Probability
Classical Probability
Grade 12
Question:
<p><strong>For Problems 13–15:</strong> An amoeba either splits into two or remains the same or eventually dies out immediately after completion of every second with probabilities, respectively, 1/2, 1/4, and 1/4. Let the initial amoeba be called as mother amoeba and after every second, the amoeba, if it is distinct from the previous one, be called as 2nd, 3rd, ... generations.</p><p>The probability that immediately after completion of 2 s all the amoeba population dies out is</p>
<p>(1) 9/32</p>
<p>(2) 11/32</p>
<p>(3) 1/2</p>
<p>(4) 3/32</p>
Step-by-Step Solution
Key Concept: Track all possible population states after each second using conditional probability: after 1s the mother either splits (→2 amoebas), stays (→1 amoeba), or dies (→0). After 2s, calculate extinction probability by considering what each state leads to.
<p><strong>Step 1: Classify outcomes after 1 second</strong></p><p>Mother amoeba (1st generation) can:</p><ul><li>Split into 2 amoebas: probability 1/2</li><li>Remain as 1 amoeba: probability 1/4</li><li>Die: probability 1/4</li></ul><p><strong>Step 2: Trace paths to extinction by t=2s</strong></p><p><strong>Case 1:</strong> Mother dies at t=1s → Population = 0 at t=2s</p><p>Probability = 1/4</p><p><strong>Case 2:</strong> Mother remains at t=1s → Single 2nd gen amoeba at t=1s</p><p>This amoeba must then die in the next second to achieve extinction by t=2s</p><p>Probability = (1/4) × (1/4) = 1/16</p><p><strong>Case 3:</strong> Mother splits at t=1s → Two 2nd gen amoebas at t=1s</p><p>Both amoebas must independently die by t=2s for extinction</p><p>Probability = (1/2) × (1/4) × (1/4) = 1/32</p><p><strong>Step 3: Sum all extinction paths</strong></p><p>Total probability = 1/4 + 1/16 + 1/32</p><p>= 8/32 + 2/32 + 1/32 = 11/32</p><p>∴ Answer: A (11/32)</p>
Correct Answer: A