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(C) Let the letters be $L_1, L_2, L_3$ and their corresponding envelopes be $E_1, E_2, E_3$. The total number of ways to arrange $3$ letters in $3$ en

Vedclass · Accepted Answer

$2/3$

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(C) Let the letters be $L_1, L_2, L_3$ and their corresponding envelopes be $E_1, E_2, E_3$. The total number of ways to arrange $3$ letters in $3$ envelopes is $3! = 6$. These arrangements are: $(L_1E_1, L_2E_2, L_3E_3)$ - All $3$ correct $(L_1E_1, L_2E_3, L_3E_2)$ - $L_1$ correct $(L_2E_2, L_1E_1, L_3E_3)$ - $L_3$ correct $(L_3E_3, L_2E_2, L_1E_1)$ - $L_2$ correct $(L_1E_2, L_2E_3, L_3E_1)$ - None correct $(L_1E_3, L_2E_1, L_3E_2)$ - None correct There are $4$ arrangements where at least one letter is in its proper envelope. Thus,the required probability is $\frac{4}{6} = \frac{2}{3}$.

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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Notation : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$

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