Probability
Geometric Distribution
Grade 12
Question:
<p>Two fair dice are thrown till outcome is 12. What is the probability that one has to do 20 throws for this?</p>
<p>(a) \(\left(\dfrac{35}{36}\right)^{18}\left(\dfrac{1}{36}\right)^{2}\)</p>
<p>(b) \(\left(\dfrac{35}{36}\right)^{19}\left(\dfrac{1}{36}\right)\)</p>
<p>(c) \(\left(\dfrac{1}{36}\right)^{19}\left(\dfrac{35}{36}\right)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Getting a sum of 12 requires both dice showing 6 (probability 1/36), so this is a geometric distribution problem where we need exactly 19 failures followed by 1 success on the 20th throw.
<p><strong>Step 1:</strong> Identify the probability of getting sum 12 in a single throw of two fair dice.</p><p>Sum = 12 occurs only when both dice show 6: (6,6)</p><p>P(success) = 1/36</p><p><strong>Step 2:</strong> Identify probability of NOT getting sum 12.</p><p>P(failure) = 1 - 1/36 = 35/36</p><p><strong>Step 3:</strong> For exactly 20 throws needed, we need 19 consecutive failures followed by success on the 20th throw.</p><p>P(first success on 20th throw) = (35/36)^19 × (1/36)</p><p><strong>Step 4:</strong> Simplify the expression.</p><p>P = (35/36)^19 × (1/36) = (35^19)/(36^20)</p><p>∴ Answer: B</p>
Correct Answer: B