Step 1:
Let the envelope be denoted by E1, E2, E3 and the corresponding letters by L1, L2, L3
At least 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) (E1L1,E2L3,E3L2),(E1L3,E2L2,E3L1),(E1L2,E2L1,E3L3)
(ii) Two letter in correct envelope.
(ie) (E1L1,E2L2,E3L3)
∴ No of favorable outcomes =4
Step 3:
Total no of outcomes =3!=3×2×1=6
∴ Required probability =n(E)/n(S)
⇒4/6
⇒2/3
Hence (A) is the correct answer.