GATE CS 2015 Set 1 Q29 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

For a frame to be sent in the first slot without any collision, exactly one of the four stations must transmit, while the other three stations must not transmit.

Let $p_i$ be the probability that station $S_i$ transmits in a slot, and $q_i = 1 - p_i$ be the probability that it does not transmit. The given probabilities are:

• $p_1 = 0.1 \implies q_1 = 0.9$
• $p_2 = 0.2 \implies q_2 = 0.8$
• $p_3 = 0.3 \implies q_3 = 0.7$
• $p_4 = 0.4 \implies q_4 = 0.6$

Since the transmission attempts of the stations are independent, the probability of exactly one station transmitting is the sum of the probabilities of the following four mutually exclusive cases:

1. Only $S_1$ transmits: $$P_1 = p_1 \times q_2 \times q_3 \times q_4 = 0.1 \times 0.8 \times 0.7 \times 0.6 = 0.0336$$
2. Only $S_2$ transmits: $$P_2 = q_1 \times p_2 \times q_3 \times q_4 = 0.9 \times 0.2 \times 0.7 \times 0.6 = 0.0756$$
3. Only $S_3$ transmits: $$P_3 = q_1 \times q_2 \times p_3 \times q_4 = 0.9 \times 0.8 \times 0.3 \times 0.6 = 0.1296$$
4. Only $S_4$ transmits: $$P_4 = q_1 \times q_2 \times q_3 \times p_4 = 0.9 \times 0.8 \times 0.7 \times 0.4 = 0.2016$$

Summing these individual probabilities gives the total probability of a successful transmission without collision:

Thus, the probability of sending a frame in the first slot without any collision is 0.4404.