This is a four-part question dressed up as one, which I did not fully appreciate until I was halfway through the variance calculation and realized I hadn't even thought about the adjacent pairs part yet.
First, identify the distributions: Option A is Binomial(100, 0.04) and Option B is deterministic with exactly 4 ads. Compute expected values and variances using standard formulas. Then, for Option A, approximate the probability of exceeding 8 ads using normal approximation with continuity correction; for Option B, the probability is 0. Finally, compute the expected number of adjacent ad pairs for each option using linearity of expectation and compare the results to assess which option is more likely to produce long runs.
Pro tip: When comparing ad placement strategies, emphasize that while both have the same expected number of ads, the variance and clustering behavior differ significantly, which impacts user experience and ad effectiveness. This shows you understand the trade-offs beyond just the mean.
Recognize that Option A follows a Binomial distribution with n=100 and p=0.04, while Option B is deterministic with exactly 4 ads (one per 25 posts).
For Option A, use E[X]=np=4 and Var(X)=np(1-p)=3.84. For Option B, E[X]=4 and Var(X)=0.
For Option A, find P(X > 8) using normal approximation with continuity correction: P(X ≥ 9) ≈ 1 - Φ((8.5-4)/√3.84) ≈ 0.011. For Option B, P(X > 8) = 0.
For Option A, each adjacent pair has probability p^2=0.0016, so expected pairs = 99 * 0.0016 = 0.1584. For Option B, ads are placed at positions 25, 50, 75, 100, so there are no adjacent ads; expected pairs = 0.
Option A has a small chance of exceeding 8 ads and can produce adjacent ads, while Option B never exceeds 4 ads and has no adjacent ads. Thus, Option A is more likely to produce long runs of consecutive ads, though the probability is low.
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