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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?

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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 $$1-e^{-1}\left(1+\frac{1}{1!}+\frac{1^2}{2!}+\frac{1^3}{3!}\right).$$

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