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Three letters A, B, C placed in three envelopes; P(none in correct envelope) is
A1/6
B2/3
C1/2
D1/3
Answer & Solution
Correct answer: D. 1/3
1. Total arrangements = 3! = 6.
2. Derangements of 3 elements: D3 = 2 (BCA and CAB).
3. P(derangement) = 2/6.
4. = 1/3.
_Source: ICAI BoS Foundation Paper 3, Ch 15 'Probability', p.12_
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