Probability
Binomial Distribution
Grade 12
Question:
<p>One hundred identical coins, each with probability \(p\) of showing up head, are tossed. If \(0 < p < 1\) and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins then the value of \(p\) is</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{49}{101}\)</p>
<p>(c) \(\frac{50}{101}\)</p>
<p>(d) \(\frac{51}{101}\)</p>
Step-by-Step Solution
Key Concept: The probability of getting exactly k heads follows a binomial distribution. The condition that P(X=50) > P(X=49) determines the range of p by analyzing the ratio of consecutive binomial probabilities.
<p><strong>Step 1:</strong> For X ~ Binomial(n=100, p), we have:</p><p>P(X=k) = C(100,k)p^k(1-p)^(100-k)</p><p><strong>Step 2:</strong> Form the ratio P(X=50)/P(X=49):</p><p>P(X=50)/P(X=49) = [C(100,50)p^50(1-p)^50]/[C(100,49)p^49(1-p)^51]</p><p><strong>Step 3:</strong> Simplify:</p><p>= [C(100,50)/C(100,49)] × [p/(1-p)]</p><p>= [100!/(50!×50!)] × [49!×51!/100!] × [p/(1-p)]</p><p>= [51/50] × [p/(1-p)]</p><p><strong>Step 4:</strong> Apply condition P(X=50) > P(X=49):</p><p>[51/50] × [p/(1-p)] > 1</p><p>51p > 50(1-p)</p><p>51p > 50 - 50p</p><p>101p > 50</p><p>p > 50/101</p><p><strong>Step 5:</strong> Combined with 0 < p < 1, we get:</p><p>50/101 < p < 1</p><p>∴ Answer: D</p>
Correct Answer: D