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Question: There are 6 letters and 6 self addressed envelopes.What is the probability that atleast 1 is placed correctly??

Ans: 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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