Answer:
The probability that more than 6 samples are needed to have 2 mutations is P=0.866.
Step-by-step explanation:
This question can be analized with the binomial distribution.
We need to calculate the probability that more of 6 samples are needed to have 2 samples with mutations.
This is equal to the proability of taking 2 samples and getting one mutation or less.
Then we have a binomial distribution with n=6 and p=0.11:
[tex]P(X\leq1)=P(0)+P(1)\\\\P(X\leq 1)=(1-p)^6+6p(1-p)^5\\\\P(X\leq 1)=0.89^6+6*0.11*0.89^5=0.497+0.369=0.866[/tex]
