Question

Samples of rejuvenated mitochondria are mutated (defective) in 2% of cases. Suppose 12 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities.
(a) No samples are mutated.
(b) At most one sample is mutated.
(c) More than half the samples are mutated.
(c) is 0.00
Round your answers to two decimal places

Answer 1

Answer:

a) 0.7847 = 78.47% probability that no samples are mutated.

b) 0.9769 = 97.69% 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 they are mutated, or they are 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

2% of cases.

This means that $p = 0.02$

12 samples are studied

This means that $n = 12$.

(a) No samples are mutated.

This is $P(X = 0)$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.02)^{0}.(0.98)^{12} = 0.7847$

0.7847 = 78.47% probability that no samples are mutated.

(b) At most one sample is mutated.

This is:

$P(X \geq 1) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.02)^{0}.(0.98)^{12} = 0.7847$

$P(X = 1) = C_{12,1}.(0.02)^{1}.(0.98)^{11} = 0.1922$

$P(X \geq 1) = P(X = 0) + P(X = 1) = 0.7847 + 0.1922 = 0.9769$

0.9769 = 97.69% probability that at most one sample is mutated.

(c) More than half the samples are mutated.

This is:

$P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,7}.(0.02)^{7}.(0.98)^{5} \approx 0$

So the others(greater than 7) wil be 0 too

0% probability that more than half the samples are mutated.