# Poisson distribution question

The question says when an $8$-bit word is being transmitted. There is a chance of $0.1$ for an error in each bit independently. Every bit is transmitted $3$ times. When decoding, choose the bit that appeared more times.

(a) Find the probability that an $8$-bit word would be decoded correctly.

(b) In a file that contains $1000$ words, find the distribution of the number of words that were decoded incorrectly

For part a, I use Poisson distribution:

$f (k) = λ^k * e^{-λ} / k!$, and $λ = np$, where $n=8$, and $p=0.1$

so $f(k)= 0.8^8*e ^{-0.8} / 8! = 1.8697e-6$

$1-f(k)=0.99999813$

but I found I did not use the "$3$ times" information in the question, so I quite doubt about my solution...

And for part b, I am thinking it is just using the probability of decoding a word correctly * $1000$, right?

I appreciate a lot if someone could give me a clue about question a, thank you!

• "For part a, I use Poisson distribution:" In fact, you used the Poisson Approximation of a Binomial Distribution. Don't! Just use Binomial Distributions. – Graham Kemp Jan 31 '16 at 6:26

a) Look at a single bit, say $0$. It is transmitted as $000$. The probability that two or more $0$'s will be received, and hence the bit will be decoded correctly, is $(0.9)^3+\binom{3}{1}(0.9)^2(0.1)$. Call this probability $a$. We used a simple case of the binomial distribution.
Now suppose we have an $8$-bit word to transmit. The probability it will be transmitted correctly is the probability that all $8$ bits, after the encoding, will be transmitted correctly. This probability is $a^8$.
b) Let $X$ be the number of words decoded incorrectly. The probability a particular word is decoded wrongly is $1-a^8$. Thus $X$ has binomial distribution, $n=1000$, $p=1-a^8$.