<p>A die is thrown 7 times. What is the chance that an odd number turns up (i) exactly 4 times, (ii) at least 4 times?</p>
Step-by-Step Solution
Key Concept: This is a binomial probability problem where n=7 trials, p=1/2 (probability of odd number), and we need P(X=4) for part (i) and P(X≥4) for part (ii). Use the binomial formula B(n,k) = C(n,k)·p^k·(1-p)^(n-k).
<p><strong>Step 1 (Part i - Exactly 4 times):</strong> Use binomial formula with n=7, k=4, p=1/2, q=1/2</p><p>P(X=4) = C(7,4) × (1/2)^4 × (1/2)^3 = C(7,4) × (1/2)^7</p><p>C(7,4) = 7!/(4!×3!) = 35</p><p>P(X=4) = 35/128 ✓</p><p><strong>Step 2 (Part ii - At least 4 times):</strong> Find P(X≥4) = P(X=4) + P(X=5) + P(X=6) + P(X=7)</p><p>P(X=5) = C(7,5) × (1/2)^7 = 21/128</p><p>P(X=6) = C(7,6) × (1/2)^7 = 7/128</p><p>P(X=7) = C(7,7) × (1/2)^7 = 1/128</p><p>P(X≥4) = (35 + 21 + 7 + 1)/128 = 64/128 = 1/2 ✓</p>
Correct Answer: (i) 35/128, (ii) 1/2