Question

Each time we sequence a single nucleotide of DNA using a next-generation technique, there is a .007 probability that the wrong nucleotide is predicted. Each nucleotide prediction is independent. If we are sequencing a single strand of DNA comprised of 200 nucleotides, what is the probability that our result will have 4 or more incorrect predictions?

Answer 1

Answer

we have probability p = 0.007

and sample size is n = 200

we have to find the probability of having 4 or more incorrect predictions.

This is a binomial problem, so we can write it as

Probability(4 or more incorrect prediction) = 1-(Probability of 3 or less predictions)...............equation 1

because we know that the total probability is always equal to 1

We have the foormula

where p is probability = 0.007, n is total sample size = 200and r is selected value

P(0) = =

P(1) = =

P(2) = $inom{200}{2}*(0.007)^2*(1-0.007)^{200-2}$ =

P(3) = =

now adding all these probabilities, we get

P(3 or less) = 0.2454+0.3460+0.2427+0.1129 = 0.9469

So, required probability of 4 or more incorrect prediction = 1- P(3 or less) = 1 - 0.9469 = 0.0531