Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the enveloped at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the enveloped at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
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.
Total number of ways of putting three letters into three envelopes is
3P3 = 3! (ways) = 6.
The number of ways in which none of the letter is put into proper envelope is 2.
∴ P(none) = 2/6 = 1/3
∴ p(atleast one letter is in its proper envelop) = 1 – p (none)
= 1 – 1/3 = 2/3