Probability of an error going undetected with a single parity bit

Question

Problem:
A byte (which is $8$ bits) is being transmitted over an unreliable computer network. After the last bit is transmitted a parity bit is sent. The probability that a given bit is sent correctly is $0.9$. What is the probability that in transmitting a byte (including the parity bit) an error will occur and it will go undetected?
Answer:
If exactly one bit is transmitted incorrectly then the network will generate a parity error. That is, the system will detect the problem. If exactly three bits are wrong, the system will generate a parity error. However, if there are exactly two bits wrong or exactly four bits wrong or exactly six bits wrong no parity error will be detected exactly eight bits wrong no parity error will be detected. What we have here is a binomial distribution. Let $p_e$ be the probability we seek. \begin{align*} n &= 9 \\ p &= \dfrac{ 9 } { 10 } \\ q &= \dfrac{ 1 } { 10 } \\ p_e &= { 9 \choose 2 }p^7 q^2 + { 9 \choose 4 }p^5 q^4 + { 9 \choose 6 }p^3 q^6 + { 9 \choose 8 } p q^8 \\ { 9 \choose 2 } &= \dfrac{ 9(8) } {2} = 36 \\ { 9 \choose 4 } &= \dfrac{ 9 } { 4 } = \dfrac{ 9(8)(7)(6) } {4(3)(2)} \\ { 9 \choose 4 } &= \dfrac{ 9(7)(6) } { 3 } = 3(7)(6) = 126 \\ { 9 \choose 6 } &= \dfrac{ 9(8)(7) } { 3(2) } = 3(4)(7) = 84 \\ { 9 \choose 8 } &= \dfrac{ 9 } { 1 } = 9 \\ % p_e &= (45){\left( \dfrac{ 9 } { 10 }\right) }^7 {\left( \dfrac{ 1 } { 10 }\right) }^2 \\ &+ (126){\left( \dfrac{ 9 } { 10 }\right) }^5 {\left( \dfrac{ 1 } { 10 }\right) }^4 + (84){\left( \dfrac{ 9 } { 10 }\right) }^3 {\left( \dfrac{ 1 } { 10 }\right) }^6 + 9\left( \dfrac{ 9 } { 10 }\right) \left( {\dfrac{ 1 } { 10 }} \right) ^8\\ % \left( 10^9 \right) p_e &= 36(9^7) + 126(9^5) + 84(9^3) + 9(9) \\ \end{align*} \begin{align*} \left( 10^9 \right) p_e &= 36(4782969) + 126( 59049) + 84(729) + 81 \\ \left( 10^9 \right) p_e &= 179688375 \\ p_e &= \dfrac{ 179688375 }{ 10^9 } \\ p_e &= 0.179688375 \end{align*} Is my solution correct?

Answer 1

As requested in comments:

Your calculation of the probability of a positive even number of bit errors looks correct, assuming these are independent:

$$\sum\limits_{m=2,4,6,8} {9 \choose m}\left(\frac{1}{10}\right)^m\left(\frac{9}{10}\right)^{9-m}= 0.179688375$$

and is the probability that there is an error and it goes undetected.

The conditional probability that errors go undetected given that there is one or more errors would need to divide that by $1-\left(\frac{9}{10}\right)^{9}$ to get about $0.293331$