Question

Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 17 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. Round your answers to two decimal places

Answer 1

Answer:

a) 59.58%

b) 90.90%

c) 0%

Step-by-step explanation:

Let p be the probability a sample is mutated. Then q= 1-p is the probability the sample IS NOT mutated.

So p=0.03 (3%) and q=0.97 (97%).

The experiment of finding a mutated sample out of 17 clearly has a binomial distribution, where the probability of finding k mutated samples out of 17 is given by

$P(17;k)=\binom{17}{k}(0.03)^k(0.97)^{17-k}$

a)  

We want P(17;0)  

$P(17;0)=\binom{17}{0}(0.03)^0(0.97)^{17}=(0.97)^{17}=0.5958=59.58\%$

b)

We want P(17;0)+P(17;1)

$P(17;0)+P(17;1)=0.5958+\binom{17}{1}(0.03)(0.97)^{18}=0.5958+0.3132=0.9090=90.90\%$

c)

We want P(17;9)+P(17;10)+...+P(17;17)

Computing this number, we get  

$3.8448*10^{-34}$

This figure is zero if we round to 2 decimals

Answer 2

The probability for No samples is mutated, At most one sample is mutated, More than half the samples are mutated are 59.58%, 90.90%, and 0% respectively.

What is probability?

Probability means possibility. It deals with the occurrence of a random event. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

Let p be the probability a sample is mutated. and q be the probability of a sample not mutated. Then

$\rm p = 0.09 (3\%) \ \ and \ \ q = 0.97 (97\%)$

The experiment of finding a mutated sample out of 17 clearly has a binomial distribution, where the probability of finding k mutated sample out of 17 is given by

$\rm P(17, k) = (^{17}_k)0.03^k * 0.97^{17-k}$

a.  We want P(17, 0)

$\rm P(17, 0) = (^{17}_k)0.03^0 * 0.97^{17-0} = 0.97^17 = 0.5958 = 59.58\%$

b.  We want P(17, 0) + P(17, 1)

$\rm P(17, 0) + P(17, 1) = 0.5958 + (^{17}_1)0.03^1 * 0.97^{18} = 0.9090 = 90.90\%$

c.  We want P(17, 9) + P(17, 10) + ....... + P(17, 17)

$\rm P(17, 9) +P(17, 10) + ..... +P(17, 17)= \Sigma ^{17} _{x =9}(^{17}_k)0.03^k * 0.97^{17-k}$

On solving, we have

$\rm P(17, 9) +P(17, 10) + ..... +P(17, 17) = 3.8448 * 10^{-34}$

This will be zero if we round to 2 decimals.

More about the probability link is given below.

https://brainly.com/question/795909