = 0.4$.
Using the Bernoulli variance shortcut:
$$\text{Var}(X) = p \times q = 0.6 \times 0.4 = 0.24$$
Let's verify with the general expectation formula to be sure:
$E(X) = \sum x_i p_i = (0 \times 0.4) + (1 \times 0.6) = 0.6$.
$E(X^2) = \sum x_i^2 p_i = (0^2 \times 0.4) + (1^2 \times 0.6) = 0.6$.
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$
$$\text{Var}(X) = 0.6 - (0.6)^2$$
$$\text{Var}(X) = 0.6 - 0.36 = 0.24$$
Step 4: Final Answer:
The variance of $X$ is $0.24$, matching option (B).
