DNA sequence - probability Your DNA code is composed of a series of four nucleotides: adenine, guanine, thymidine and cytosine (A, G, T and C, respectively).  
a)  What is the probability an individual has the following nucleotide sequence: “TATATA” at any particular position?  You may assume independence.                            
My answer:  The probability of getting "TATATA" is (1/4)(1/4)(1/4)(1/4)(1/4)(1/4).
b)  What is the probability that an individual has k T’s in their DNA code at any particular position? (k can be any integer and you may assume independence).      Here we're looking for the probability of k consecutive Ts.
My answer:  The probability of k consecutive "T"s (in fact any k specific nucleotides) is (1/4)^k.
Thoughts please.
 A: Depends on the length of DNA.
For example, if the length of DNA is, say 3 units, then answer to your first question will be zero.
Let the DNA be an N unit closed chain.
Part (a)
$$P(sequence\ occur\ somewhere)=1-P(sequence\ does\ not\ occur\ anywhere)$$
Now,
\begin{align}P(sequence\ does\ not\ occur\ at\ position\ i)&=1-P(sequence\ occur\ at\ position\ i)\\
P(sequence\ does\ not\ occur\ at\ position\ i)&=1-\frac{1}{4^6}\end{align}
Multiplying the probability that sequence does not occur at any position 1, 2, 3, ... N we get,
$$P(sequence\ does\ not\ occur\ anywhere)=\left(1-\frac{1}{4^6}\right)^N$$
Therefore,
$$P(sequence\ occur\ somewhere)=1-\left(1-\frac{1}{4^6}\right)^N$$
For Part (b) replace $4^6$ with $4^k$ in the above.
A: Your answers are correct. I have tried disproving either however I was unable to do so. Since it is one out of the four possibilities, #$1$ is correct. The second one is also correct because to get T, it is a $\frac{1}{4}$ of it happening. Thus, if you want to get it $k$ amounts of times, it would be ${\frac{1}{4}}^k$.
A: You should take into account that the two probabilities that you are looking for are strictly affected by the total length of the nucleotide sequence that you are considering.
For instance, as regards your first question, your answer is correct only if you consider a single sequence of 6 nucleotides, since the probability of finding a TATATA clearly increases if you consider longer sequences. If you consider the whole  DNA of an individual (as you state in your question), the probability is very near to 1 because of the high number of nucleotide pairs (over 3 milliards) of the human genome.
