Question

3. A data packet containing 18 bits is sent over a noisy communication channel. The probability that the k-th bit is received correctly at the destination is 1 - p, and, therefore, it is erased with probability p. Erasures are independent over time. (a) 1 pt What is the probability that exactly 7 bits are erased from this packet? (b) 1 pt What is the maximum value of p such that the probability that the packet is received correctly (i.e., all 18 bits are received correctly) is greater than 0.5?

Answer 1

Let x = bit k is erased (a) P (x = 4) = 18 c_7 P^7 (1 - p)^18 - 7 = 18 c_7. P^7 (1 - p)^11 (b) P (x = 0) = 18 c_18 P degree (1 - 1)^18 = 1.(1 - 9)^18 implies 0.5 (1 - p)^18 > 0.5 1 - p > (0.5)^1/18 > 0.9622 P < 1 - 0.9622 p < 0.0377 H 6 P 12 P(x = 2) in header) = 6 c_2 p^2 (1 - p)^2/18 c_2 p^7 (1 - p)^16 = 6 times 5 p^2 (1 - p)^4/18 times 17 p^2 (1 - p)^16 = 5/51 times 1/(1 - p)^12 (d) lessthanorequalto A rightarrow Erasure in channel x rightarrow erasure in header P(A intersection x) = 6c_2 p^8 (1 - p)^4 P(X/A) = P(A intersect x)/P(A) = 6C_2 p^2 (1 - P)^2/18 C_2 p^2 (1 - P)^4 = 5/51 1/(1 - p)^12 (C) P(X/A) = P(A intersection x)/P(A) = 6 c_2 p^2 (1 - p)^4/18 C_7 p^2 (1 - p)^4 = 15/31824 times 1/(1 - p)^12