$\begin{array}{1 1}(A)\;\large\frac{2}{3}\\(B)\;\large\frac{5}{3}\\(C)\;\large\frac{7}{3}\\(D)\;\text{None of these}\end{array} $

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- Required probability =$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$

Step 1:

Let the envelope be denoted by $E_1,E_2,E_3$ and the corresponding letters by $L_1,L_2,L_3$

Atleast one letter should be in right envelope.

Let us consider all the favorable outcomes

Step 2:

(i) 1 letter in correct envelope and 2 in wrong envelope.

(ie) $(E_1L_1,E_2L_3,E_3L_2),(E_1L_3,E_2L_2,E_3L_1),(E_1L_2,E_2L_1,E_3L_3)$

(ii) Two letter in correct envelope.

(ie) $(E_1L_1,E_2L_2,E_3L_3)$

$\therefore$ No of favorable outcomes =4

Step 3:

Total no of outcomes =$3!=3\times 2\times 1=6$

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{4}{6}$

$\Rightarrow \large\frac{2}{3}$

Hence (A) is the correct answer.

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