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There are 6 letters and 6 self addressed envelopes.What is the probability that atleast 1 is placed correctly??
If we let A be the event that letter A is in the correct envelope, and similarly B is event that letter B is in the correct envelope, then P(A) = 1/5 P(A and B) = 1/5 x 1/4 Now use the inclusion, exclusion formula to get the probability that A or B or C .... or E are correctly placed. P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E) - P(A and B) - P(B and C) -.... + P(A and B and C) + P(B and C and D) + .... - P(A and B and C and D) - P(...) -...... + P(A and B and C and D and E) = 5 x (1/5) - 5C2 x (1/5)(1/4) + 5C3 x (1/5)(1/4)(1/3) - 5C4 x (1/5)(1/4)(1/3)(1/2) + (1/5)(1/4)(1/3)(1/2)(1/1) 5x4 1 5x4x3 1 5x4x3x2 1 1 1 - ---- x ---- + ------ x ----- - --------- x ------- + -------- 1x2 5x4 1x2x3 5x4x3 1x2x3x4 5x4x3x2 5x4x3x2x1 = 1 - 1/2! + 1/3! - 1/4! + 1/5! This is the probability that at least one letter is correctly placed.
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