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Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
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Let the envelope be denoted by $A, B, C$ and the corresponding letters by $a, b, c$. 1 letter in correct envelope and 2 in wrong envelope may be put as $(A a, B c, C b),(A c, B b, C a),(A b, B a, C c)$.
Two letters or consequently all the letters are in correct envelope may be put up in one way i.e. $(\mathrm{Aa}, \mathrm{Bb}, \mathrm{Ca})$.
$\therefore$ Number of exhaustive cases $=3 !=6$ ways
Number of favourable cases $=4$
Probability that alteast one letter is in its proper envelope $=\frac{4}{6}=\frac{2}{3}$
Two letters or consequently all the letters are in correct envelope may be put up in one way i.e. $(\mathrm{Aa}, \mathrm{Bb}, \mathrm{Ca})$.
$\therefore$ Number of exhaustive cases $=3 !=6$ ways
Number of favourable cases $=4$
Probability that alteast one letter is in its proper envelope $=\frac{4}{6}=\frac{2}{3}$
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