Question 1200597
Certainly, let's calculate the probabilities.

**1. Probability that the applicant has seen the job advertised in paper A, given only one reply:**

* **Define Probabilities:**
    * Let P(A), P(B), and P(C) be the probabilities of an undergraduate seeing the ad in papers A, B, and C, respectively.
    * Let R(A), R(B), and R(C) be the probabilities of replying to the ad in papers A, B, and C, respectively.

* **Given:**
    * P(A) = 2/6 = 1/3
    * P(B) = 3/6 = 1/2
    * P(C) = 1/6
    * R(A) = 0.002
    * R(B) = 0.001
    * R(C) = 0.005

* **Calculate the probability of receiving a reply from each paper:**
    * P(reply from A) = P(A) * R(A) = (1/3) * 0.002 = 0.000667
    * P(reply from B) = P(B) * R(B) = (1/2) * 0.001 = 0.0005
    * P(reply from C) = P(C) * R(C) = (1/6) * 0.005 = 0.000833

* **Calculate the probability that the applicant has seen the job advertised in paper A, given only one reply:**
    * P(A | one reply) = P(reply from A) / (P(reply from A) + P(reply from B) + P(reply from C))
    * P(A | one reply) = 0.000667 / (0.000667 + 0.0005 + 0.000833) 
    * P(A | one reply) = 0.000667 / 0.002 
    * P(A | one reply) = 1/3 

**Therefore, the probability that the applicant has seen the job advertised in paper A, given only one reply, is 1/3.**

**2. Probability that both applicants saw the job advertised in paper A, given two replies:**

* Assuming the replies are independent events:
    * P(both replies from A) = P(reply from A) * P(reply from A)
    * P(both replies from A) = 0.000667 * 0.000667 
    * P(both replies from A) = 0.000000444889

* **Calculate the probability that both applicants saw the job advertised in paper A, given two replies:**
    * P(both A | two replies) = P(both replies from A) / (P(reply from any paper) * P(reply from any paper))
    * P(both A | two replies) = 0.000000444889 / (0.002 * 0.002) 
    * P(both A | two replies) = 0.111222222 

**Therefore, the probability that both applicants saw the job advertised in paper A, given two replies, is approximately 0.1112.**