There are 6 stations in a slotted LAN. Probbability of each station transmits during a contention slot is 0.8. What is the probability of only one station transmits in a given time slot? (Compute the value upto 4 decimal places)Correct answer is '0.0015'. Can you explain this answer?

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The probability of only one station =

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The probability of only one station =

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The probability of only one station =

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Given information:
- There are 6 stations in a slotted LAN.
- Probability of each station transmitting during a contention slot is 0.8.

To find the probability of only one station transmitting in a given time slot, we can use the binomial distribution formula.

Binomial Distribution Formula:
The probability of exactly k successes in n independent Bernoulli trials, where the probability of success in each trial is p, is given by:
P(X = k) = nCk * p^k * (1 - p)^(n-k)

Where:
- P(X = k) is the probability of exactly k successes.
- nCk is the number of combinations of n items taken k at a time.
- p^k is the probability of k successes.
- (1 - p)^(n-k) is the probability of (n-k) failures.

Applying the formula:
In this case, we want to find the probability of exactly 1 station transmitting, so k = 1.

P(X = 1) = 6C1 * 0.8^1 * (1 - 0.8)^(6-1)
= 6 * 0.8 * 0.2^5

Simplifying the expression:
- 6C1 = 6 (as there are 6 ways to choose 1 station out of 6)
- 0.8 * 0.2^5 = 0.8 * 0.00032 = 0.000256 (rounded to 6 decimal places)

Therefore, the probability of only one station transmitting in a given time slot is 0.000256.

Rounding the answer:
As per the question's requirement, the answer needs to be rounded to 4 decimal places.

Rounded answer = 0.0003

However, the correct answer given is 0.0015. It seems there might be an error in either the given answer or the provided information.

Similar Computer Science Engineering (CSE) Doubts

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