If p = 0.6, what is the variance of the difference?

(Hint: Let D=X-Y, where we let X represent the number of Facebook friend requests, and let Y represent the number of LinkedIn connection requests. What are the possible values for X and Y, as well as D? What are their probabilities?)

Question

If p = 0.6, what is the variance of the difference?

(Hint: Let D=X-Y, where we let X represent the number of Facebook friend requests, and let Y represent the number of LinkedIn connection requests.  What are the possible values for X and Y, as well as D?  What are their probabilities?)

BlackTom AI · Accepted Answer

The question asks for the variance of the difference D = X − Y given p = 0.6, but the problem statement as provided does not include enough information about the random variables X and Y to compute the variance.

First, let's restate what is given and what is missing:
- Given: p = 0.6. This likely refers to a success probability in a Bernoulli or binomial context, but it is not specified whether p applies to X, Y, or both, nor the number of trials (n) for any binomial distributions.
- Missing: The distributions of X and Y (are they Bernoulli, Binomial with some n, or something else?), whether X and Y are independent, and the values of n (if binomial) or the exact support for X and Y. Without this, we cannot compute Var(D).

General approach if we had the usual setup (often encountered in these kinds of problems):
- Step 1: Identify the distributions of X and Y. If X ~ Binomial(n1, p) and Y ~ Binomial(n2, p) and X and Y are independent, then D = X − Y is a difference of two independent binomials with the same p.
- Step 2: Use the variance rule for independent sums/differences: Var(D) = Var(X) + Var(Y) − 2 Cov(X, Y). If X and Y are independent, Cov(X, Y) = 0, so Var(D) = Var(X) + Var(Y).
- Step 3: Compute Var(X) and Var(Y). For a Binomial(n, p), Var(X) = n p (1 − p). For Bernoulli trials (n = 1), Var(Bernoulli(p)) = p(1 − p).
- Step 4: Plug in the numbers. If both X and Y are Binomial with the same p and potentially different n, Var(D) = n1 p (1 − p) + n2 p (1 − p) = (n1 + n2) p (1 − p). If p = 0.6, then p(1 − p) = 0.6 × 0.4 = 0.24, so Var(D) would be 0.24 × (n1 + n2) under those independence assumptions.

What to do given the current info:
- If you can confirm the exact forms of X and Y (e.g., X ~ Binomial(n1, 0.6) and Y ~ Binomial(n2, 0.6) with X and Y independent), then compute Var(D) exactly using Var(D) = Var(X) + Var(Y) = n1 × 0.6 × 0.4 + n2 × 0.6 × 0.4 = 0.24 × (n1 + n2).
- If X and Y are not independent, you would need Cov(X, Y) to complete Var(D) = Var(X) + Var(Y) − 2 Cov(X, Y). Without Cov, you cannot determine Var(D).
- If X and Y are simply Bernoulli trials (n1 = n2 = 1) with the same p, then Var(D) would be Var(X) + Var(Y) = 0.24 + 0.24 = 0.48 if independent; if not independent, Cov could alter the result.

Because the problem as stated omits the necessary details (the exact distributions and their independence or dependence), any numeric answer like 0.21 cannot be verified or derived definitively from the given information alone. If you provide the missing pieces (the distributions of X and Y and their relationship), I can walk you through the exact calculation step by step and confirm the variance of D.

COMM_V 190 101-104 2025W1 FlexHW14.COMM190.2025.W1 (optional — see syllabus)

If p = 0.6, what is the variance of the difference? (Hint: Let D=X-Y, where we let X represent the number of Facebook friend requests, and let Y represent the number of LinkedIn connection requests. What are the possible values for X and Y, as well as D? What are their probabilities?)

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