Respuesta :
Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
Given percentage of rejuvenated mitochondria defective is 1%, and sample size is 18.
Binomial distribution is the probability of exactly x successes on n repeated trials and X can have two outcomes.
P(X=x)=[tex]C_{n,x} p^{x}(1-p)^{n-x}[/tex]
percentage of defective rejuvendated mitochondria=1%
p=0.01
Sample size=18
n=18
a) No samples are mutated
This means P(X=0)=[tex]C_{18,0}(0.01)^{0} (0.99)^{18}[/tex]
=0.83
b) At most one sample is mutated.
P(X<=1)=P(X=0)+P(X=1)
so,
P(X=0)=[tex]C_{18,0} (0.01)^{0} (0.99)^{18}[/tex]
=0.83
P(X=1)=[tex]C_{18,1}(0.01)^{1} (0.99)^{17}[/tex]=
=0.1512
P(X<=1)=0.83+0.1512
=0.9812
c) More than half the samples are mutated.
P(X>9)=P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)
Using two decimals digits precision all will be 0.
Hence Probability that no samples are mutated is 0.83, probability that at most one sample is mutated is 0.9812 and probability that more than half the samples are mutated is 0.
Learn more about probability at https://brainly.com/question/24756209
#SPJ4
