I have difficulties with the following question

We consider two DNA sequences with the letters A, G, T, C. The letters are independent (on different locations) and appear with probability $P(A) = P(G) = P(C) = 1/6$, $P(T) = 1/2$

The sequences have both infinite length (=idealization) and the two sequences are independent. When the letters are read by a sequence processor, an error is made with probability $10^{−4}$ . Whenever an error is made, the “wrong” letters appears with equal probability: e.g. if G is read wrongly, then the letters A, T, C occur each with probability 1/3. Errors occur independently and independent of the letter which is read.

What is the distribution the number of times G's read in the reading of the first 10000 symbols of the first sequence. And what are the parameters.

This is how far I got:

It's binomial distributed with parameter $n = 10000$ en $p^*$ to find the latter I came up with this

Let $K := \text{ G is being read }$


To calculate $P(K|F)$ I used $P(K|F) = P(K|F,G)P(G)+P(K|F,G^c)P(G^c)= 0 +1/3*1/4$ I'm not sure if this is correct? The other probabilities look easy to v.


For each element of the sequence, the probability to read a $G$ is defined as follows:

$$p = (1-10^{-4})\frac{1}{6}+10^{-4}\left(1-\frac{1}{6}\right)\frac{1}{3}$$


  1. $(1-10^{-4})$ is the probability to read a given element correctly;
  2. $\frac{1}{6}$ is the probability to find a G in the sequence;
  3. $10^{-4}$ is the probability to read a given element wrongly;
  4. $\left(1-\frac{1}{6}\right)$ is the probability to find a symbol different from G in the sequence;
  5. $\frac{1}{3}$ is the probability to read a G given that the sequence do not contain a G.

Moreover, given this $p$, we can say that the number $X$ of G read by the machine follows a binomial distribution:

$$P(X = k) = {10000 \choose k} p^k (1-p)^{10000-k}.$$

  • $\begingroup$ thanks I got the same :). Do you know the answer to this one aswell? The distribution of the number of places in the first 10000 letters of both sequences where the letters are read identical, but are not truly identical. $\endgroup$ May 29 '18 at 21:26

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