Respuesta :
Answer:
a) 0.6730 = 67.30% probability that no samples are mutated.
b) 0.9436 = 94.36% probability that at most one sample is mutated.
c) 0% probability that more than half the samples are mutated.
Step-by-step explanation:
For each sample, there are only two possible outcomes. Either it is mutated, or it is not. The probability of a sample being mutated is independent of any other sample, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases.
This means that [tex]p = 0.03[/tex]
13 samples are studied
This means that [tex]n = 13[/tex]
(a) No samples are mutated.
This is P(X = 0). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{13,0}.(0.03)^{0}.(0.97)^{13} = 0.6730[/tex]
0.6730 = 67.30% probability that no samples are mutated.
(b) At most one sample is mutated.
This is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{13,0}.(0.03)^{0}.(0.97)^{13} = 0.6730[/tex]
[tex]P(X = 1) = C_{13,1}.(0.03)^{1}.(0.97)^{12} = 0.2706[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.6730 + 0.2706 = 0.9436[/tex]
0.9436 = 94.36% probability that at most one sample is mutated.
(c) More than half the samples are mutated.
This is:
[tex]P(X > 6.5) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{13,7}.(0.03)^{7}.(0.97)^{6} \approx 0[/tex]
Using two decimal digits precision, all will be 0. So
0% probability that more than half the samples are mutated.
