Consider a LAN with four nodes $S_1, S_2, S_3$ and $S_4$. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by $S_1, S_2, S_3$ and $S_4$ are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _____________.

A successful transmission without collision in a slot occurs when exactly one of the four nodes transmits while the remaining three nodes do not transmit.

Let the transmission probabilities be:

• $P(S_1) = 0.1$, so $P(S_1') = 1 - 0.1 = 0.9$

• $P(S_2) = 0.2$, so $P(S_2') = 1 - 0.2 = 0.8$

• $P(S_3) = 0.3$, so $P(S_3') = 1 - 0.3 = 0.7$

• $P(S_4) = 0.4$, so $P(S_4') = 1 - 0.4 = 0.6$

The probability of sending a frame without collision is the sum of probabilities of mutually exclusive events where only one node transmits:

$$\begin{aligned} P(\text{Success}) &= P(S_1 \text{ only}) + P(S_2 \text{ only}) + P(S_3 \text{ only}) + P(S_4 \text{ only}) \[6pt] P(S_1 \text{ only}) &= 0.1 \times 0.8 \times 0.7 \times 0.6 = 0.0336 \[4pt] P(S_2 \text{ only}) &= 0.9 \times 0.2 \times 0.7 \times 0.6 = 0.0756 \[4pt] P(S_3 \text{ only}) &= 0.9 \times 0.8 \times 0.3 \times 0.6 = 0.1296 \[4pt] P(S_4 \text{ only}) &= 0.9 \times 0.8 \times 0.7 \times 0.4 = 0.2016 \end{aligned}$$

Adding these probabilities:

Thus, the probability of sending a frame without collision is approximately 0.44 (accepted range: 0.40 to 0.46).