I'm not quite sure how to get the correct probability for this question.
Q: The probability of error in the transmission of a binary digit over a communication channel is 1/10^3. Write an expression for the exact probability of more than 3 errors when transmitting a block of 10^3 bits. What is its approximate value? Assume independence.
Since it's approximate, I'm using Poisson's distribution formula.
The expected value (for λ) is 1, I believe.
K would be 3.
However, by plugging these values in and obtaining the result, I'm only getting the answer for exactly 3 errors. I need to get the probability of receiving more than three errors.
Initially, I thought it would be: 1 – P(0 errors) – P(1 error) – P(2 errors) – P(3 errors)
However, I'm not sure if this is correct or if there's a possibility of overlap. Can anyone help in what the probability should be or if I'm going about this problem in the correct way?
Best Answer
Under our assumptions, the exact distribution is binomial, $n=10^3$, $p=\frac{1}{10^3}$. The exact probability of more than $3$ errors is $1$ minus the probability of $3$ or fewer errors. So our exact probability is $$1-\binom{n}{0}p^0(1-p)^n-\binom{n}{1}p^1(1-p)^{n-1}-\binom{n}{2}p^2(1-p)^{n-2}-\binom{n}{3}p^3(1-p)^{n-3}.$$
For an approximation, you are doing it in the correct way. We do use the Poisson distribution with $\lambda=np=1$, and the approximate probability is $$1-e^{-1}\left(1+\frac{1}{1!}+\frac{1^2}{2!}+\frac{1^3}{3!}\right).$$