<p>A fair coin is tossed 99 times. Let \(X\) be the number of times heads occurs. Then \(P(X = r)\) is maximum when \(r\) is</p>
Step-by-Step Solution
Key Concept: For a binomial distribution with parameters n and p=0.5, the probability P(X=r) is maximized when r is closest to the mode, which equals ⌊(n+1)p⌋ when (n+1)p is not an integer. Here with n=99 and p=0.5, the mode is ⌊50⌋=50.
<p><strong>Step 1:</strong> For X ~ Binomial(n=99, p=0.5), we use the ratio test to find where P(X=r) is maximum.</p><p><strong>Step 2:</strong> Consider the ratio: P(X=r)/P(X=r-1) = [C(99,r)/C(99,r-1)] = (99-r+1)/r = (100-r)/r</p><p><strong>Step 3:</strong> P(X=r) > P(X=r-1) when (100-r)/r > 1, which gives 100-r > r, so r < 50.</p><p><strong>Step 4:</strong> P(X=r) > P(X=r+1) when (100-r)/r > (99-r)/(r+1), which simplifies to (100-r)(r+1) > r(99-r), giving r+1 > 50, so r ≥ 50.</p><p><strong>Step 5:</strong> Since P(X=r) increases for r < 50 and decreases for r > 50, the maximum occurs at r = 50.</p><p>∴ Answer: D (r = 50)</p>
Correct Answer: D