The number of independent errors in a transmitted message of n symbols follows a Binomial distribution. The probability of an error in transmission is .001. If 2,000 symbols were sent,
a) Find the expected number of errors.
b) Write out the formula for the distribution of errors in a 2000 symbol message.
c) Find the probability of an error-free message.
d) Find the probability of exactly 1 error.
e) Find the probability of at least 1 error.
p=.001, n=2000
a)Expected number of errors=np=2
b)P(k)=2000Ck*.001^k*.999^(2000-k)
c)P(0)=.999^2000
d)P(1)=2000*.001*.999^1999
e)P(k>=1)=1-P(0)
a) The expected number of errors can be found by multiplying the number of symbols sent (2,000) by the probability of an error in transmission (.001).
Expected number of errors = 2,000 * 0.001 = 2
b) The formula for the distribution of errors in a 2000 symbol message follows a Binomial distribution. The probability mass function (PMF) of the Binomial distribution is given by:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- X is the random variable representing the number of errors,
- n is the number of symbols sent (2000 in this case),
- k is the number of errors,
- p is the probability of an error in transmission (0.001 in this case),
- C(n, k) represents the combination function, which calculates the number of ways to choose k items out of n items.
c) The probability of an error-free message can be found by using the formula for the distribution of errors and setting k = 0 (since an error-free message means there are 0 errors).
P(X = 0) = C(2000, 0) * (0.001)^0 * (1-0.001)^(2000-0)
Since C(2000, 0) = 1 and any number raised to the power of 0 is 1, the equation simplifies to:
P(X = 0) = (1) * (0.999)^2000 ≈ 0.819
Therefore, the probability of an error-free message is approximately 0.819 or 81.9%.
d) The probability of exactly 1 error can be found by setting k = 1 in the formula for the distribution of errors.
P(X = 1) = C(2000, 1) * (0.001)^1 * (1-0.001)^(2000-1)
Since C(2000, 1) = 2000, the equation becomes:
P(X = 1) = 2000 * (0.001) * (0.999)^(1999) ≈ 0.270
Therefore, the probability of exactly 1 error is approximately 0.270 or 27.0%.
e) The probability of at least 1 error can be calculated by summing the probabilities of having 1 error, 2 errors, 3 errors, and so on, up to the maximum possible number of errors (which is 2000 in this case).
P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + ... + P(X = 2000)
This can be a very tedious calculation to perform manually since there are 2000 terms to sum. However, you can use software or a calculator that has statistical functions to calculate it easily.