<p>An ordinary coin is tossed <em>n</em> times. If the probability of getting 7 heads is equal to that of 9 heads then the probability of getting 2 heads is</p>
<p>A. \(\dfrac{15}{2^{13}}\)</p>
<p>B. \(\dfrac{2}{15}\)</p>
<p>C. \(\dfrac{15}{2^{8}}\)</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: Use the binomial probability formula P(X=k) = C(n,k)·(1/2)^n. Since P(7 heads) = P(9 heads), we get C(n,7) = C(n,9), which implies n = 16. Then calculate P(2 heads) using n=16.
<p><strong>Step 1:</strong> Write binomial probabilities. P(7 heads in n tosses) = C(n,7)·(1/2)^n and P(9 heads) = C(n,9)·(1/2)^n</p><p><strong>Step 2:</strong> Set them equal: C(n,7)·(1/2)^n = C(n,9)·(1/2)^n, which gives C(n,7) = C(n,9)</p><p><strong>Step 3:</strong> Use the property that C(n,r) = C(n,s) implies r+s = n. Therefore 7+9 = n, so <strong>n = 16</strong></p><p><strong>Step 4:</strong> Calculate P(2 heads): P(2) = C(16,2)·(1/2)^16 = (16·15/2)·(1/2)^16 = 120·(1/2)^16 = <strong>120/2^16 = 15/2^12</strong></p><p>∴ Answer: A</p>
Correct Answer: A