We send a frame of 256 bits. If the probability that a bit changes in transmission is 0.001 and each bit is independent.
a. What is the probability that exactly 6 bits change?
b. What is the probability that exactly 0 bits change ?
To solve these problems, we can use the concept of the binomial distribution. The binomial distribution calculates the probability of getting a certain number of successes in a fixed number of trials, given a specific probability of success in each trial.
The formula for the binomial distribution is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes,
- n is the number of trials,
- k is the number of successes,
- p is the probability of success in each trial, and
- C(n, k) is the binomial coefficient, calculated as C(n, k) = n! / (k! * (n - k)!)
Now let's apply this formula to the given problem:
a. To find the probability that exactly 6 bits change (k = 6), with a frame of 256 bits (n = 256), and a probability of 0.001 that a bit changes in transmission (p = 0.001), we can calculate:
P(X = 6) = C(256, 6) * (0.001)^6 * (1 - 0.001)^(256 - 6)
b. Likewise, to find the probability that exactly 0 bits change (k = 0), we have:
P(X = 0) = C(256, 0) * (0.001)^0 * (1 - 0.001)^(256 - 0)
To calculate these probabilities, let me quickly compute the binomial coefficients and solve the equations.